\displaystyle \textbf{Question 1: }\text{The points }A(4,5,10),\ B(2,3,4)\text{ and }C(1,2,-1)\text{ are three vertices} \\ \text{of a parallelogram}  ABCD. \text{Find vector and cartesian equations for the sides }AB\text{ and } \\ BC\text{ and find the coordinates of }D.\quad [\text{CBSE 2010}]
\displaystyle \text{Answer:}
\displaystyle  \text{The line }AB\text{ passes through }A(4,5,10)\text{ and }B(2,3,4)\text{ having position vectors}
\displaystyle \overrightarrow{a}=4\widehat{i}+5\widehat{j}+10\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}+3\widehat{j}+4\widehat{k}\text{ respectively. So, vector equation of }AB\text{ is}
\displaystyle \overrightarrow{r}=\overrightarrow{a}+\lambda(\overrightarrow{b}-\overrightarrow{a})
\displaystyle \text{or, }\overrightarrow{r}=4\widehat{i}+5\widehat{j}+10\widehat{k}+\lambda\{(2\widehat{i}+3\widehat{j}+4\widehat{k})-(4\widehat{i}+5\widehat{j}+10\widehat{k})\}
\displaystyle \text{or, }\overrightarrow{r}=4\widehat{i}+5\widehat{j}+10\widehat{k}+\lambda(-2\widehat{i}-2\widehat{j}-6\widehat{k})
\displaystyle \text{or, }\overrightarrow{r}=4\widehat{i}+5\widehat{j}+10\widehat{k}+\mu(\widehat{i}+\widehat{j}+3\widehat{k}),\ \text{where }\mu=-2\lambda\qquad \ldots(i)
\displaystyle \text{The Cartesian equations of line (i) are}
\displaystyle \frac{x-4}{1}=\frac{y-5}{1}=\frac{z-10}{3}
\displaystyle \text{Since }BC\text{ passes through the points }B(2,3,4)\text{ and }C(1,2,-1)\text{ having position vector}
\displaystyle \overrightarrow{b}=2\widehat{i}+3\widehat{j}+4\widehat{k}\text{ and }\overrightarrow{c}=\widehat{i}+2\widehat{j}-\widehat{k}\text{ respectively. Therefore, vector equation of }BC\text{ is}
\displaystyle \overrightarrow{r}=\overrightarrow{b}+\lambda(\overrightarrow{c}-\overrightarrow{b})
\displaystyle \text{or, }\overrightarrow{r}=2\widehat{i}+3\widehat{j}+4\widehat{k}+\mu\{(\widehat{i}+2\widehat{j}-\widehat{k})-(2\widehat{i}+3\widehat{j}+4\widehat{k})\}
\displaystyle \text{or, }\overrightarrow{r}=2\widehat{i}+3\widehat{j}+4\widehat{k}+\mu(-\widehat{i}-\widehat{j}-5\widehat{k})
\displaystyle \text{or, }\overrightarrow{r}=2\widehat{i}+3\widehat{j}+4\widehat{k}+v(\widehat{i}+\widehat{j}+5\widehat{k}),\ \text{where }v=-\mu\qquad \ldots(ii)
\displaystyle \text{The Cartesian equations of line (ii) are}
\displaystyle \frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{5}
\displaystyle \text{Suppose the coordinates of }D\text{ are }(x,y,z).\ \text{Since }ABCD\text{ is a parallelogram, the} \\ \text{diagonals }AC\text{ and }BD\text{ bisect each other. Therefore, }AC\text{ and }BD\text{ must have the} \\ \text{same mid-point.}
\displaystyle \text{The coordinates of the mid-point of }AC\text{ are }P\left(\frac{4+1}{2},\ \frac{5+2}{2},\ \frac{10-1}{2}\right)\text{ i.e. }P\left(\frac{5}{2},\ \frac{7}{2},\ \frac{9}{2}\right)
\displaystyle \text{and the coordinates of the mid-point of }BD\text{ are }Q\left(\frac{2+x}{2},\ \frac{3+y}{2},\ \frac{4+z}{2}\right).
\displaystyle \text{Since }P\text{ and }Q\text{ coincide. Therefore,}
\displaystyle \frac{2+x}{2}=\frac{5}{2},\ \frac{3+y}{2}=\frac{7}{2},\ \frac{4+z}{2}=\frac{9}{2}\Rightarrow x=3,\ y=4,\ z=5
\displaystyle \text{Thus, the coordinates of }D\text{ are }(3,4,5).

\displaystyle \textbf{Question 2: }\text{Find the vector equation of a line passing through a point with} \\ \text{position vector }2\widehat{i}-\widehat{j}+\widehat{k}, \text{and parallel to the line joining the points } -\widehat{i}+4\widehat{j}+\widehat{k} \text{ and }\widehat{i}+2\widehat{j}+2\widehat{k}.\ \text{Also, find the cartesian equivalent of this equation.}\quad [\text{CBSE 2003}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }A,\ B\text{ and }C\text{ be the points with position vectors }2\widehat{i}-\widehat{j}+\widehat{k},\ -\widehat{i}+4\widehat{j}+\widehat{k}\text{ and }\widehat{i}+2\widehat{j}+2\widehat{k}\text{ respectively.}
\displaystyle \text{We have to find the equation of a line passing through the point }A\text{ and parallel to }\overrightarrow{BC}.
\displaystyle \text{Clearly, }\overrightarrow{BC}=(\widehat{i}+2\widehat{j}+2\widehat{k})-(-\widehat{i}+4\widehat{j}+\widehat{k})=2\widehat{i}-2\widehat{j}+\widehat{k}
\displaystyle \text{We know that the equation of a line passing through a point }\overrightarrow{a}\text{ and parallel to }\overrightarrow{b}\text{ is }\overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}.
\displaystyle \text{Here, }\overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}-2\widehat{j}+\widehat{k}.\ \text{So, the vector equation of the required line is}
\displaystyle \overrightarrow{r}=(2\widehat{i}-\widehat{j}+\widehat{k})+\lambda(2\widehat{i}-2\widehat{j}+\widehat{k})\qquad \ldots(i)
\displaystyle \text{Reduction to cartesian form: Putting }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in (i), we obtain}
\displaystyle x\widehat{i}+y\widehat{j}+z\widehat{k}=(2+2\lambda)\widehat{i}+(-1-2\lambda)\widehat{j}+(1+\lambda)\widehat{k}
\displaystyle \Rightarrow x=2+2\lambda,\ y=-1-2\lambda,\ z=1+\lambda
\displaystyle \Rightarrow \frac{x-2}{2}=\frac{y+1}{-2}=\frac{z-1}{1},\ \text{which is the cartesian equivalent of equation (i).}

\displaystyle \textbf{Question 3: }\text{The cartesian equations of a line are }6x-2=3y+1=2z-2.\ \text{Find its direction ratios and also find vector equation of the line.}\quad [\text{CBSE 2003}]
\displaystyle \text{Answer:}
\displaystyle   \text{Recall that in the symmetrical form of a line the coefficients of }x,\ y\text{ and }z \\ \text{ are unity.} \text{Therefore, to put the given line in symmetrical form, we must make the} \\ \text{coefficients of }x,\ y\text{ and }z\text{ as unity.}
\displaystyle \text{We have, }6x-2=3y+1=2z-2
\displaystyle \Rightarrow 6\left(x-\frac{1}{3}\right)=3\left(y+\frac{1}{3}\right)=2(z-1)
\displaystyle \Rightarrow \frac{x-\frac{1}{3}}{1}=\frac{y+\frac{1}{3}}{2}=\frac{z-1}{3}\qquad [\text{Dividing throughout by the l.c.m. of }6,3,2\text{ i.e. }6]
\displaystyle \text{This shows that the given line passes through }\left(\frac{1}{3},-\frac{1}{3},1\right)\text{ and has direction} \\ \text{ratios proportional to }1,2,3.
\displaystyle \text{In vector form this means that the line passes through the point having position vector } \\ \overrightarrow{a}=\frac{1}{3}\widehat{i}-\frac{1}{3}\widehat{j}+\widehat{k}\text{ and is parallel to the vector }\overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k}.
\displaystyle \text{Therefore, its vector equation is}
\displaystyle \overrightarrow{r}=\left(\frac{1}{3}\widehat{i}-\frac{1}{3}\widehat{j}+\widehat{k}\right)+\lambda(\widehat{i}+2\widehat{j}+3\widehat{k}).

\displaystyle \textbf{Question 4: }\text{Find the point on the line }\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2} \\ \text{ at a distance of }3\sqrt{2}\text{ from the point }(1,2,3).\quad [\text{CBSE 2008}]
\displaystyle \text{Answer:}
\displaystyle  \text{The coordinates of any point on the line }\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}\text{ are given by}
\displaystyle \frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda
\displaystyle \text{or, }x+2=3\lambda,\ y+1=2\lambda,\ z-3=2\lambda\text{ or, }x=3\lambda-2,\ y=2\lambda-1,\ z=2\lambda+3\qquad \ldots(i)
\displaystyle \text{So, let the coordinates of the desired point be }(3\lambda-2,\ 2\lambda-1,\ 2\lambda+3). \\ \text{The distance between this point and }(1,2,3)\text{ is }3\sqrt{2}.
\displaystyle \therefore \sqrt{(3\lambda-2-1)^{2}+(2\lambda-1-2)^{2}+(2\lambda+3-3)^{2}}=3\sqrt{2}
\displaystyle \Rightarrow 9(\lambda-1)^{2}+(2\lambda-3)^{2}+4\lambda^{2}=18\Rightarrow 17\lambda^{2}-30\lambda=0\Rightarrow \lambda=0,\ \lambda=\frac{30}{17}
\displaystyle \text{Substituting the values of }\lambda\text{ in (i), we obtain that the coordinates of the desired} \\ \text{point are }(-2,-1,3)\text{ and }\left(\frac{56}{17},\ \frac{43}{17},\ \frac{111}{17}\right).

\displaystyle \textbf{Question 5: }\text{Find the Cartesian equations of the line passing through the point }(-1,3,-2)\text{ and}
\displaystyle \text{perpendicular to the lines }\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\text{ and }\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}.\quad [\text{CBSE 2005, 2012}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let the direction ratios of the required line be proportional to }a,b,c.
\displaystyle \text{Since it is perpendicular to the two given lines. Therefore,}
\displaystyle a+2b+3c=0\qquad \ldots(i)
\displaystyle \text{and}\ -3a+2b+5c=0\qquad \ldots(ii)
\displaystyle \text{Solving (i) and (ii) by cross-multiplication, we get}
\displaystyle \frac{a}{4}=\frac{b}{-14}=\frac{c}{8}\text{ or, }\frac{a}{2}=\frac{b}{-7}=\frac{c}{4}=k\ (\text{say})
\displaystyle \text{Thus, the required line passes through }(-1,3,-2)\text{ and has direction ratios proportional to }2,-7,4.
\displaystyle \text{So, its Cartesian equations are}
\displaystyle \frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}

\displaystyle \textbf{Question 6: }\text{Show that the line }\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\text{ and }\frac{x-4}{5}=\frac{y-1}{2}=z\text{ intersect.} \text{Find their point of intersection. }\quad [\text{CBSE 2004, 2005}]
\displaystyle \text{Answer:}
\displaystyle   \text{The coordinates of any point on first line are given by}
\displaystyle \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda\ (\text{say})
\displaystyle \text{or, }x=2\lambda+1,\ y=3\lambda+2,\ z=4\lambda+3
\displaystyle \text{So, the coordinates of a general point on first line are }(2\lambda+1,3\lambda+2,4\lambda+3).
\displaystyle \text{The coordinates of any point on second line are given by}
\displaystyle \frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{1}=\mu\ (\text{say})
\displaystyle \text{or, }x=5\mu+4,\ y=2\mu+1,\ z=\mu
\displaystyle \text{So, the coordinates of a general point on second line are }(5\mu+4,2\mu+1,\mu).
\displaystyle \text{If the lines intersect, then they have a common point.} \\ \text{So, for some values of }\lambda\text{ and }\mu, \text{we must have}
\displaystyle 2\lambda+1=5\mu+4,\ 3\lambda+2=2\mu+1\text{ and }4\lambda+3=\mu
\displaystyle \text{or, }2\lambda-5\mu=3,\ 3\lambda-2\mu=-1,\ 4\lambda-\mu=-3.
\displaystyle \text{Solving first two of these two equations, we get }\lambda=-1\text{ and }\mu=-1.
\displaystyle \text{Clearly, }\lambda=-1\text{ and }\mu=-1\text{ satisfy the third equation. So, the given lines intersect.}
\displaystyle \text{Putting }\lambda=-1\text{ in }(2\lambda+1,3\lambda+2,4\lambda+3),\text{ the coordinates of the required} \\ \text{point of intersection are }(-1,-1,-1).

\displaystyle \textbf{Question 7: }\text{Show that the lines }\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}\text{ and }\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}\text{ do not intersect.}\quad [\text{CBSE 2002}]
\displaystyle \text{Answer:}
\displaystyle   \text{The coordinates of any point on first line are given by}
\displaystyle \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}=\lambda\ (\text{say})
\displaystyle \text{or, }x=3\lambda+1,\ y=2\lambda-1,\ z=5\lambda+1
\displaystyle \text{So, the coordinates of any point on this line are }(3\lambda+1,2\lambda-1,5\lambda+1).
\displaystyle \text{The coordinates of any point on the second line are given by}
\displaystyle \frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}=\mu\ (\text{say})
\displaystyle \text{or, }x=4\mu-2,\ y=3\mu+1,\ z=-2\mu-1
\displaystyle \text{So, the coordinates of any point on second line are }(4\mu-2,3\mu+1,-2\mu-1).
\displaystyle \text{If the lines intersect, then they have a common point. So, for some values of } \\ \lambda\text{ and }\mu, \text{we must have}
\displaystyle 3\lambda+1=4\mu-2,\ 2\lambda-1=3\mu+1\text{ and }5\lambda+1=-2\mu-1
\displaystyle \Rightarrow 3\lambda-4\mu=-3\qquad \ldots(i)
\displaystyle 2\lambda-3\mu=2\qquad \ldots(ii)
\displaystyle 5\lambda+2\mu=-2\qquad \ldots(iii)
\displaystyle \text{Solving (i) and (ii), we obtain }\lambda=-17\text{ and }\mu=-12.
\displaystyle \text{These values of }\lambda\text{ and }\mu\text{ do not satisfy the third equation. Hence, the given lines} \\ \text{do not intersect.}

\displaystyle \textbf{Question 8: }\text{Show that the lines }\overrightarrow{r}=(\widehat{i}+\widehat{j}-\widehat{k})+\lambda(3\widehat{i}-\widehat{j})\text{ and }
\displaystyle \overrightarrow{r}=(4\widehat{i}-\widehat{k})+\mu(2\widehat{i}+3\widehat{k})\text{ intersect. Find their point of intersection.}\quad [\text{CBSE 2014}]
\displaystyle \text{Answer:}
\displaystyle  \text{The position vectors of arbitrary points on the given lines are}
\displaystyle (\widehat{i}+\widehat{j}-\widehat{k})+\lambda(3\widehat{i}-\widehat{j})=(3\lambda+1)\widehat{i}+(1-\lambda)\widehat{j}-\widehat{k}
\displaystyle \text{and }(4\widehat{i}-\widehat{k})+\mu(2\widehat{i}+3\widehat{k})=(2\mu+4)\widehat{i}+0\widehat{j}+(3\mu-1)\widehat{k}\text{ respectively.}
\displaystyle \text{If the lines intersect, then they have a common point.} \\ \text{So, for some values of }\lambda\text{ and }\mu,\text{ we must have}
\displaystyle (3\lambda+1)\widehat{i}+(1-\lambda)\widehat{j}-\widehat{k}=(2\mu+4)\widehat{i}+0\widehat{j}+(3\mu-1)\widehat{k}
\displaystyle \Rightarrow 3\lambda+1=2\mu+4,\ 1-\lambda=0\text{ and }-1=3\mu-1
\displaystyle \text{Solving last two of these equations, we get }\lambda=1\text{ and }\mu=0.
\displaystyle \text{These values of }\lambda\text{ and }\mu\text{ satisfy the first equation. So, the given lines intersect.}
\displaystyle \text{Putting }\lambda=1\text{ in first line, we get }\overrightarrow{r}=(\widehat{i}+\widehat{j}-\widehat{k})+(3\widehat{i}-\widehat{j})=4\widehat{i}+0\widehat{j}-\widehat{k}.
\displaystyle \text{Thus, the coordinates of the point of intersection are }(4,0,-1).

\displaystyle \textbf{Question 9: }\text{Find the image of the point } (1,6,3) \text{ in the line } \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}. \text{ Also, write the equation of the line joining the given point and its image and} \\ \text{find the length of the segment joining the given point and its image.}\quad [\text{CBSE 2010}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }Q\text{ be the image of point }P(1,6,3)\text{ in the line }\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} \\ \text{ and }M\text{ be the foot of perpendicular drawn from }P\text{ to this line. Then, }PM=MQ.
\displaystyle \text{Let the coordinates of }M\text{ be given by }\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=r.
\displaystyle \text{Let the coordinates of }M\text{ be }(r,\ 2r+1,\ 3r+2).
\displaystyle \text{The direction ratios of }PM\text{ are proportional to }r-1,\ 2r-5,\ 3r-1.
\displaystyle \text{Since }PM\text{ is perpendicular to the given line. Therefore,}
\displaystyle 1(r-1)+2(2r-5)+3(3r-1)=0\Rightarrow 14r-14=0\Rightarrow r=1.
\displaystyle \text{So, the coordinates of }M\text{ are }(1,3,5).
\displaystyle \text{Let }Q(x_{1},y_{1},z_{1})\text{ be the coordinates of }Q.\ \text{Since }M\text{ is the mid-point of }PQ.
\displaystyle \therefore \frac{x_{1}+1}{2}=1,\ \frac{y_{1}+6}{2}=3,\ \frac{z_{1}+3}{2}=5\Rightarrow x_{1}=1,\ y_{1}=0,\ z_{1}=7.
\displaystyle \text{Thus, the coordinates of }Q\text{ are }(1,0,7).
\displaystyle \text{So, the cartesian equations of }PQ\text{ are }\frac{x-1}{1-1}=\frac{y-6}{0-6}=\frac{z-3}{7-3}\text{ or, }\frac{x-1}{0}=\frac{y-6}{-6}=\frac{z-3}{4}.
\displaystyle \text{and, }PQ=\sqrt{(1-1)^{2}+(6-0)^{2}+(3-7)^{2}}=2\sqrt{13}.

\displaystyle \textbf{Question 10: }\text{Find the shortest distance between the lines whose vector equations are}
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+4\widehat{k})\text{ and }\overrightarrow{r}=(2\widehat{i}+4\widehat{j}+5\widehat{k})+\mu(4\widehat{i}+6\widehat{j}+8\widehat{k}).\quad [\text{CBSE 2008, 2015}]
\displaystyle \text{Answer:}
\displaystyle  \text{The vector equations of given lines are}
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+4\widehat{k})\qquad \ldots(i)
\displaystyle \text{and }\overrightarrow{r}=(2\widehat{i}+4\widehat{j}+5\widehat{k})+2\mu(2\widehat{i}+3\widehat{j}+4\widehat{k})\qquad \ldots(ii)
\displaystyle \text{Equation (ii) can be re-written as }
\displaystyle \overrightarrow{r}=(2\widehat{i}+4\widehat{j}+5\widehat{k})+\mu(2\widehat{i}+3\widehat{j}+4\widehat{k})\qquad \ldots(iii)
\displaystyle \text{These two lines pass through the points having position vectors }\overrightarrow{a_{1}}=\widehat{i}+2\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{a_{2}}=2\widehat{i}+4\widehat{j}+5\widehat{k}\text{ respectively and both are parallel to the vector }\overrightarrow{b}=2\widehat{i}+3\widehat{j}+4\widehat{k}.
\displaystyle \text{So, the shortest distance between them is given by}
\displaystyle \text{S.D.}=\frac{|(\overrightarrow{a_{2}}-\overrightarrow{a_{1}})\times\overrightarrow{b}|}{|\overrightarrow{b}|}\qquad \ldots(iv)
\displaystyle \text{Now, }(\overrightarrow{a_{2}}-\overrightarrow{a_{1}})\times\overrightarrow{b}=(\widehat{i}+2\widehat{j}+2\widehat{k})\times(2\widehat{i}+3\widehat{j}+4\widehat{k})=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&2&2\\2&3&4\end{vmatrix}=2\widehat{i}-0\widehat{j}-\widehat{k}.
\displaystyle \therefore |(\overrightarrow{a_{2}}-\overrightarrow{a_{1}})\times\overrightarrow{b}|=\sqrt{4+0+1}=\sqrt{5}\text{ and }|\overrightarrow{b}|=\sqrt{4+9+16}=\sqrt{29}.
\displaystyle \text{Substituting these values in (iv), we obtain }\text{S.D.}=\frac{\sqrt{5}}{\sqrt{29}}.

\displaystyle \textbf{Question 11. }\text{Distance of the point }(p,q,r)\text{ from }X\text{-axis is} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }q \qquad \text{(b) }|q| \qquad  \text{(c) }|q|+|r| \qquad \text{(d) }\sqrt{p^{2}+r^{2}}
\displaystyle \text{Answer:}
\displaystyle \text{1. (d) Given, the point is }(p,q,r).
\displaystyle \text{Any point on the Y-axis is of the form }(0,q,0).
\displaystyle \therefore \text{Distance of }(p,q,r)\text{ from Y-axis is}
\displaystyle \sqrt{(0-p)^2+(q-q)^2+(0-r)^2}=\sqrt{p^2+r^2}
\displaystyle \text{Hence, distance of }(p,q,r)\text{ from Y-axis is }\sqrt{p^2+r^2}.
\\

\displaystyle \textbf{Question 12. }\text{The point }(x,y,0)\text{ on the }XY\text{-plane divides the line segment joining}
\displaystyle \text{the points }(1,2,3)\text{ and }(3,2,1)\text{ in the}   \text{ratio} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }1:2\text{ internally} \qquad \text{(b) }2:1\text{ internally}
\displaystyle \text{(c) }3:1\text{ internally} \qquad \text{(d) }3:1\text{ externally}
\displaystyle \text{Answer:}
\displaystyle \text{2. (d) Let the point }P(x,y,0)\text{ on the XY-plane divide}
\displaystyle \text{the line segment joining the points }A(1,2,3)\text{ and}
\displaystyle B(3,2,1)\text{ in the ratio }k:1.
\displaystyle P(x,y,0)=\left(\frac{3k+1}{k+1},\frac{2k+2}{k+1},\frac{k+3}{k+1}\right)
\displaystyle \therefore \frac{k+3}{k+1}=0
\displaystyle \Rightarrow k+3=0
\displaystyle \Rightarrow k=-3
\displaystyle \text{Hence, the ratio is }3:1\text{ externally.}
\\

\displaystyle \textbf{Question 13. }\text{The angle between the lines }2x=3y=-z\text{ and}
\displaystyle 6x=-y=-4z\text{ is} \hspace{5.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }0^\circ \qquad \text{(b) }30^\circ  \qquad \text{(c) }45^\circ \qquad \text{(d) }90^\circ
\displaystyle \text{Answer:}
\displaystyle \text{3. (d) The given equations of lines can be rewritten as}
\displaystyle \frac{x-0}{3}=\frac{y-0}{2}=\frac{z-0}{-6}
\displaystyle \text{and }\frac{x-0}{2}=\frac{y-0}{-12}=\frac{z-0}{-3}.
\displaystyle \text{Here, direction ratios of first line are }(3,2,-6)
\displaystyle \text{and direction ratios of second line are }(2,-12,-3).
\displaystyle \text{Let }\theta\text{ be the angle between the lines.}
\displaystyle \text{Then, }\cos\theta=\frac{|3\times2+2\times(-12)+(-6)\times(-3)|}{\sqrt{3^2+2^2+(-6)^2}\sqrt{2^2+(-12)^2+(-3)^2}}
\displaystyle =\frac{|6-24+18|}{\sqrt{49}\sqrt{157}}=0
\displaystyle \Rightarrow \theta=\frac{\pi}{2}=90^\circ
\displaystyle \text{Hence, the angle between the lines is }90^\circ.
\\

\displaystyle \textbf{Question 14. }\text{If a line makes angles of }90^\circ,\ 135^\circ\text{ and }45^\circ\text{ with the } X,\ Y\text{ and }Z
\displaystyle \text{axes respectively, then its direction }   \text{cosines are} \hspace{0.2cm}\text{[CBSE 2023; CBSE 2019]}
\displaystyle \text{(a) }0,\ -\frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}}
\displaystyle \text{(b) }-\frac{1}{\sqrt{2}},\ 0,\ \frac{1}{\sqrt{2}}
\displaystyle \text{(c) }\frac{1}{\sqrt{2}},\ 0,\ -\frac{1}{\sqrt{2}}
\displaystyle \text{(d) }0,\ \frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}}
\displaystyle \text{Answer:}
\displaystyle \text{4. (a) Let direction cosines of the line be }l,m\text{ and }n.
\displaystyle \text{Given, }\alpha=90^\circ,\ \beta=135^\circ\text{ and }\gamma=45^\circ
\displaystyle \text{Then, }l=\cos\alpha=\cos90^\circ=0
\displaystyle \text{and }m=\cos\beta=\cos135^\circ=-\frac{1}{\sqrt2}
\displaystyle \text{and }n=\cos\gamma=\cos45^\circ=\frac{1}{\sqrt2}
\displaystyle \text{Hence, the direction cosines of a line are }0,\ -\frac{1}{\sqrt2},\ \frac{1}{\sqrt2}.
\\

\displaystyle \textbf{Question 15. }\text{The value of }\lambda\text{ for which the angle between the lines}
\displaystyle \overrightarrow{r}=\widehat{i}+\widehat{j}+\widehat{k}+p(2\widehat{i}+\widehat{j}+2\widehat{k})\text{ and}   \overrightarrow{r}=(1+q)\widehat{i}+(1+q\lambda)\widehat{j}+(1+q)\widehat{k}\text{ is }\frac{\pi}{2},
\displaystyle \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }-4 \qquad \text{(b) }4  \qquad \text{(c) }2 \qquad \text{(d) }-2
\displaystyle \text{Answer:}
\displaystyle \text{5. (a) The given lines can be rewritten as}
\displaystyle \overrightarrow{r}=\hat{i}+\hat{j}+\hat{k}+p(2\hat{i}+\hat{j}+2\hat{k})
\displaystyle \text{and}
\displaystyle \overrightarrow{r}=(\hat{i}+\hat{j}+\hat{k})+q(\hat{i}+\lambda\hat{j}+\hat{k})
\displaystyle \text{Let }\theta\text{ be the angle between these lines.}
\displaystyle \text{Then, }\cos\theta=\frac{(2\hat{i}+\hat{j}+2\hat{k})\cdot(\hat{i}+\lambda\hat{j}+\hat{k})}{\sqrt{2^2+1^2+2^2}\sqrt{1+\lambda^2+1}}
\displaystyle \Rightarrow \cos\frac{\pi}{2}=\frac{2+\lambda+2}{3\sqrt{\lambda^2+2}}
\displaystyle \Rightarrow \frac{4+\lambda}{3\sqrt{\lambda^2+2}}=0
\displaystyle \Rightarrow 4+\lambda=0
\displaystyle \Rightarrow \lambda=-4
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\displaystyle \textbf{Question 16. }\text{The direction cosines of vector }\overrightarrow{BA},\text{ where coordinates}
\displaystyle \text{of }A\text{ and }B\text{ are }(1,2,-1)\text{ and }(3,4,0)\text{ respectively, are} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }-2,\ -2,\ -1
\displaystyle \text{(b) }-\frac{2}{3},\ -\frac{2}{3},\ -\frac{1}{3}
\displaystyle \text{(c) }2,\ 2,\ 1
\displaystyle \text{(d) }\frac{2}{3},\ \frac{2}{3},\ \frac{1}{3}
\displaystyle \text{Answer:}
\displaystyle \text{6. (b) The given points are }A(1,2,-1)\text{ and }B(3,4,0).
\displaystyle \overrightarrow{BA}=(1-3)\hat{i}+(2-4)\hat{j}+(-1-0)\hat{k}=-2\hat{i}-2\hat{j}-\hat{k}
\displaystyle |\overrightarrow{BA}|=\sqrt{(-2)^2+(-2)^2+(-1)^2}=\sqrt{4+4+1}=\sqrt9=3
\displaystyle \text{Hence, the direction cosines of }\overrightarrow{BA}\text{ are}
\displaystyle -\frac23,\ -\frac23,\ -\frac13.
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\displaystyle \textbf{Question 17. }\text{Assertion (A) The lines }\overrightarrow{r}=\overrightarrow{a}_{1}+\lambda\overrightarrow{b}_{1}\text{ and }\overrightarrow{r}=\overrightarrow{a}_{2}+\mu\overrightarrow{b}_{2}
\displaystyle \text{are perpendicular, when }\overrightarrow{b}_{1}\cdot\overrightarrow{b}_{2}=0. \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Reason (R) The angle }\theta\text{ between the lines }\overrightarrow{r}=\overrightarrow{a}_{1}+\lambda\overrightarrow{b}_{1}\text{ and}
\displaystyle \overrightarrow{r}=\overrightarrow{a}_{2}+\mu\overrightarrow{b}_{2}\text{ is given by }\cos\theta=\frac{\overrightarrow{b}_{1}\cdot\overrightarrow{b}_{2}}{|\overrightarrow{b}_{1}||\overrightarrow{b}_{2}|}.
\displaystyle \text{(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).}
\displaystyle \text{(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).}
\displaystyle \text{(c) (A) is correct but (R) is incorrect.}
\displaystyle \text{(d) Both (A) and (R) are incorrect.}
\displaystyle \text{Answer:}
\displaystyle \text{11. (a) Assertion: The given lines are }
\displaystyle \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and }\overrightarrow{r}=\overrightarrow{a_2}+\lambda\overrightarrow{b_2}.
\displaystyle \text{Let }\theta\text{ be the angle between these lines.}
\displaystyle \cos\theta=\frac{\overrightarrow{b_1}\cdot\overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}
\displaystyle \text{When }\overrightarrow{b_1}\cdot\overrightarrow{b_2}=0
\displaystyle \text{Then, }\cos\theta=0
\displaystyle \Rightarrow \theta=\frac{\pi}{2}
\displaystyle \text{Hence, the given lines are perpendicular.}
\displaystyle \text{Both Assertion and Reason are true and Reason is a}
\displaystyle \text{correct explanation of Assertion.}
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\displaystyle \textbf{Question 18. }\text{Find the coordinates of points on line }   \frac{x}{1}=\frac{y-1}{2}=\frac{z+1}{2}
\displaystyle \text{which are at a distance of }\sqrt{11}\text{ units from origin.} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{28. Given, equation of line is}
\displaystyle \frac{x}{1}=\frac{y-1}{2}=\frac{z+1}{2}=\lambda\ (\text{Let})
\displaystyle \Rightarrow x=\lambda,\ y=2\lambda+1\text{ and }z=2\lambda-1
\displaystyle \therefore \text{Point is }(\lambda,2\lambda+1,2\lambda-1) \qquad ...(i)
\displaystyle \text{Given, distance of this point from origin is }\sqrt{11}\text{ units.}
\displaystyle \therefore \sqrt{(\lambda-0)^2+(2\lambda+1-0)^2+(2\lambda-1-0)^2}=\sqrt{11}
\displaystyle \text{Squaring on both sides, we get}
\displaystyle \lambda^2+4\lambda^2+1+4\lambda+4\lambda^2+1-4\lambda=11
\displaystyle 9\lambda^2+2=11\Rightarrow 9\lambda^2=9\Rightarrow \lambda^2=1\Rightarrow \lambda=\pm1
\displaystyle \text{From Eq. (i), we have }(1,3,1)\text{ and }(-1,-1,-3).
\displaystyle \text{Hence, the required points are }(1,3,1)\text{ and }(-1,-1,-3).
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\displaystyle \textbf{Question 19. }\text{Find the vector and the cartesian equations of a line passes through}
\displaystyle \text{the point }A(4,2,-1)\text{ and parallel to the }  \text{line }5x-25=14-7y=35z. \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{29. Given, line can be written as}
\displaystyle \frac{x-5}{1/15}=\frac{y-2}{-1/7}=\frac{z-0}{1/35}
\displaystyle \Rightarrow \frac{x-5}{7}=\frac{y-2}{-5}=\frac{z-0}{1} \qquad ...(i)
\displaystyle \text{Hence, parallel vector of given line}
\displaystyle \text{i.e., }\overrightarrow{b}=7\hat{i}-5\hat{j}+\hat{k}
\displaystyle \text{Since, required line is parallel to given line (i).}
\displaystyle \overrightarrow{b}=7\hat{i}-5\hat{j}+\hat{k}\text{ will also be parallel vector of required}
\displaystyle \text{line which passes through }A(1,2,-1).
\displaystyle \therefore \text{Required vector equation of line is}
\displaystyle \overrightarrow{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda(7\hat{i}-5\hat{j}+\hat{k})
\displaystyle \text{For cartesian equation, write}
\displaystyle \overrightarrow{r}=(x\hat{i}+y\hat{j}+z\hat{k})
\displaystyle \Rightarrow (x\hat{i}+y\hat{j}+z\hat{k})=(\hat{i}+2\hat{j}-\hat{k})+\lambda(7\hat{i}-5\hat{j}+\hat{k})
\displaystyle \text{On comparing the coefficient of }\hat{i},\hat{j}\text{ and }\hat{k},\text{ we get}
\displaystyle x=1+7\lambda,\ y=2-5\lambda\text{ and }z=-1+\lambda
\displaystyle \Rightarrow \frac{x-1}{7}=\frac{y-2}{-5}=\frac{z+1}{1}=\lambda
\displaystyle \therefore \frac{x-1}{7}=\frac{y-2}{-5}=\frac{z+1}{1}
\displaystyle \text{which is the required cartesian equation of line.}
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\displaystyle \textbf{Question 20. }\text{Find the value of }p,\text{ so that lines }   \frac{x-1}{-2}=\frac{y-4}{3p}=\frac{z-3}{4}
\displaystyle \text{and }\frac{x-2}{4p}=\frac{y-5}{2}=\frac{1-z}{7}\text{ are perpendicular to each }   \text{other.} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{30. The given lines are written as}
\displaystyle \frac{x-1}{-2}=\frac{y-4}{3p}=\frac{z-3}{4}
\displaystyle \text{and }\frac{x-2}{4p}=\frac{y-5}{2}=\frac{z-1}{-7}
\displaystyle \text{Here, direction ratios of above lines are }(-2,3p,4)
\displaystyle \text{and }(4p,2,-7).
\displaystyle \text{We know that two lines are perpendicular, if}
\displaystyle a_1a_2+b_1b_2+c_1c_2=0
\displaystyle \Rightarrow -2\times4p+3p\times2+4\times(-7)=0
\displaystyle \Rightarrow -8p+6p-28=0
\displaystyle \Rightarrow -2p=28
\displaystyle \Rightarrow p=-14
\displaystyle \text{Hence, the value of }p\text{ is }-14.

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\displaystyle \textbf{Question 21. }\text{Check whether the lines given by equations }x=2\lambda+2,
\displaystyle y=7\lambda+1,\ z=-3\lambda-3\text{ and }x=-\mu-2,\ y=2\mu+8,
\displaystyle z=4\mu+5\text{ are perpendicular or not.} \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{31. The given equations of lines can be written as}
\displaystyle \frac{x-2}{2}=\frac{y-1}{7}=\frac{z+3}{-3}=\lambda
\displaystyle \text{and }\frac{x+2}{-1}=\frac{y-8}{2}=\frac{z-5}{4}=\mu
\displaystyle \Rightarrow \frac{x-2}{2}=\frac{y-1}{7}=\frac{z+3}{-3}
\displaystyle \text{and }\frac{x+2}{-1}=\frac{y-8}{2}=\frac{z-5}{4}
\displaystyle \text{We know that if given lines are perpendicular to each}
\displaystyle \text{other, then }a_1a_2+b_1b_2+c_1c_2=0
\displaystyle \Rightarrow 2\times(-1)+7\times2+(-3)\times4
\displaystyle \Rightarrow -2+14-12=0
\displaystyle \text{Hence, the given equations of lines are perpendicular}
\displaystyle \text{to each other.}
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\displaystyle \textbf{Question 22. }\text{Find the distance between the lines}
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}-4\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+6\widehat{k}),   \overrightarrow{r}=(3\widehat{i}+3\widehat{j}-5\widehat{k})+\mu(4\widehat{i}+6\widehat{j}+12\widehat{k}).
\displaystyle \hspace{2.2cm}\text{[CBSE 2023; CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{37. Given, equation of lines are}
\displaystyle \overrightarrow{r}=(\hat{i}+2\hat{j}-4\hat{k})+\lambda(2\hat{i}+3\hat{j}+6\hat{k})
\displaystyle \text{and}
\displaystyle \overrightarrow{r}=(3\hat{i}+3\hat{j}-5\hat{k})+\mu(4\hat{i}+6\hat{j}+12\hat{k})
\displaystyle \text{On comparing with }\overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b},\text{ we get}
\displaystyle \overrightarrow{a_1}=(\hat{i}+2\hat{j}-4\hat{k}),\ \overrightarrow{b_1}=(2\hat{i}+3\hat{j}+6\hat{k})
\displaystyle \text{and }\overrightarrow{a_2}=(3\hat{i}+3\hat{j}-5\hat{k}),\ \overrightarrow{b_2}=(4\hat{i}+6\hat{j}+12\hat{k})
\displaystyle \text{Now, }\overrightarrow{a_2}-\overrightarrow{a_1}=(3\hat{i}+3\hat{j}-5\hat{k})-(\hat{i}+2\hat{j}-4\hat{k})
\displaystyle =2\hat{i}+\hat{j}-\hat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&6\\4&6&12\end{vmatrix}
\displaystyle =\hat{i}(36-36)-\hat{j}(24-24)+\hat{k}(12-12)=0
\displaystyle \text{So, both given lines are parallel and }\overrightarrow{b}=2\hat{i}+3\hat{j}+6\hat{k}
\displaystyle \text{Then, }\overrightarrow{b}\times(\overrightarrow{a_2}-\overrightarrow{a_1})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&6\\2&1&-1\end{vmatrix}
\displaystyle =\hat{i}(-3-6)-\hat{j}(-2-12)+\hat{k}(2-6)
\displaystyle =-9\hat{i}+14\hat{j}-4\hat{k}
\displaystyle \text{Now, required distance between given lines is}
\displaystyle d=\frac{\left|\overrightarrow{b}\times(\overrightarrow{a_2}-\overrightarrow{a_1})\right|}{|\overrightarrow{b}|}
\displaystyle =\frac{\sqrt{81+196+16}}{\sqrt{4+9+36}}
\displaystyle =\frac{\sqrt{293}}{\sqrt{49}}=\frac{\sqrt{293}}{7}\text{ units}.
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\displaystyle \textbf{Question 23. }\text{Find the coordinates of the foot of the perpendicular}
\displaystyle \text{drawn from the point }P(0,2,3)\text{ to the line }   \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}. \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{38. Given, equation of line is }\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}
\displaystyle \text{Let }\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda\text{ (say)}
\displaystyle \Rightarrow \frac{x+3}{5}=\lambda,\ \frac{y-1}{2}=\lambda\text{ and }\frac{z+4}{3}=\lambda
\displaystyle \Rightarrow x=5\lambda-3,\ y=2\lambda+1\text{ and }z=3\lambda-4
\displaystyle \therefore \text{Coordinates of point }L\text{ are }(5\lambda-3,2\lambda+1,3\lambda-4)\qquad ...(i)
\displaystyle \text{Now, DR's of line }PL
\displaystyle =(5\lambda-3-0,\ 2\lambda+1-2,\ 3\lambda-4-3)
\displaystyle =(5\lambda-3,\ 2\lambda-1,\ 3\lambda-7)
\displaystyle \text{DR's of line }AB\text{ are }(5,2,3).
\displaystyle \because PL\perp AB
\displaystyle \therefore a_1a_2+b_1b_2+c_1c_2=0\qquad ...(ii)
\displaystyle \text{where }a_1=5\lambda-3,\ b_1=2\lambda-1,\ c_1=3\lambda-7
\displaystyle \text{and }a_2=5,\ b_2=2,\ c_2=3
\displaystyle \text{From Eq. (ii), we get}
\displaystyle 5(5\lambda-3)+2(2\lambda-1)+3(3\lambda-7)=0
\displaystyle \Rightarrow 25\lambda-15+4\lambda-2+9\lambda-21=0
\displaystyle \Rightarrow 38\lambda-38=0
\displaystyle \Rightarrow 38\lambda=38\Rightarrow \lambda=1
\displaystyle \therefore \text{Foot of perpendicular }L=(5\lambda-3,2\lambda+1,3\lambda-4)=(2,3,-1)
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\displaystyle \textbf{Question 24. }\text{Find the coordinates of the foot of the perpendicular}
\displaystyle \text{drawn from point }P(5,7,3)\text{ to the line }   \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}. \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the foot of the perpendicular be }Q(x,y,z).
\displaystyle \text{The given line is } \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}=t
\displaystyle \therefore Q(15+3t,\ 29+8t,\ 5-5t)
\displaystyle \text{Direction ratios of the line are }(3,8,-5).
\displaystyle \overrightarrow{PQ}=(15+3t-5,\ 29+8t-7,\ 5-5t-3)
\displaystyle =(10+3t,\ 22+8t,\ 2-5t)
\displaystyle \text{Since }PQ\perp \text{ line, } \overrightarrow{PQ}\cdot(3,8,-5)=0
\displaystyle 3(10+3t)+8(22+8t)-5(2-5t)=0
\displaystyle 30+9t+176+64t-10+25t=0
\displaystyle 196+98t=0
\displaystyle t=-2
\displaystyle \therefore Q=(15+3(-2),\ 29+8(-2),\ 5-5(-2))
\displaystyle =(9,\ 13,\ 15)
\displaystyle \therefore \text{The coordinates of the foot of the perpendicular are }(9,\ 13,\ 15).
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\displaystyle \textbf{Question 25. }\text{Find the image of the point }(2,-1,5)\text{ in the line}
\displaystyle \frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}. \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \textbf{66. Let }T\text{ be the image of the point }P(2,-1,5). \text{ }Q\text{ is the}
\displaystyle \text{foot of perpendicular drawn from }P\text{ on the line }AB.
\displaystyle \text{Given, equation of line }AB\text{ is}
\displaystyle \frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}\qquad ...(i)
\displaystyle \text{Let }\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}=\lambda
\displaystyle \Rightarrow x=10\lambda+11,\ y=-4\lambda-2,\ z=-11\lambda-8
\displaystyle \text{Then, coordinates of }Q\text{ are}
\displaystyle (10\lambda+11,\ -4\lambda-2,\ -11\lambda-8)
\displaystyle \text{Now, D.R.'s of line }PQ
\displaystyle =(10\lambda+11-2,\ -4\lambda-2+1,\ -11\lambda-8-5)
\displaystyle =(10\lambda+9,\ -4\lambda-1,\ -11\lambda-13)
\displaystyle \text{Since, line }PQ\perp AB
\displaystyle \therefore a_1a_2+b_1b_2+c_1c_2=0
\displaystyle \text{where }a_1=10\lambda+9,\ b_1=-4\lambda-1,\ c_1=-11\lambda-13
\displaystyle \text{and }a_2=10,\ b_2=-4,\ c_2=-11
\displaystyle \therefore (10\lambda+9)(10)+(-4\lambda-1)(-4)+(-11\lambda-13)(-11)=0
\displaystyle \Rightarrow 100\lambda+90+16\lambda+4+121\lambda+143=0
\displaystyle \Rightarrow 237\lambda+237=0
\displaystyle \Rightarrow \lambda=-1
\displaystyle \text{On putting }\lambda=-1\text{ in Eq. (i), we get}
\displaystyle Q=(10(-1)+11,\ -4(-1)-2,\ -11(-1)-8)
\displaystyle =(1,2,3)
\displaystyle \text{Let image of point }P\text{ be }T(x,y,z). \text{ Then, }Q\text{ will be}
\displaystyle \text{the mid-point of }PT.
\displaystyle \text{By using mid-point formula,}
\displaystyle Q=\text{Mid-point of }P(2,-1,5)\text{ and }T(x,y,z)
\displaystyle \Rightarrow \left(\frac{x+2}{2},\frac{y-1}{2},\frac{z+5}{2}\right)=(1,2,3)
\displaystyle \text{On equating corresponding coordinates, we get}
\displaystyle \frac{x+2}{2}=1,\ \frac{y-1}{2}=2,\ \frac{z+5}{2}=3
\displaystyle \Rightarrow x=0,\ y=5,\ z=1
\displaystyle \therefore \text{Coordinates of }T=(0,5,1)
\displaystyle \text{Hence, the coordinate image of point }P(2,-1,5)\text{ is}
\displaystyle T(0,5,1).
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\displaystyle \textbf{Question 26. }\text{Vertices }B\text{ and }C\text{ of }\triangle ABC\text{ lie on the line}
\displaystyle \frac{x+2}{2}=\frac{y-1}{1}=\frac{z}{4}.\text{ Find the area of }\triangle ABC\text{ given that } \text{point }A\text{ has coordinates }
\displaystyle (1,-1,2) \text{ and the line segment }   BC\text{ has length of }5\text{ units.} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \textbf{67. Let }h\text{ be the height of }\triangle ABC.\text{ Then, }h\text{ is the length of}
\displaystyle \text{perpendicular from }A(1,-1,2)\text{ to the line}
\displaystyle \frac{x+2}{2}=\frac{y-1}{1}=\frac{z-0}{4}
\displaystyle \text{Clearly, line }\frac{x+2}{2}=\frac{y-1}{1}=\frac{z-0}{4}\text{ passes through the}
\displaystyle \text{point }P(-2,1,0)\text{ and is parallel to the vector}
\displaystyle \overrightarrow{b}=2\hat{i}+\hat{j}+4\hat{k}
\displaystyle \text{Let }\frac{x+2}{2}=\frac{y-1}{1}=\frac{z-0}{4}=\lambda.\text{ Then, coordinates of }M
\displaystyle \text{are }(2\lambda-2,\lambda+1,4\lambda)
\displaystyle \text{Now, D.R.'s of }\overrightarrow{AM}=(2\lambda-2-1,\lambda+1+1,4\lambda-2)
\displaystyle =(2\lambda-3,\lambda+2,4\lambda-2)
\displaystyle \text{Since, }AM\perp BC
\displaystyle \therefore 2(2\lambda-3)+1(\lambda+2)+4(4\lambda-2)=0
\displaystyle \Rightarrow 4\lambda-6+\lambda+2+16\lambda-8=0
\displaystyle \Rightarrow 21\lambda-12=0\Rightarrow \lambda=\frac{4}{7}
\displaystyle \text{Thus, the coordinates of }M\text{ are}
\displaystyle \left(2\times\frac{4}{7}-2,\frac{4}{7}+1,4\times\frac{4}{7}\right)
\displaystyle =\left(-\frac{6}{7},\frac{11}{7},\frac{16}{7}\right)
\displaystyle \text{i.e., }h=|AM|
\displaystyle =\sqrt{\left(-\frac{6}{7}-1\right)^2+\left(\frac{11}{7}+1\right)^2+\left(\frac{16}{7}-2\right)^2}
\displaystyle =\sqrt{\left(-\frac{13}{7}\right)^2+\left(\frac{18}{7}\right)^2+\left(\frac{2}{7}\right)^2}
\displaystyle =\sqrt{\frac{169+324+4}{49}}=\sqrt{\frac{497}{49}}=\sqrt{\frac{71}{7}}
\displaystyle \text{It is given that length of }BC\text{ is }5\text{ units.}
\displaystyle \therefore \text{Area of }\triangle ABC=\frac{1}{2}(BC\times h)
\displaystyle =\frac{1}{2}\times5\times\sqrt{\frac{71}{7}}
\displaystyle =\sqrt{\frac{1775}{28}}\ \text{sq units}
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\displaystyle \textbf{Question 27. }\text{Show that the following lines do not intersect each other}
\displaystyle \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}\text{ and }\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}. \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \textbf{68. Given, equations of lines are}
\displaystyle \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}=\lambda\qquad ...(i)
\displaystyle \text{and}
\displaystyle \frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}=\mu\qquad ...(ii)
\displaystyle \text{The coordinates of points on lines (i) and (ii) will be}
\displaystyle (3\lambda+1,\ 2\lambda-1,\ 5\lambda+1)\text{ and }(4\mu-2,\ 3\mu+1,\ -2\mu-1)
\displaystyle \text{If the lines intersect, then they have a common point.}
\displaystyle \text{So, }3\lambda+1=4\mu-2\Rightarrow 3\lambda-4\mu=-3\qquad ...(iii)
\displaystyle \text{and }2\lambda-1=3\mu+1\Rightarrow 2\lambda-3\mu=2\qquad ...(iv)
\displaystyle \text{and }5\lambda+1=-2\mu-1\Rightarrow 5\lambda+2\mu=-2\qquad ...(v)
\displaystyle \text{On solving Eqs. (iii) and (iv), we get}
\displaystyle \lambda=-17\text{ and }\mu=-12
\displaystyle \text{On putting the value of }\lambda\text{ and }\mu\text{ in Eq. (v), we get}
\displaystyle 5(-17)+2(-12)=-2
\displaystyle \Rightarrow -85-24=-2
\displaystyle \Rightarrow -109\neq-2
\displaystyle \text{Since, the value of }\lambda\text{ and }\mu\text{ does not satisfy}
\displaystyle \text{Eq. (v), so the lines do not intersect each other.}
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\displaystyle \textbf{Question 28. }\text{Find the angle between the lines } 2x=3y=-z\text{ and }
\displaystyle 6x=-y=-4z. \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \textbf{69. Given, equations of lines can be written as}
\displaystyle \frac{x-0}{1/2}=\frac{y-0}{1/3}=\frac{z-0}{-1}
\displaystyle \text{and}
\displaystyle \frac{x}{1/6}=\frac{y-0}{-1}=\frac{z-0}{-1/4}
\displaystyle \text{Now, D.R.'s of lines are }\left(\frac12,\frac13,-1\right)\text{ and }\left(\frac16,-1,-\frac14\right)
\displaystyle \text{Let }\theta\text{ be the angle between the given lines.}
\displaystyle \cos\theta=\frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}
\displaystyle =\frac{\left(\frac12\right)\left(\frac16\right)+\left(\frac13\right)(-1)+(-1)\left(-\frac14\right)}{\sqrt{\left(\frac12\right)^2+\left(\frac13\right)^2+(-1)^2}\sqrt{\left(\frac16\right)^2+(-1)^2+\left(-\frac14\right)^2}}
\displaystyle =\frac{\frac1{12}-\frac13+\frac14}{\sqrt{\frac14+\frac19+1}\sqrt{\frac1{36}+1+\frac1{16}}}
\displaystyle =0
\displaystyle \Rightarrow \cos\theta=\cos\frac{\pi}{2}
\displaystyle \Rightarrow \theta=\frac{\pi}{2}
\displaystyle \text{Hence, the required angle between the given lines is }\frac{\pi}{2}.
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\displaystyle \textbf{Question 29. }\text{Find the value of }b,\text{ so that the lines}
\displaystyle \frac{x-1}{2}=\frac{y-b}{3}=\frac{z-3}{4}\text{ and }\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{2} \text{are intersecting lines. }
\displaystyle \text{Also, find the point of intersection } \text{of these given lines.} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{70. Given, equations of lines are}
\displaystyle \frac{x-1}{2}=\frac{y-b}{3}=\frac{z-3}{4}=\lambda\text{ (say)}\qquad ...(i)
\displaystyle \text{and}
\displaystyle \frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1}=\mu\text{ (say)}\qquad ...(ii)
\displaystyle \text{The coordinates of points on line (i) and (ii) will be}
\displaystyle (2\lambda+1,3\lambda+b,4\lambda+3)\text{ and }(5\mu+4,2\mu+1,\mu).
\displaystyle \text{If the lines intersect, then point will be common.}
\displaystyle \text{So, }2\lambda+1=5\mu+4\Rightarrow 2\lambda-5\mu=3\qquad ...(iii)
\displaystyle 3\lambda+b=2\mu+1\Rightarrow 3\lambda-2\mu=1-b\qquad ...(iv)
\displaystyle 4\lambda+3=\mu\Rightarrow 4\lambda-\mu=-3\qquad ...(v)
\displaystyle \text{On solving Eqs. (iii) and (v), we get}
\displaystyle \lambda=-1\text{ and }\mu=-1
\displaystyle \text{Since, given lines are intersect to each other.}
\displaystyle \therefore \text{The value of }\lambda\text{ and }\mu\text{ satisfies the Eq. (iv), we get}
\displaystyle 3(-1)-2(-1)=1-b
\displaystyle \Rightarrow -3+2=1-b
\displaystyle \Rightarrow b+1=1\Rightarrow b=2
\displaystyle \text{Now, putting }\lambda=-1\text{ and }b=2\text{ in coordinates of points}
\displaystyle \text{on line (i), we get}
\displaystyle =(2(-1)+1,3(-1)+2,4(-1)+3)
\displaystyle =(-1,-1,-1)
\displaystyle \text{Hence, the intersection point is }(-1,-1,-1).
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\displaystyle \textbf{Question 30. }\text{The equations of motion of a rocket are }x=2t, y=-4t,\
\displaystyle z=4t,  \text{ where time }t\text{ is in seconds and the } \text{coordinates of a moving point in kilometre.}
\displaystyle  \hspace{2.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{72. Given, equations of motion of a rocket are}
\displaystyle x=2t,\quad y=-4t\text{ and }z=4t
\displaystyle \text{or }\frac{x}{2}=t,\ \frac{y}{-4}=t\text{ and }\frac{z}{4}=t
\displaystyle \Rightarrow \frac{x}{2}=\frac{y}{-4}=\frac{z}{4}\qquad ...(i)
\displaystyle \text{which is the equation of line passing through the origin}
\displaystyle \text{and having direction ratios }2,-4,4.
\displaystyle \text{The line (i) is the path of the rocket.}
\displaystyle \text{At }t\text{ s rocket will be at the point }(2t,-4t,4t).
\displaystyle \therefore \text{At }t=10\text{ s rocket will be at the point }P(20,-40,40).
\displaystyle \text{Now, the distance of point }P(20,-40,40)\text{ from}
\displaystyle \text{starting point }O(0,0,0)\text{ is given by}
\displaystyle OP=\sqrt{(20)^2+(-40)^2+(40)^2}
\displaystyle =\sqrt{400+1600+1600}
\displaystyle =\sqrt{3600}=60\text{ km}
\displaystyle \text{Now, the distance of point }P(20,-40,40)\text{ from the given line}
\displaystyle \overrightarrow{r}=20\widehat{i}-10\widehat{j}+40\widehat{k}+\mu(10\widehat{i}-20\widehat{j}+10\widehat{k})\text{ is}
\displaystyle =\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\times\overrightarrow{b}|}{|\overrightarrow{b}|}
\displaystyle =\frac{|-30\widehat{j}\times(10\widehat{i}-20\widehat{j}+10\widehat{k})|}{|10\widehat{i}-20\widehat{j}+10\widehat{k}|}
\displaystyle =\frac{|-300\widehat{i}+300\widehat{k}|}{\sqrt{(10)^2+(-20)^2+(10)^2}}
\displaystyle =\frac{\sqrt{(-300)^2+(300)^2}}{\sqrt{100+400+100}}
\displaystyle =\frac{300\sqrt2}{\sqrt{600}}=\frac{300\sqrt2}{10\sqrt6}=10\sqrt3\text{ km}
\\

\displaystyle \textbf{Question 31. }\text{The two lines }x=ay+b,\ z=cy+d\text{ and }x=a'y+b',
\displaystyle z=c'y+d'\text{ are perpendicular to each other, if} \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }\frac{a}{a'}+\frac{c}{c'}=1 \qquad \text{(b) }\frac{a}{a'}+\frac{c}{c'}=-1
\displaystyle \text{(c) }aa'+cc'=1 \qquad \text{(d) }aa'+cc'=-1
\displaystyle \text{Answer:}
\displaystyle \text{8. (d) We have,}
\displaystyle x=ay+b,\ z=cy+d
\displaystyle \text{and}
\displaystyle x=a'y+b',\ z=c'y+d'
\displaystyle \therefore \frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}
\displaystyle \text{and }\frac{x-b'}{a'}=\frac{y}{1}=\frac{z-d'}{c'}
\displaystyle \text{Since, these lines are perpendicular,}
\displaystyle aa'+1+c\,c'=0
\displaystyle \left[\text{two lines are perpendicular, if }a_1a_2+b_1b_2+c_1c_2=1\right]
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\displaystyle \textbf{Question 32. }\text{The lines }\frac{x-1}{k}=\frac{y-3}{1}=\frac{4-z}{k}\text{ and } \frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{-2}
\displaystyle  \text{are mutually perpendicular, if the } \text{value of }k\text{ is} \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }-\frac{2}{3} \qquad \text{(b) }\frac{2}{3}
\displaystyle \text{(c) }-2 \qquad \text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \text{9. (a) We have,}
\displaystyle \frac{x-2}{1}=\frac{y-3}{1}=\frac{4-z}{k}
\displaystyle \text{and}
\displaystyle \frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{-2}
\displaystyle \text{or }\frac{x-2}{1},\ \frac{y-3}{1},\ \frac{z-4}{-k}
\displaystyle \text{and}
\displaystyle \frac{x-1}{k},\ \frac{y-4}{2},\ \frac{z-5}{-2}
\displaystyle \text{Since, the given lines are perpendicular,}
\displaystyle (1)(k)+(1)(2)+(-k)(-2)=0
\displaystyle \Rightarrow k+2+2k=0
\displaystyle \Rightarrow 3k+2=0
\displaystyle \Rightarrow k=-\frac23
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\displaystyle \textbf{Question 33. }\text{Find the shortest distance between two skew lines is } \text{to both the lines.}
\displaystyle  \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{15. For skew lines, the line of the shortest distance will be}
\displaystyle \text{perpendicular to both the lines and it is unique also.}
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\displaystyle \textbf{Question 34. }\text{Find the vector equation of the line which passes } \text{through the point }
\displaystyle (3,4,5)\text{ and is parallel to the vector }   2\widehat{i}+2\widehat{j}-3\widehat{k}. \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{16. Equation of a line passing through a point with}
\displaystyle \text{position vector }\overrightarrow{a}\text{ and parallel to a vector }\overrightarrow{b}\text{ is}
\displaystyle \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}
\displaystyle \text{Since, line passes through }(3,4,5)
\displaystyle \therefore \overrightarrow{a}=3\hat{i}+4\hat{j}+5\hat{k}
\displaystyle \text{Since, line is parallel to }2\hat{i}+2\hat{j}-3\hat{k}
\displaystyle \therefore \overrightarrow{b}=2\hat{i}+2\hat{j}-3\hat{k}
\displaystyle \text{Equation of line is }\overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}\text{ i.e.}
\displaystyle \overrightarrow{r}=(3\hat{i}+4\hat{j}+5\hat{k})+\lambda(2\hat{i}+2\hat{j}-3\hat{k})
\displaystyle \text{which is the required vector equation.}
\\

\displaystyle \textbf{Question 35. }\text{Find the vector and cartesian equations of the line which passes through}
\displaystyle \text{the point }(1,1,1)\text{ and is perpendicular }  \text{to the lines with equations}
\displaystyle \frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}\text{ and }\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}.
\displaystyle \text{Also, find the angle between the given lines.} \hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{74. Any line through the point }(1,1,1)\text{ is given by}
\displaystyle \frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c}\qquad ...(i)
\displaystyle \text{where }a,b\text{ and }c\text{ are the D.R.'s of line (i).}
\displaystyle \text{Now, the line (i) is perpendicular to the lines}
\displaystyle \frac{x+2}{1}=\frac{y-3}{-2}=\frac{z+1}{4}
\displaystyle \text{and }\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}
\displaystyle \text{whose D.R.'s are }(1,-2,4)\text{ and }(2,3,4),\text{ respectively.}
\displaystyle \therefore a-2b+4c=0\qquad ...(ii)
\displaystyle \text{and }2a+3b+4c=0\qquad ...(iii)
\displaystyle \text{By cross-multiplication method, we get}
\displaystyle \frac{a}{8-12}=\frac{b}{8-4}=\frac{c}{3-(-4)}
\displaystyle \Rightarrow \frac{a}{-4}=\frac{b}{4}=\frac{c}{-1}
\displaystyle \therefore \text{D.R.'s of line (i) are }-4,4,-1.
\displaystyle \text{The required cartesian equation of line (i) is}
\displaystyle \frac{x-1}{-4}=\frac{y-1}{4}=\frac{z-1}{-1}
\displaystyle \text{and vector equation is}
\displaystyle \overrightarrow{r}=\widehat{i}+\widehat{j}+\widehat{k}+\lambda(-4\widehat{i}+4\widehat{j}-\widehat{k}).
\displaystyle \text{Again, let }\theta\text{ be the angle between the given lines.}
\displaystyle \cos\theta=\frac{(1\times2)+(-2\times3)+(4\times4)}{\sqrt{1+4+16}\sqrt{4+9+16}}
\displaystyle =\frac{24}{\sqrt{21}\sqrt{29}}=\frac{24}{\sqrt{609}}
\displaystyle \therefore \theta=\cos^{-1}\left(\frac{24}{\sqrt{609}}\right)
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\displaystyle \textbf{Question 36. }\text{A line passes through the point with position vector}
\displaystyle 2\widehat{i}-\widehat{j}+4\widehat{k}\text{ and is in the direction of the vector }   \widehat{i}+\widehat{j}-2\widehat{k}.\text{ Find the equation of}
\displaystyle \text{the line in cartesian form.} \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{17. The given line passes through the point }A\text{ having}
\displaystyle \text{position vector }\overrightarrow{a}=2\hat{i}-\hat{j}+4\hat{k}\text{ and is parallel to the}
\displaystyle \text{vector }\overrightarrow{b}=\hat{i}+\hat{j}-2\hat{k}.
\displaystyle \therefore \text{The equation of the given line is}
\displaystyle \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}
\displaystyle \Rightarrow \overrightarrow{r}=(2\hat{i}-\hat{j}+4\hat{k})+\lambda(\hat{i}+\hat{j}-2\hat{k}) \qquad ...(i)
\displaystyle \text{For cartesian equation, put }\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\text{ in Eq. (i), we get}
\displaystyle (x\hat{i}+y\hat{j}+z\hat{k})=(2\hat{i}-\hat{j}+4\hat{k})+\lambda(\hat{i}+\hat{j}-2\hat{k})
\displaystyle x\hat{i}+y\hat{j}+z\hat{k}=(2+\lambda)\hat{i}+(\lambda-1)\hat{j}+(4-2\lambda)\hat{k}
\displaystyle \Rightarrow x=2+\lambda,\ y=\lambda-1,\ z=4-2\lambda
\displaystyle \therefore \frac{x-2}{1}=\frac{y+1}{1}=\frac{z-4}{-2}=\lambda
\displaystyle \text{Hence, }\frac{x-2}{1}=\frac{y+1}{1}=\frac{z-4}{-2}\text{ is the required equation of}
\displaystyle \text{the given line in cartesian form.}
\\

\displaystyle \textbf{Question 37. }\text{Find the value of }\lambda,\text{ so that the lines}
\displaystyle \frac{1-x}{3}=\frac{7y-14}{2}=\frac{z-3}{2}\text{ and }\frac{7-7x}{3\lambda}=\frac{y-5}{1}=\frac{6-z}{5}
\displaystyle \text{are at right angle. Also, find whether the lines are }   \text{intersecting or not.} \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{43. Given, equation of lines can be written in standard form as}
\displaystyle \frac{x-1}{-3}=\frac{y-2}{\lambda/7}=\frac{z-3}{2}=r_1\text{ (let)}\qquad ...(i)
\displaystyle \text{and}
\displaystyle \frac{x-1}{\frac{-3\lambda}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}=r_2\text{ (let)}\qquad ...(ii)
\displaystyle \text{These lines will intersect at right angle, if}
\displaystyle -3\left(\frac{-3\lambda}{7}\right)+\frac{\lambda}{7}(1)+2(-5)=0
\displaystyle \left[\because \text{ two lines with D.R.'s }a_1,b_1,c_1\text{ and }a_2,b_2,c_2\text{ are}\right.
\displaystyle \left.\text{perpendicular if }a_1a_2+b_1b_2+c_1c_2=0\right]
\displaystyle \Rightarrow \frac{9\lambda}{7}+\frac{\lambda}{7}=10
\displaystyle \Rightarrow \frac{10\lambda}{7}=10
\displaystyle \Rightarrow \lambda=7,
\displaystyle \text{which is the required value of }\lambda.
\displaystyle \text{Now, let us check whether the lines are intersecting or not.}
\displaystyle \text{Coordinates of any point on line (i) are}
\displaystyle (-3r_1+1,\ r_1+2,\ 2r_1+3)
\displaystyle \text{and coordinates of any point on line (ii) are}
\displaystyle (-3r_2+1,\ r_2+5,\ -5r_2+6)
\displaystyle \text{Clearly, the lines will intersect if}
\displaystyle (-3r_1+1,\ r_1+2,\ 2r_1+3)=(-3r_2+1,\ r_2+5,\ -5r_2+6)
\displaystyle \text{For some }r_1,r_2\in R
\displaystyle \Rightarrow -3r_1+1=-3r_2+1,\ r_1+2=r_2+5,\ 2r_1+3=-5r_2+6
\displaystyle \Rightarrow r_1=r_2,\ r_1-r_2=3,\ 2r_1+5r_2=3
\displaystyle \text{which is not possible simultaneously for any }r_1,r_2\in R.
\displaystyle \text{Hence, the lines are not intersecting.}
\\

\displaystyle \textbf{Question 38. }\text{If the lines }\frac{x-1}{3}=\frac{y-2}{2}=\frac{z-3}{2}\text{ and}
\displaystyle \frac{x-1}{3\lambda}=\frac{y-1}{2}=\frac{z-6}{-5}\text{ are perpendicular, find the value } \text{of }\lambda.\text{ Hence, find whether the }
\displaystyle \text{lines are intersecting or }   \text{not.} \hspace{2.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Direction ratios of the first line are }(-3,\ 2\lambda,\ 2).
\displaystyle \text{Direction ratios of the second line are }(3\lambda,\ 2,\ -5).
\displaystyle \text{Since the lines are perpendicular,}
\displaystyle (-3)(3\lambda)+(2\lambda)(2)+2(-5)=0
\displaystyle -9\lambda+4\lambda-10=0
\displaystyle -5\lambda-10=0
\displaystyle \lambda=-2
\displaystyle \text{Now, the lines become}
\displaystyle \frac{x-1}{-3}=\frac{y-2}{-4}=\frac{z-3}{2}=r
\displaystyle \text{and }\frac{x-1}{-6}=\frac{y-1}{2}=\frac{z-6}{-5}=s
\displaystyle \therefore x=1-3r,\ y=2-4r,\ z=3+2r
\displaystyle \text{and }x=1-6s,\ y=1+2s,\ z=6-5s
\displaystyle \text{For intersection, }1-3r=1-6s\Rightarrow r=2s
\displaystyle \text{Also, }2-4r=1+2s
\displaystyle 1=4r+2s
\displaystyle 1=8s+2s
\displaystyle s=\frac{1}{10},\quad r=\frac{1}{5}
\displaystyle \text{Now, }3+2r=3+\frac{2}{5}=\frac{17}{5}
\displaystyle \text{and }6-5s=6-\frac{5}{10}=\frac{11}{2}
\displaystyle \text{Since }\frac{17}{5}\neq\frac{11}{2},\text{ the lines do not intersect.}
\displaystyle \therefore \lambda=-2\text{ and the lines are not intersecting.}
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\displaystyle \textbf{Question 39. }\text{Find the shortest distance between the lines}
\displaystyle \overrightarrow{r}=(4\widehat{i}-\widehat{j})+\lambda(\widehat{i}+2\widehat{j}-3\widehat{k})   \text{ and }\overrightarrow{r}=(\widehat{i}-\widehat{j}+2\widehat{k})+\mu(2\widehat{i}+4\widehat{j}-5\widehat{k}). \hspace{0.2cm}\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{45. Given, equation of lines are}
\displaystyle \overrightarrow{r}=(4\widehat{i}-\widehat{j})+\lambda(\widehat{i}+2\widehat{j}-3\widehat{k})\qquad ...(i)
\displaystyle \text{and}
\displaystyle \overrightarrow{r}=(\widehat{i}-\widehat{j}+2\widehat{k})+\mu(2\widehat{i}+4\widehat{j}-5\widehat{k})\qquad ...(ii)
\displaystyle \text{On comparing Eqs. (i) and (ii) with }\overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1}\text{ and}
\displaystyle \overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2}\text{ respectively, we get}
\displaystyle \overrightarrow{a_1}=4\widehat{i}-\widehat{j},\quad \overrightarrow{b_1}=\widehat{i}+2\widehat{j}-3\widehat{k}
\displaystyle \text{and}
\displaystyle \overrightarrow{a_2}=\widehat{i}-\widehat{j}+2\widehat{k},\quad \overrightarrow{b_2}=2\widehat{i}+4\widehat{j}-5\widehat{k}
\displaystyle \text{Here, }\overrightarrow{a_2}-\overrightarrow{a_1}=-3\widehat{i}+2\widehat{k}
\displaystyle \text{and }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&2&-3\\2&4&-5\end{vmatrix}
\displaystyle =\widehat{i}(-10+12)-\widehat{j}(-5+6)+\widehat{k}(4-4)=2\widehat{i}-\widehat{j}
\displaystyle \Rightarrow |\overrightarrow{b_1}\times\overrightarrow{b_2}|=\sqrt{2^2+(-1)^2}=\sqrt5
\displaystyle \text{Now, the shortest distance between the given lines is given by}
\displaystyle d=\frac{|(\overrightarrow{b_1}\times\overrightarrow{b_2})\cdot(\overrightarrow{a_2}-\overrightarrow{a_1})|}{|\overrightarrow{b_1}\times\overrightarrow{b_2}|}
\displaystyle =\frac{|(2\widehat{i}-\widehat{j})\cdot(-3\widehat{i}+2\widehat{k})|}{\sqrt5}
\displaystyle =\frac{|-6|}{\sqrt5}=\frac{6}{\sqrt5}\text{ units}
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\displaystyle \textbf{Question 40. }\text{Find the shortest distance between the lines}
\displaystyle \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\text{ and }\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}.   \hspace{2.2cm}\text{[CBSE 2018C]}
\displaystyle \text{Answer:}
\displaystyle \text{46. Given, lines are }\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}
\displaystyle \text{and }\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}
\displaystyle \text{On comparing the given equations of lines with}
\displaystyle \frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}
\displaystyle \text{and }\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2},\text{ we get}
\displaystyle x_1=1,\ y_1=2,\ z_1=3,\ a_1=2,\ b_1=3,\ c_1=4
\displaystyle \text{and }x_2=2,\ y_2=4,\ z_2=5,\ a_2=3,\ b_2=4,\ c_2=5
\displaystyle \text{On putting these values in}
\displaystyle \begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix},\text{ we get}
\displaystyle \begin{vmatrix}1&2&2\\2&3&4\\3&4&5\end{vmatrix}
\displaystyle =1(15-16)-2(10-12)+2(8-9)
\displaystyle =-1+4-2=1
\displaystyle \text{Now, }\sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}
\displaystyle =\sqrt{(3\times5-4\times4)^2+(4\times3-5\times2)^2+(2\times4-3\times3)^2}
\displaystyle =\sqrt{(15-16)^2+(12-10)^2+(8-9)^2}
\displaystyle =\sqrt{(-1)^2+(2)^2+(-1)^2}=\sqrt6
\displaystyle \therefore \text{Shortest distance}=\frac{\left|\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}\right|}{\sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}}
\displaystyle =\frac{1}{\sqrt6}\text{ unit, which is the required shortest distance.}
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\displaystyle \textbf{Question 41. }\text{Find the vector equation of the line passing through}
\displaystyle \text{the point }A(1,2,-1)\text{ and parallel to the line }   5x-25=14-7y=35z. \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{35. Given, line is }5x-25=14-7y=35z
\displaystyle \Rightarrow \frac{x-5}{1/5}=\frac{2-y}{1/7}=\frac{z}{1/35}
\displaystyle \Rightarrow \frac{x-5}{1/5}=\frac{y-2}{-1/7}=\frac{z-0}{1/35} \qquad ...(i)
\displaystyle \Rightarrow \frac{x-5}{7}=\frac{y-2}{-5}=\frac{z-0}{1}
\displaystyle \text{Hence, parallel vector of given line i.e.}
\displaystyle \overrightarrow{b}=7\hat{i}-5\hat{j}+\hat{k}
\displaystyle \text{Since, required line is parallel to given line (i).}
\displaystyle \overrightarrow{b}=7\hat{i}-5\hat{j}+\hat{k}\text{ will also be parallel vector of}
\displaystyle \text{required line which passes through }A(1,2,-1).
\displaystyle \therefore \text{The required vector equation is}
\displaystyle \overrightarrow{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda(7\hat{i}-5\hat{j}+\hat{k})
\\

\displaystyle \textbf{Question 42. }\text{Find the vector and cartesian equations of a line}
\displaystyle \text{passing through }(1,2,-4)\text{ and perpendicular to the two}
\displaystyle \text{lines } \frac{x-8}{3}=\frac{y+9}{-16}=\frac{z-10}{7}
\displaystyle \text{and }\frac{x-15}{3}=\frac{y-29}{8}=\frac{z+5}{-5}. \hspace{2.2cm}\text{[CBSE 2017, 12]}
\displaystyle \text{Answer:}
\displaystyle \text{75. Any line through }(1,2,-4)\text{ can be written as}
\displaystyle \frac{x-1}{a}=\frac{y-2}{b}=\frac{z+4}{c}\qquad ...(i)
\displaystyle \text{where }a,b\text{ and }c\text{ are D.R.'s of line (i).}
\displaystyle \text{Now, the line (i) be perpendicular to the lines}
\displaystyle \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}
\displaystyle \text{and }\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}
\displaystyle \text{The D.R.'s of above lines are }(3,-16,7)\text{ and }(3,8,-5),
\displaystyle \text{respectively, which are perpendicular to the Eq. (i).}
\displaystyle \therefore 3a-16b+7c=0\qquad ...(ii)
\displaystyle \text{and }3a+8b-5c=0\qquad ...(iii)
\displaystyle \text{By cross-multiplication, we get}
\displaystyle \frac{a}{80-56}=\frac{b}{21+15}=\frac{c}{24+48}
\displaystyle \Rightarrow \frac{a}{24}=\frac{b}{36}=\frac{c}{72}
\displaystyle \Rightarrow \frac{a}{2}=\frac{b}{3}=\frac{c}{6}=\lambda\text{ (say)}
\displaystyle \Rightarrow a=2\lambda,\ b=3\lambda,\ c=6\lambda
\displaystyle \text{The equation of required line in cartesian form is}
\displaystyle \frac{x-1}{2\lambda}=\frac{y-2}{3\lambda}=\frac{z+4}{6\lambda}
\displaystyle \text{or }\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}
\displaystyle \text{and in vector form is}
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}-4\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+6\widehat{k}).
\\

\displaystyle \textbf{Question 43. }\text{Find the vector and cartesian equation of the line}
\displaystyle \text{through the point }(1,2,-4)\text{ and perpendicular to the } \text{two lines}
\displaystyle \overrightarrow{r}=(8\widehat{i}-19\widehat{j}+10\widehat{k})+\lambda(3\widehat{i}-16\widehat{j}+7\widehat{k})
\displaystyle \text{and }\overrightarrow{r}=(15\widehat{i}+29\widehat{j}+5\widehat{k})+\mu(3\widehat{i}+8\widehat{j}-5\widehat{k}). \hspace{2.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{48. Given, equations of lines are}
\displaystyle \overrightarrow{r}=(8\widehat{i}-19\widehat{j}+10\widehat{k})+\lambda(3\widehat{i}-16\widehat{j}+7\widehat{k})
\displaystyle \text{and}
\displaystyle \overrightarrow{r}=(15\widehat{i}+29\widehat{j}+5\widehat{k})+\mu(3\widehat{i}+8\widehat{j}-5\widehat{k})
\displaystyle \text{On comparing with vector form of equation of a line,}
\displaystyle \text{i.e., }\overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b},\text{ we get}
\displaystyle \overrightarrow{b_1}=3\widehat{i}-16\widehat{j}+7\widehat{k}\text{ and }\overrightarrow{b_2}=3\widehat{i}+8\widehat{j}-5\widehat{k}
\displaystyle \text{Now, we determine unit vector }\overrightarrow{b}\text{ as}
\displaystyle \overrightarrow{b}=\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&-16&7\\3&8&-5\end{vmatrix}
\displaystyle =\widehat{i}(80-56)-\widehat{j}(-15-21)+\widehat{k}(24+48)
\displaystyle =24\widehat{i}+36\widehat{j}+72\widehat{k}=12(2\widehat{i}+3\widehat{j}+6\widehat{k})
\displaystyle \text{Since, the required line is perpendicular to the given lines.}
\displaystyle \text{So, it is parallel to }\overrightarrow{b_1}\times\overrightarrow{b_2}.\text{ Now, the equation of}
\displaystyle \text{a line passing through the point }(1,2,-4)\text{ and parallel}
\displaystyle \text{to }24\widehat{i}+36\widehat{j}+72\widehat{k}\text{ or }(2\widehat{i}+3\widehat{j}+6\widehat{k})\text{ is}
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}-4\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+6\widehat{k})
\displaystyle \text{which is required vector equation of a line.}
\displaystyle \text{For cartesian equation, put }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k},\text{ we get}
\displaystyle x\widehat{i}+y\widehat{j}+z\widehat{k}=(\widehat{i}+2\widehat{j}-4\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+6\widehat{k})
\displaystyle \text{On comparing the coefficients of }\widehat{i},\widehat{j}\text{ and }\widehat{k},\text{ we get}
\displaystyle x=1+2\lambda,\ y=2+3\lambda\text{ and }z=-4+6\lambda
\displaystyle \Rightarrow \frac{x-1}{2}=\lambda,\ \frac{y-2}{3}=\lambda\text{ and }\frac{z+4}{6}=\lambda
\displaystyle \therefore \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}
\displaystyle \text{which is required cartesian equation of a line.}
\\

\displaystyle \textbf{Question 44. }\text{The equations of a line is }5x-3=15y+7=3-10z.
\displaystyle \text{Write the direction cosines of the line.} \hspace{2.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{19. Given equations of a line is}
\displaystyle 5x-3=15y+7=3-10z \qquad ...(i)
\displaystyle \text{Let us first convert the equation in standard form}
\displaystyle \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}
\displaystyle \text{Let us divide Eq. (i) by LCM (coefficients of }x,y\text{ and }z)\text{ i.e. LCM }(5,15,10)=30
\displaystyle \text{Now, the Eq. (i) becomes}
\displaystyle \frac{5x-3}{30}=\frac{15y+7}{30}=\frac{3-10z}{30}
\displaystyle \Rightarrow \frac{5\left(x-\frac35\right)}{30}=\frac{15\left(y+\frac7{15}\right)}{30}=\frac{-10\left(z-\frac3{10}\right)}{30}
\displaystyle \Rightarrow \frac{x-\frac35}{6}=\frac{y+\frac7{15}}{2}=\frac{z-\frac3{10}}{-3}
\displaystyle \text{On comparing the above equation with standard form, we get}
\displaystyle 6,\ 2,\ -3\text{ are the D.R.'s of the given line.}
\displaystyle \therefore \text{The direction cosines of given line are}
\displaystyle \frac6{\sqrt{6^2+2^2+(-3)^2}},\ \frac2{\sqrt{6^2+2^2+(-3)^2}},\ \frac{-3}{\sqrt{6^2+2^2+(-3)^2}}
\displaystyle \text{i.e. }\frac67,\ \frac27,\ -\frac37.
\\

\displaystyle \textbf{Question 45. }\text{Write the distance of a point }P(a,b,c)\text{ from }X\text{-axis.} \hspace{0.2cm}\text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{20. Given, point is }P(a,b,c).
\displaystyle \text{Then, the coordinates of the point on X-axis be }(a,0,0)
\displaystyle \left[\because \text{ x-coordinate of both points will be same}\right]
\displaystyle \therefore \text{Required distance}=\sqrt{(a-a)^2+(0-b)^2+(0-c)^2}
\displaystyle =\sqrt{0+b^2+c^2}=\sqrt{b^2+c^2}.
\\

\displaystyle \textbf{Question 46. }\text{If the cartesian equation of a line is } \frac{3-x}{5}=\frac{y+4}{7}=\frac{2z-6}{4},
\displaystyle \text{ then write the vector equation }   \text{of the line.} \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{21. Given, cartesian equation of a line is}
\displaystyle \frac{3-x}{5}=\frac{y+4}{7}=\frac{2z-6}{4}
\displaystyle \text{On rewriting the given equation in standard form,}
\displaystyle \text{we get}
\displaystyle \frac{x-3}{-5}=\frac{y+4}{7}=\frac{z-3}{2}=\lambda\ (\text{let})
\displaystyle \Rightarrow x=-5\lambda+3,\ y=7\lambda-4\text{ and }z=2\lambda+3
\displaystyle \text{Now, }x\hat{i}+y\hat{j}+z\hat{k}=(-5\lambda+3)\hat{i}+(7\lambda-4)\hat{j}+(2\lambda+3)\hat{k}
\displaystyle \Rightarrow \overrightarrow{r}=(3\hat{i}-4\hat{j}+3\hat{k})+\lambda(-5\hat{i}+7\hat{j}+2\hat{k})
\displaystyle \text{which is the required equation of line in vector form.}
\\

\displaystyle \textbf{Question 47. }\text{Write the equation of the straight line through the point}
\displaystyle (\alpha,\beta,\gamma)\text{ and parallel to }Z\text{-axis.} \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{22. The vector equation of a line parallel to Z-axis is}
\displaystyle \overrightarrow{m}=0\hat{i}+0\hat{j}+\hat{k}. \text{ Then, the required line passes through}
\displaystyle \text{the point }A(\alpha,\beta,\gamma),\text{ whose position vector is}
\displaystyle \overrightarrow{r_1}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}\text{ and is parallel to the vector}
\displaystyle \overrightarrow{m}=(0\hat{i}+0\hat{j}+\hat{k})
\displaystyle \therefore \text{The equation is }\overrightarrow{r}=\overrightarrow{r_1}+\lambda\overrightarrow{m}
\displaystyle \Rightarrow \overrightarrow{r}=(\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k})+\lambda(0\hat{i}+0\hat{j}+\hat{k})
\displaystyle \Rightarrow \overrightarrow{r}=(\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k})+\lambda\hat{k}
\\

\displaystyle \textbf{Question 48. }\text{Show that the lines }   \overrightarrow{r}=(\widehat{i}+\widehat{j}-\widehat{k})+\lambda(3\widehat{i}-\widehat{j})\text{ and}
\displaystyle \overrightarrow{r}=(4\widehat{i}-\widehat{k})+\mu(2\widehat{i}+3\widehat{k})\text{ intersect. Also, find their}   \text{point of intersection.} \hspace{0.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{49. Given, lines can be rewritten as}
\displaystyle \overrightarrow{r}=(3\lambda+1)\widehat{i}+(1-\lambda)\widehat{j}-\widehat{k}\qquad ...(i)
\displaystyle \text{and}
\displaystyle \overrightarrow{r}=(4+2\mu)\widehat{i}+0\widehat{j}+(3\mu-1)\widehat{k}\qquad ...(ii)
\displaystyle \text{Clearly, any point on line (i) is of the form}
\displaystyle P(3\lambda+1,\ 1-\lambda,\ -1)
\displaystyle \text{and any point on line (ii) is of the form}
\displaystyle Q(4+2\mu,\ 0,\ 3\mu-1).
\displaystyle \text{If lines (i) and (ii) intersect, then these points must}
\displaystyle \text{coincide for some }\lambda\text{ and }\mu.
\displaystyle \text{Consider,}
\displaystyle 3\lambda+1=4+2\mu
\displaystyle \Rightarrow 3\lambda-2\mu=3\qquad ...(iii)
\displaystyle \text{and}
\displaystyle 1-\lambda=0\qquad ...(iv)
\displaystyle 3\mu-1=-1\qquad ...(v)
\displaystyle \text{From Eq. (iv), we get }\lambda=1\text{ and put the value of }\lambda\text{ in Eq. (iii), we get}
\displaystyle 3(1)-2\mu=3\Rightarrow -2\mu=3-3\Rightarrow \mu=0
\displaystyle \text{On putting the value of }\mu\text{ in Eq. (v), we get}
\displaystyle 3(0)-1=-1\Rightarrow 0-1=-1
\displaystyle \Rightarrow -1=-1,\text{ which is true.}
\displaystyle \text{Hence, both lines intersect each other.}
\displaystyle \text{The point of intersection of both lines can be obtained}
\displaystyle \text{by putting }\lambda=1\text{ in coordinates of }P.
\displaystyle \text{So, the point of intersection is }(3+1,\ 1-1,\ -1),\text{ i.e.}
\displaystyle (4,0,-1).
\\

\displaystyle \textbf{Question 49. }\text{Find the direction cosines of the line}
\displaystyle \frac{x+2}{2}=\frac{2y-7}{6}=\frac{5-z}{6}. \text{ Also, find the vector equation of the line through the point}
\displaystyle  A(-1,2,3)\text{ and parallel to}   \text{the given line.} \hspace{2.2cm}\text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{50. Given, equation of line is}
\displaystyle \frac{x+2}{2}=\frac{2y-7}{6}=\frac{5-z}{6}
\displaystyle \text{This equation can be written as}
\displaystyle \frac{x+2}{2}=\frac{y-\frac{7}{2}}{3}=\frac{z-5}{-6}
\displaystyle \text{So, direction ratios of line are }(2,3,-6).
\displaystyle \text{Now, direction cosines of a line are}
\displaystyle l=\frac{2}{\sqrt{2^2+3^2+(-6)^2}},
\displaystyle m=\frac{3}{\sqrt{2^2+3^2+(-6)^2}},
\displaystyle n=\frac{-6}{\sqrt{2^2+3^2+(-6)^2}}
\displaystyle \left[\because l=\frac{a}{\sqrt{a^2+b^2+c^2}},\ m=\frac{b}{\sqrt{a^2+b^2+c^2}},\ n=\frac{c}{\sqrt{a^2+b^2+c^2}}\right]
\displaystyle \Rightarrow l=\frac{2}{\sqrt{49}},\ m=\frac{3}{\sqrt{49}},\ n=\frac{-6}{\sqrt{49}}
\displaystyle \text{So, direction cosines of given line are }\left(\frac{2}{7},\frac{3}{7},-\frac{6}{7}\right).
\displaystyle \text{Here, D.R.'s of a line parallel to given line are }(2,3,-6).
\displaystyle \text{So, the required equation of line passes through the point }A(-1,2,3)
\displaystyle \text{and parallel to given line is}
\displaystyle \frac{x+1}{2}=\frac{y-2}{3}=\frac{z-3}{-6}.
\\

\displaystyle \textbf{Question 50. }\text{Find the angle between the lines}
\displaystyle \overrightarrow{r}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda(3\widehat{i}+2\widehat{j}+6\widehat{k})
\displaystyle \text{and }\overrightarrow{r}=7\widehat{i}-6\widehat{j}-6\widehat{k}+\mu(\widehat{i}+2\widehat{j}+2\widehat{k}). \hspace{2.2cm}\text{[CBSE 2014; CBSE 2008C]}
\displaystyle \text{Answer:}
\displaystyle \text{51. Given, equations of lines are}
\displaystyle \overrightarrow{r}=(2\widehat{i}-5\widehat{j}+\widehat{k})+\lambda(3\widehat{i}+2\widehat{j}+6\widehat{k})\qquad ...(i)
\displaystyle \text{and}
\displaystyle \overrightarrow{r}=(7\widehat{i}-6\widehat{j}-6\widehat{k})+\mu(\widehat{i}+2\widehat{j}+2\widehat{k})\qquad ...(ii)
\displaystyle \text{On comparing Eqs. (i) and (ii) with vector form of}
\displaystyle \text{equation of line i.e. }\overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b},\text{ we get}
\displaystyle \overrightarrow{a_1}=2\widehat{i}-5\widehat{j}+\widehat{k},\quad \overrightarrow{b_1}=3\widehat{i}+2\widehat{j}+6\widehat{k}
\displaystyle \text{and }\overrightarrow{a_2}=7\widehat{i}-6\widehat{j}-6\widehat{k},\quad \overrightarrow{b_2}=\widehat{i}+2\widehat{j}+2\widehat{k}
\displaystyle \text{We know that the angle between two lines is given by}
\displaystyle \cos\theta=\left|\frac{\overrightarrow{b_1}\cdot\overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}\right|
\displaystyle \therefore \cos\theta=\left|\frac{(3\widehat{i}+2\widehat{j}+6\widehat{k})\cdot(\widehat{i}+2\widehat{j}+2\widehat{k})}{\sqrt{(3)^2+(2)^2+(6)^2}\sqrt{(1)^2+(2)^2+(2)^2}}\right|
\displaystyle =\left|\frac{3+4+12}{\sqrt{49}\sqrt9}\right|
\displaystyle =\frac{19}{7\times3}=\frac{19}{21}
\displaystyle \text{Hence, the angle between given two lines is }
\displaystyle \theta=\cos^{-1}\left(\frac{19}{21}\right).
\\

\displaystyle \textbf{Question 51. }\text{Show that the lines }
\displaystyle \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}\text{ and }\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}
\displaystyle \text{intersect. Also, find their point of intersection.} \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{52. Given, lines are}
\displaystyle \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=\lambda\text{ (let)}\qquad ...(i)
\displaystyle \text{and }\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}=\mu\text{ (let)}\qquad ...(ii)
\displaystyle \text{Then, any point on line (i) is of the form}
\displaystyle P(3\lambda-1,\ 5\lambda-3,\ 7\lambda-5)\qquad ...(iii)
\displaystyle \text{and any point on line (ii) is of the form}
\displaystyle Q(\mu+2,\ 3\mu+4,\ 5\mu+6)\qquad ...(iv)
\displaystyle \text{If lines (i) and (ii) intersect, then these points must}
\displaystyle \text{coincide for some }\lambda\text{ and }\mu.
\displaystyle \text{Consider,}
\displaystyle 3\lambda-1=\mu+2
\displaystyle \Rightarrow 3\lambda-\mu=3\qquad ...(v)
\displaystyle 5\lambda-3=3\mu+4
\displaystyle \Rightarrow 5\lambda-3\mu=7\qquad ...(vi)
\displaystyle \text{and}
\displaystyle 7\lambda-5=5\mu+6
\displaystyle \Rightarrow 7\lambda-5\mu=11\qquad ...(vii)
\displaystyle \text{On multiplying Eq. (v) by }3\text{ and then subtracting Eq. (vi), we get}
\displaystyle 9\lambda-3\mu-5\lambda+3\mu=9-7
\displaystyle \Rightarrow 4\lambda=2\Rightarrow \lambda=\frac12
\displaystyle \text{On putting the value of }\lambda\text{ in Eq. (v), we get}
\displaystyle 3\times\frac12-\mu=3
\displaystyle \Rightarrow \frac32-\mu=3
\displaystyle \Rightarrow \mu=-\frac32
\displaystyle \text{On putting the values of }\lambda\text{ and }\mu\text{ in Eq. (vii), we get}
\displaystyle 7\times\frac12-5\left(-\frac32\right)=11
\displaystyle \Rightarrow \frac72+\frac{15}{2}=11
\displaystyle \Rightarrow \frac{22}{2}=11
\displaystyle \Rightarrow 11=11,\text{ which is true.}
\displaystyle \text{Hence, lines (i) and (ii) intersect and their point of}
\displaystyle \text{intersection is}
\displaystyle P\left(3\times\frac12-1,\ 5\times\frac12-3,\ 7\times\frac12-5\right)
\displaystyle \left[\text{put }\lambda=\frac12\text{ in Eq. (iii)}\right]
\displaystyle \text{i.e. }P\left(\frac12,-\frac12,-\frac32\right).
\\

\displaystyle \textbf{Question 52. }\text{Find the value of }p,\text{ so that the lines}
\displaystyle l_{1}:\frac{1-x}{3}=\frac{7y-14}{p}=\frac{z-3}{2}\text{ and } l_{2}:\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}
\displaystyle \text{ are perpendicular to each }   \text{other. Also, find the equation of a line passing through}
\displaystyle \text{a point }(3,2,-4)\text{ and parallel to line }l_{1}. \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Direction ratios of }l_{1}\text{ are }\left(-3,\frac{p}{7},2\right).
\displaystyle \text{Direction ratios of }l_{2}\text{ are }\left(-\frac{3p}{7},1,-5\right).
\displaystyle \text{Since the lines are perpendicular,}
\displaystyle (-3)\left(-\frac{3p}{7}\right)+\frac{p}{7}(1)+2(-5)=0
\displaystyle \frac{9p}{7}+\frac{p}{7}-10=0
\displaystyle \frac{10p}{7}=10
\displaystyle p=7
\displaystyle \text{Therefore, direction ratios of }l_{1}\text{ are }(-3,1,2).
\displaystyle \therefore \text{Equation of the required line is } \frac{x-3}{-3}=\frac{y-2}{1}=\frac{z+4}{2}.
\\

\displaystyle \textbf{Question 53. }\text{A line passes through the point }(2,-1,3)\text{ and is perpendicular to the}
\displaystyle \text{lines }  \overrightarrow{r}=(\widehat{i}+\widehat{j}-\widehat{k})+\lambda(2\widehat{i}-2\widehat{j}+\widehat{k}) \text{ and } \overrightarrow{r}=(2\widehat{i}-\widehat{j}-3\widehat{k})+\mu(\widehat{i}+2\widehat{j}+2\widehat{k}).
\displaystyle \text{ Obtain its } \text{equation in vector and cartesian forms.}   \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Direction vectors of the given lines are}
\displaystyle \overrightarrow{a}=2\widehat{i}-2\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}+2\widehat{j}+2\widehat{k}
\displaystyle \text{Required line is perpendicular to both the given lines.}
\displaystyle \therefore \text{Direction vector of the required line is }\overrightarrow{a}\times\overrightarrow{b}
\displaystyle \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&-2&1\\1&2&2\end{vmatrix}
\displaystyle =\widehat{i}(-4-2)-\widehat{j}(4-1)+\widehat{k}(4+2)
\displaystyle =-6\widehat{i}-3\widehat{j}+6\widehat{k}
\displaystyle =-3(2\widehat{i}+\widehat{j}-2\widehat{k})
\displaystyle \therefore \text{Direction vector of required line may be taken as }2\widehat{i}+\widehat{j}-2\widehat{k}
\displaystyle \text{Hence, vector form is}
\displaystyle \overrightarrow{r}=2\widehat{i}-\widehat{j}+3\widehat{k}+t(2\widehat{i}+\widehat{j}-2\widehat{k})
\displaystyle \text{Cartesian form is}
\displaystyle \frac{x-2}{2}=\frac{y+1}{1}=\frac{z-3}{-2}
\\

\displaystyle \textbf{Question 54. }\text{Find the shortest distance between the lines whose vector equations are}
\displaystyle \overrightarrow{r}=\widehat{i}+\widehat{j}+\lambda(2\widehat{i}-\widehat{j}+\widehat{k})   \text{ and }\overrightarrow{r}=2\widehat{i}+\widehat{j}-\widehat{k}+\mu(3\widehat{i}-5\widehat{j}+2\widehat{k}). \hspace{0.2cm}\text{[CBSE 2014; CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{55. Given, equations of lines are}
\displaystyle \overrightarrow{r}=(\widehat{i}+\widehat{j})+\lambda(2\widehat{i}-\widehat{j}+\widehat{k})\qquad ...(i)
\displaystyle \text{and}
\displaystyle \overrightarrow{r}=(2\widehat{i}+\widehat{j}-\widehat{k})+\mu(3\widehat{i}-5\widehat{j}+2\widehat{k})\qquad ...(ii)
\displaystyle \text{On comparing equations with vector equation}
\displaystyle \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b},\text{ we get}
\displaystyle \overrightarrow{a_1}=\widehat{i}+\widehat{j},\ \overrightarrow{b_1}=2\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \text{and}
\displaystyle \overrightarrow{a_2}=2\widehat{i}+\widehat{j}-\widehat{k},\ \overrightarrow{b_2}=3\widehat{i}-5\widehat{j}+2\widehat{k}
\displaystyle \text{We know that the shortest distance is given by}
\displaystyle d=\left|\frac{(\overrightarrow{b_1}\times\overrightarrow{b_2})\cdot(\overrightarrow{a_2}-\overrightarrow{a_1})}{|\overrightarrow{b_1}\times\overrightarrow{b_2}|}\right|\qquad ...(iii)
\displaystyle \text{Now, }\overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&-1&1\\3&-5&2\end{vmatrix}
\displaystyle =\widehat{i}(-2+5)-\widehat{j}(4-3)+\widehat{k}(-10+3)
\displaystyle =3\widehat{i}-\widehat{j}-7\widehat{k}\qquad ...(iv)
\displaystyle \text{and }|\overrightarrow{b_1}\times\overrightarrow{b_2}|=\sqrt{3^2+(-1)^2+(-7)^2}
\displaystyle =\sqrt{59}\qquad ...(v)
\displaystyle \text{Also, }\overrightarrow{a_2}-\overrightarrow{a_1}=(2\widehat{i}+\widehat{j}-\widehat{k})-(\widehat{i}+\widehat{j})
\displaystyle =\widehat{i}-\widehat{k}\qquad ...(vi)
\displaystyle \text{From Eqs. (iii), (iv), (v) and (vi), we get}
\displaystyle d=\left|\frac{(3\widehat{i}-\widehat{j}-7\widehat{k})\cdot(\widehat{i}-\widehat{k})}{\sqrt{59}}\right|
\displaystyle =\left|\frac{3-0+7}{\sqrt{59}}\right|=\frac{10}{\sqrt{59}}
\displaystyle \text{Hence, the required shortest distance is }\frac{10}{\sqrt{59}}\text{ units.}
\\

\displaystyle \textbf{Question 55. }\text{Find the shortest distance between the two lines whose vector equations are}
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda(\widehat{i}-3\widehat{j}+2\widehat{k})   \text{ and } \\ \overrightarrow{r}=(4\widehat{i}+5\widehat{j}+6\widehat{k})+\mu(2\widehat{i}+3\widehat{j}+\widehat{k}).\hspace{0.2cm}\text{[CBSE 2014 C]}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a}_{1}=\widehat{i}+2\widehat{j}+3\widehat{k},\quad \overrightarrow{b}_{1}=\widehat{i}-3\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{a}_{2}=4\widehat{i}+5\widehat{j}+6\widehat{k},\quad \overrightarrow{b}_{2}=2\widehat{i}+3\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{a}_{2}-\overrightarrow{a}_{1}=3\widehat{i}+3\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{b}_{1}\times\overrightarrow{b}_{2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-3&2\\2&3&1\end{vmatrix}
\displaystyle =-9\widehat{i}+3\widehat{j}+9\widehat{k}
\displaystyle \text{Shortest distance}=\frac{\left|(\overrightarrow{a}_{2}-\overrightarrow{a}_{1})\cdot(\overrightarrow{b}_{1}\times\overrightarrow{b}_{2})\right|}{|\overrightarrow{b}_{1}\times\overrightarrow{b}_{2}|}
\displaystyle =\frac{|(3\widehat{i}+3\widehat{j}+3\widehat{k})\cdot(-9\widehat{i}+3\widehat{j}+9\widehat{k})|}{\sqrt{(-9)^{2}+3^{2}+9^{2}}}
\displaystyle =\frac{|-27+9+27|}{\sqrt{171}}
\displaystyle =\frac{9}{3\sqrt{19}}=\frac{3}{\sqrt{19}}
\displaystyle \therefore \text{Shortest distance}=\frac{3}{\sqrt{19}}\text{ units.}
\\

\displaystyle \textbf{Question 56. }\text{Find the shortest distance between the following lines}
\displaystyle \frac{x-3}{1}=\frac{y-5}{2}=\frac{z-7}{1}\text{ and }\frac{x+1}{1}=\frac{y+1}{-6}=\frac{z+1}{2}.   \hspace{0.2cm}\text{[CBSE 2014; CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{57. Given, equations of lines are}
\displaystyle \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\qquad ...(i)
\displaystyle \text{and}
\displaystyle \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\qquad ...(ii)
\displaystyle \text{On comparing the given equations of lines with}
\displaystyle \frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}
\displaystyle \text{and }\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2},\text{ we get}
\displaystyle a_1=1,\ b_1=-2,\ c_1=1,\ x_1=3,\ y_1=5,\ z_1=7
\displaystyle \text{and }a_2=7,\ b_2=-6,\ c_2=1,\ x_2=-1,\ y_2=-1,\ z_2=-1
\displaystyle \text{We know that the shortest distance between two lines is given by}
\displaystyle d=\frac{\left|\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}\right|}{\sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}}
\displaystyle =\frac{\left|\begin{vmatrix}-4&-6&-8\\1&-2&1\\7&-6&1\end{vmatrix}\right|}{\sqrt{(-2+6)^2+(7-1)^2+(-6+14)^2}}
\displaystyle =\frac{|-4(-2+6)+6(1-7)-8(-6+14)|}{\sqrt{16+36+64}}
\displaystyle =\frac{|-16-36-64|}{\sqrt{116}}
\displaystyle =\frac{116}{\sqrt{116}}=\sqrt{116}\text{ units.}
\\

\displaystyle \textbf{Question 57. }\text{Find the vector and cartesian equation of the line passing through}
\displaystyle \text{the point }(2,1,3)\text{ and perpendicular to } \text{the lines } \frac{x-1}{1}=\frac{y-2}{3}=\frac{z-3}{2}
\displaystyle \text{and }\frac{x}{3}=\frac{y}{2}=\frac{z}{5}.   \hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{58. Any line through the point }(2,1,3)\text{ can be written as}
\displaystyle \frac{x-2}{a}=\frac{y-1}{b}=\frac{z-3}{c}\qquad ...(i)
\displaystyle \text{where }a,b\text{ and }c\text{ are the direction ratios of line (i).}
\displaystyle \text{Now, the line (i) is perpendicular to the lines}
\displaystyle \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}
\displaystyle \text{and}
\displaystyle \frac{x-0}{-3}=\frac{y-0}{2}=\frac{z-0}{5}
\displaystyle \text{D.R.'s of these two lines are }(1,2,3)\text{ and }(-3,2,5).
\displaystyle \text{We know that if two lines are perpendicular, then}
\displaystyle a_1a_2+b_1b_2+c_1c_2=0
\displaystyle \Rightarrow a+2b+3c=0\qquad ...(ii)
\displaystyle \text{and}
\displaystyle -3a+2b+5c=0\qquad ...(iii)
\displaystyle \text{On solving Eqs. (ii) and (iii), we get}
\displaystyle \frac{a}{4}=\frac{b}{-14}=\frac{c}{8}=\lambda
\displaystyle \Rightarrow a=4\lambda,\ b=-14\lambda,\ c=8\lambda
\displaystyle \Rightarrow a:b:c=2:-7:4
\displaystyle \text{Hence, required line is}
\displaystyle \frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-3}{4}
\displaystyle \text{and vector equation is}
\displaystyle \overrightarrow{r}=2\widehat{i}+\widehat{j}+3\widehat{k}+\lambda(2\widehat{i}-7\widehat{j}+4\widehat{k}).
\\

\displaystyle \textbf{Question 58. }\text{Find the direction cosines of the line } \frac{4-x}{2}=\frac{y-1}{6}=\frac{z-2}{3}.
\displaystyle \hspace{2.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{23. Given, equation of line is}
\displaystyle \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}
\displaystyle \text{It can be rewritten in standard form as}
\displaystyle \frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}
\displaystyle \text{Here, D.R.'s of the line are }(-2,6,-3).
\displaystyle \therefore \text{Direction cosines of the line are}
\displaystyle \frac{-2}{\sqrt{(-2)^2+6^2+(-3)^2}},\ \frac{6}{\sqrt{(-2)^2+6^2+(-3)^2}},\ \frac{-3}{\sqrt{(-2)^2+6^2+(-3)^2}}
\displaystyle \text{and i.e. }-\frac27,\ \frac67,\ -\frac37.
\displaystyle \text{Thus, DC's of line are }\left(-\frac27,\frac67,-\frac37\right).
\\

\displaystyle \textbf{Question 59. }\text{Write the vector equation of a line passing through the }\text{point }(1,-1,2)
\displaystyle \text{ and parallel to the line whose equation }   \text{is }\frac{x-3}{1}=\frac{y-1}{2}=\frac{z+1}{-2}. \hspace{0.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{24. We know that the vector equation of a line passing}
\displaystyle \text{through a point with position vector }\overrightarrow{a}\text{ and parallel to a given vector }\overrightarrow{b}\text{ is }
\displaystyle \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b},\ \lambda\in R.
\displaystyle \text{Here, }\overrightarrow{a}=\hat{i}-\hat{j}+2\hat{k}\text{ and }\overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}
\displaystyle \left[\because \text{D.R.'s of given line is }1,2\text{ and }-2\right]
\displaystyle \therefore \text{Required equation of line is}
\displaystyle \overrightarrow{r}=(\hat{i}-\hat{j}+2\hat{k})+\lambda(\hat{i}+2\hat{j}-2\hat{k}),\ \lambda\in R.
\\

\displaystyle \textbf{Question 60. }\text{Find the cartesian equation of the line which passes through the point}
\displaystyle (-2,4,-5)\text{ and is parallel to the line }   \frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}. \hspace{0.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{25. Given, the required line is parallel to the line}
\displaystyle \frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}
\displaystyle \text{or}
\displaystyle \frac{x+3}{3}=\frac{y-4}{-5}=\frac{z+8}{6}
\displaystyle \therefore \text{D.R.'s of both lines are proportional to each other.}
\displaystyle \text{The required equation of the line passing through}
\displaystyle (-2,4,-5)\text{ having DR's }(3,-5,6)\text{ is}
\displaystyle \frac{x+2}{3}=\frac{y-4}{-5}=\frac{z+5}{6}.
\\

\displaystyle \textbf{Question 61. }\text{The cartesian equation of a line is } 6x-2=3y+1=2z-2.
\displaystyle \text{ Find the direction cosines of the line. Write down the cartesian and vector}
\displaystyle \text{equations of a line passing through }(2,-1,-1)\text{ which are parallel to the given line.}
\displaystyle  \hspace{2.2cm}\text{[CBSE 2013C]}
\displaystyle \text{Answer:}
\displaystyle \text{59. Given, equation of line is}
\displaystyle 6x-2=3y+1=2z-2
\displaystyle \text{or}
\displaystyle \frac{x-\frac13}{1/6}=\frac{y+\frac13}{1/3}=\frac{z-1}{1/2}
\displaystyle \Rightarrow \frac{x-\frac13}{1}=\frac{y+\frac13}{2}=\frac{z-1}{3}
\displaystyle \text{Here, D.R.'s of the line are }(1,2,3).
\displaystyle \text{D.C.'s of the line are}
\displaystyle \left(\frac1{\sqrt{14}},\frac2{\sqrt{14}},\frac3{\sqrt{14}}\right).
\displaystyle \text{The equation of a line passing through }(2,-1,-1)
\displaystyle \text{and parallel to the given line is}
\displaystyle \frac{x-2}{1}=\frac{y+1}{2}=\frac{z+1}{3}
\displaystyle \text{or }x=2+\lambda,\ y=-1+2\lambda,\ z=-1+3\lambda.
\displaystyle \text{Hence, vector equation is}
\displaystyle \overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\lambda(\widehat{i}+2\widehat{j}+3\widehat{k}).
\\

\displaystyle \textbf{Question 62. }\text{Find the shortest distance between the two lines whose }   \text{vector equations are}
\displaystyle \overrightarrow{r}=(6\widehat{i}+2\widehat{j}+2\widehat{k})+\lambda(\widehat{i}-2\widehat{j}+2\widehat{k})
\displaystyle \text{and }\overrightarrow{r}=(-4\widehat{i}-\widehat{k})+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k}). \hspace{2.2cm}\text{[CBSE 2013C; CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a}_{1}=6\widehat{i}+2\widehat{j}+2\widehat{k},\quad \overrightarrow{b}_{1}=\widehat{i}-2\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{a}_{2}=-4\widehat{i}-\widehat{k},\quad \overrightarrow{b}_{2}=3\widehat{i}-2\widehat{j}-2\widehat{k}
\displaystyle \overrightarrow{a}_{2}-\overrightarrow{a}_{1}=-10\widehat{i}-2\widehat{j}-3\widehat{k}
\displaystyle \overrightarrow{b}_{1}\times\overrightarrow{b}_{2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-2&2\\3&-2&-2\end{vmatrix}
\displaystyle =8\widehat{i}+8\widehat{j}+4\widehat{k}
\displaystyle \text{Shortest distance}=\frac{\left|(\overrightarrow{a}_{2}-\overrightarrow{a}_{1})\cdot(\overrightarrow{b}_{1}\times\overrightarrow{b}_{2})\right|}{|\overrightarrow{b}_{1}\times\overrightarrow{b}_{2}|}
\displaystyle =\frac{|(-10\widehat{i}-2\widehat{j}-3\widehat{k})\cdot(8\widehat{i}+8\widehat{j}+4\widehat{k})|}{\sqrt{8^{2}+8^{2}+4^{2}}}
\displaystyle =\frac{|-80-16-12|}{\sqrt{144}}
\displaystyle =\frac{108}{12}=9
\displaystyle \therefore \text{Shortest distance}=9\text{ units.}
\\

\displaystyle \textbf{Question 63. }\text{Show that the lines }   \overrightarrow{r}=3\widehat{i}+2\widehat{j}-4\widehat{k}+\lambda(\widehat{i}+2\widehat{j}+2\widehat{k}) \text{ and }
\displaystyle \overrightarrow{r}=5\widehat{i}-2\widehat{j}+\mu(3\widehat{i}+2\widehat{j}+6\widehat{k})   \text{ are intersecting. Hence, find their point of intersection.} \\ \hspace{0.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{For the first line, }x=3+\lambda,\ y=2+2\lambda,\ z=-4+2\lambda
\displaystyle \text{For the second line, }x=5+3\mu,\ y=-2+2\mu,\ z=6\mu
\displaystyle \text{At the point of intersection,}
\displaystyle 3+\lambda=5+3\mu \qquad ...(1)
\displaystyle 2+2\lambda=-2+2\mu \qquad ...(2)
\displaystyle -4+2\lambda=6\mu \qquad ...(3)
\displaystyle \text{From (2), }\lambda-\mu=-2
\displaystyle \lambda=\mu-2
\displaystyle \text{Substituting in (1),}
\displaystyle 3+\mu-2=5+3\mu
\displaystyle 1+\mu=5+3\mu
\displaystyle \mu=-2,\quad \lambda=-4
\displaystyle \text{These values satisfy (3). Hence, the lines intersect.}
\displaystyle \text{Point of intersection }=(3-4,\ 2+2(-4),\ -4+2(-4))
\displaystyle =(-1,\ -6,\ -12)
\displaystyle \therefore \text{The lines intersect at }(-1,\ -6,\ -12).
\\

\displaystyle \textbf{Question 64. }\text{If a line has direction ratios }(2,-1,-2),\text{ then what are}
\displaystyle \text{its direction cosines?} \hspace{2.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{26. Given, DR's of the line are }(2,-1,-2).
\displaystyle \therefore \text{Direction cosines of the line are}
\displaystyle \frac{2}{\sqrt{2^2+(-1)^2+(-2)^2}},\ \frac{-1}{\sqrt{2^2+(-1)^2+(-2)^2}},\ \frac{-2}{\sqrt{2^2+(-1)^2+(-2)^2}}
\displaystyle \left[\because l=\pm\frac{a}{\sqrt{a^2+b^2+c^2}},\ m=\pm\frac{b}{\sqrt{a^2+b^2+c^2}},\ n=\pm\frac{c}{\sqrt{a^2+b^2+c^2}}\right]
\displaystyle =\left(\frac{2}{\sqrt9},\frac{-1}{\sqrt9},\frac{-2}{\sqrt9}\right)
\displaystyle =\left(\frac23,-\frac13,-\frac23\right).
\\

\displaystyle \textbf{Question 65. }\text{Computing the shortest distance between the following pair of lines, }
\displaystyle \text{determine whether they intersect or not?}
\displaystyle \overrightarrow{r}=(\widehat{i}-\widehat{j})+\lambda(2\widehat{i}-\widehat{k})\text{ and }\overrightarrow{r}=2\widehat{i}-\widehat{j}+\mu(\widehat{i}-\widehat{j}-\widehat{k}). \hspace{2.2cm}\text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a}_{1}=\widehat{i}-\widehat{j},\quad \overrightarrow{b}_{1}=2\widehat{i}-\widehat{k}
\displaystyle \overrightarrow{a}_{2}=2\widehat{i}-\widehat{j},\quad \overrightarrow{b}_{2}=\widehat{i}-\widehat{j}-\widehat{k}
\displaystyle \overrightarrow{a}_{2}-\overrightarrow{a}_{1}=\widehat{i}
\displaystyle \overrightarrow{b}_{1}\times\overrightarrow{b}_{2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&0&-1\\1&-1&-1\end{vmatrix}
\displaystyle =-\widehat{i}+\widehat{j}-2\widehat{k}
\displaystyle \text{Shortest distance}=\frac{\left|(\overrightarrow{a}_{2}-\overrightarrow{a}_{1})\cdot(\overrightarrow{b}_{1}\times\overrightarrow{b}_{2})\right|}{|\overrightarrow{b}_{1}\times\overrightarrow{b}_{2}|}
\displaystyle =\frac{|\widehat{i}\cdot(-\widehat{i}+\widehat{j}-2\widehat{k})|}{\sqrt{(-1)^{2}+1^{2}+(-2)^{2}}}
\displaystyle =\frac{1}{\sqrt{6}}
\displaystyle \text{Since the shortest distance is non-zero, the lines do not intersect.}
\displaystyle \therefore \text{Shortest distance}=\frac{1}{\sqrt{6}}\text{ units and the lines are non-intersecting (skew lines).}
\\

\displaystyle \textbf{Question 66. }\text{Find the equation of the line passing through the } \text{point }(-1,3,-2)
\displaystyle \text{ and perpendicular to the lines } \frac{x}{1}=\frac{y-2}{2}=\frac{z}{3} \text{ and } \frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}.
\displaystyle  \hspace{2.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Direction ratios of the given lines are }(1,2,3)\text{ and }(-3,2,5).
\displaystyle \text{Direction ratios of required line }=(1,2,3)\times(-3,2,5)
\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&2&3\\-3&2&5\end{vmatrix}
\displaystyle =4\widehat{i}-14\widehat{j}+8\widehat{k}=2(2\widehat{i}-7\widehat{j}+4\widehat{k})
\displaystyle \therefore \text{Required line is }\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}.
\\

\displaystyle \textbf{Question 67. }\text{Find the coordinates of the foot of perpendicular and the length of the }
\displaystyle \text{perpendicular drawn from the point } P(5,4,2)\text{ to the line }
\displaystyle \overrightarrow{r}=\widehat{i}+3\widehat{j}+\widehat{k}+\lambda(2\widehat{i}+3\widehat{j}-\widehat{k}).   \text{Also, find the image of }P\text{ in this line.} \hspace{0.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{64. (i) Do same as Q. No. 38.}
\displaystyle \text{Ans. Foot of perpendicular is }(1,6,0).
\displaystyle \text{The image of }P\text{ is }(-3,8,-2).
\displaystyle \text{(ii) Length of perpendicular}
\displaystyle =\sqrt{(5-1)^2+(4-6)^2+(2-0)^2}
\displaystyle =\sqrt{4^2+(-2)^2+2^2}
\displaystyle =\sqrt{24}=2\sqrt6\text{ units.}
\\

\displaystyle \textbf{Question 68. }\text{Write the direction cosines of the line joining the } \text{points }(1,0,0)
\displaystyle \text{ and }(0,1,1). \hspace{2.2cm}\text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{27. Clearly, the direction ratios of the line joining the points}
\displaystyle (1,0,0)\text{ and }(0,1,1)\text{ are }(0-1),(1-0)\text{ and }(1-0)\text{ i.e. }-1,1,1.
\displaystyle \therefore \text{Direction cosines are}
\displaystyle \frac{-1}{\sqrt{(-1)^2+1^2+1^2}},\ \frac{1}{\sqrt{(-1)^2+1^2+1^2}}\text{ and }\frac{1}{\sqrt{(-1)^2+1^2+1^2}}
\displaystyle \text{i.e. }-\frac1{\sqrt3},\ \frac1{\sqrt3}\text{ and }\frac1{\sqrt3}.
\\

\displaystyle \textbf{Question 69. }\text{Find the angle between the following pair of lines}
\displaystyle \frac{-x+2}{-2}=\frac{y-1}{7}=\frac{z+3}{-3}\text{ and }\frac{x+2}{-1}=\frac{2y-8}{4}=\frac{z-5}{4}
\displaystyle \text{and check whether the lines are parallel or not.} \hspace{2.2cm}\text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \textbf{65. Given, equations of two lines are}
\displaystyle \frac{x+2}{-2}=\frac{y-1}{7}=\frac{z+3}{-3}
\displaystyle \text{and}
\displaystyle \frac{x+2}{-1}=\frac{2y-8}{4}=\frac{z-5}{4}
\displaystyle \text{Above equations can be written as}
\displaystyle \frac{x+2}{-2}=\frac{y-1}{7}=\frac{z+3}{-3}\qquad ...(i)
\displaystyle \text{and}
\displaystyle \frac{x+2}{-1}=\frac{y-4}{2}=\frac{z-5}{4}\qquad ...(ii)
\displaystyle \text{On comparing Eqs. (i) and (ii) with one point form of}
\displaystyle \text{equation of line } \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c},\text{ we get}
\displaystyle a_1=2,\ b_1=7,\ c_1=-3
\displaystyle \text{and}
\displaystyle a_2=-1,\ b_2=2,\ c_2=4
\displaystyle \text{We know that angle between two lines is given by}
\displaystyle \cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}
\displaystyle \therefore \cos\theta=\frac{(2)(-1)+(7)(2)+(-3)(4)}{\sqrt{2^2+7^2+(-3)^2}\sqrt{(-1)^2+2^2+4^2}}
\displaystyle =\frac{-2+14-12}{\sqrt{62}\times\sqrt{21}}=0
\displaystyle \Rightarrow \cos\theta=\cos\frac{\pi}{2}
\displaystyle \Rightarrow \theta=\frac{\pi}{2}
\displaystyle \text{Hence, the angle between the lines is }\frac{\pi}{2}. \text{ Therefore, the}
\displaystyle \text{given pair of lines are perpendicular to each other.}
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