\displaystyle \textbf{Question 1. }\text{The equation of }X\text{-axis in space is}
\displaystyle \text{(a) }x=0,\ y=0 \qquad \text{(b) }x=0,\ z=0
\displaystyle \text{(c) }x=0 \qquad \text{(d) }y=0,\ z=0
\displaystyle \text{Answer:}
\displaystyle \text{On }X\text{-axis, the }y\text{-coordinate and }z\text{-coordinate are }0.
\displaystyle \therefore y=0,\ z=0
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 2. }\text{The coordinates of a point on the line }\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2} \\ \text{at a distance of }\frac{6}{\sqrt{2}}\text{ from the point }(1,2,3)\text{ is}
\displaystyle \text{(a) }(56,43,111) \qquad \text{(b) }\left(\frac{56}{17},\frac{43}{17},\frac{111}{17}\right)
\displaystyle \text{(c) }(2,1,3) \qquad \text{(d) }(-2,-1,-3)
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda
\displaystyle \therefore x=3\lambda-2,\ y=2\lambda-1,\ z=2\lambda+3
\displaystyle \text{Required point is }(3\lambda-2,\ 2\lambda-1,\ 2\lambda+3)
\displaystyle \text{Its distance from }(1,2,3)\text{ is }\frac{6}{\sqrt{2}}.
\displaystyle \therefore (3\lambda-2-1)^{2}+(2\lambda-1-2)^{2}+(2\lambda+3-3)^{2}=\left(\frac{6}{\sqrt{2}}\right)^{2}
\displaystyle (3\lambda-3)^{2}+(2\lambda-3)^{2}+(2\lambda)^{2}=18
\displaystyle 9\lambda^{2}-18\lambda+9+4\lambda^{2}-12\lambda+9+4\lambda^{2}=18
\displaystyle 17\lambda^{2}-30\lambda=0
\displaystyle \lambda(17\lambda-30)=0
\displaystyle \therefore \lambda=0\text{ or }\lambda=\frac{30}{17}
\displaystyle \text{For }\lambda=\frac{30}{17},
\displaystyle x=3\left(\frac{30}{17}\right)-2=\frac{56}{17}
\displaystyle y=2\left(\frac{30}{17}\right)-1=\frac{43}{17}
\displaystyle z=2\left(\frac{30}{17}\right)+3=\frac{111}{17}
\displaystyle \therefore \text{Required point is }\left(\frac{56}{17},\frac{43}{17},\frac{111}{17}\right)
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 3. }\text{The planes }2x-y+4z=5\text{ and }5x-2.5y+10z=6\text{ are}
\displaystyle \text{(a) perpendicular} \qquad \text{(b) parallel}
\displaystyle \text{(c) intersect along }Y\text{-axis} \qquad \text{(d) passes through }\left(0,0,\frac{5}{4}\right)
\displaystyle \text{Answer:}
\displaystyle \text{For the plane }2x-y+4z=5,\ \overrightarrow{n_1}=(2,-1,4)
\displaystyle \text{For the plane }5x-2.5y+10z=6,\ \overrightarrow{n_2}=(5,-2.5,10)
\displaystyle \overrightarrow{n_2}=2.5\,\overrightarrow{n_1}
\displaystyle \therefore \overrightarrow{n_1}\parallel\overrightarrow{n_2}
\displaystyle \text{Hence the planes are either parallel or coincident.}
\displaystyle \text{Multiplying the first equation by }2.5,
\displaystyle 5x-2.5y+10z=12.5
\displaystyle \text{But the second plane is }5x-2.5y+10z=6
\displaystyle \therefore \text{the planes are distinct and parallel.}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 4. }\text{The value of }k,\text{ such that }\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\text{ lies in the plane } \\ 2x-4y+z=7\text{ is}
\displaystyle \text{(a) }7 \qquad \text{(b) }-7
\displaystyle \text{(c) }4 \qquad \text{(d) No real value}
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}
\displaystyle \therefore \text{A point on the line is }(4,2,k)
\displaystyle \text{Since the line lies in the plane }2x-4y+z=7,
\displaystyle \text{the point }(4,2,k)\text{ must satisfy the plane.}
\displaystyle \therefore 2(4)-4(2)+k=7
\displaystyle \therefore 8-8+k=7
\displaystyle \therefore k=7
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 5. }\text{The sine of the angle between the straight line } \\ \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\text{ and the plane }2x-2y+z=5\text{ is}
\displaystyle \text{(a) }\frac{10}{6\sqrt{5}} \qquad \text{(b) }\frac{4}{5\sqrt{2}}
\displaystyle \text{(c) }\frac{2\sqrt{3}}{5} \qquad \text{(d) }\frac{\sqrt{2}}{10}
\displaystyle \text{Answer:}
\displaystyle \text{Direction ratios of the line are }3,4,5
\displaystyle \therefore \overrightarrow{d}=3\widehat{i}+4\widehat{j}+5\widehat{k}
\displaystyle \text{Normal to the plane }2x-2y+z=5\text{ is }
\displaystyle \overrightarrow{n}=2\widehat{i}-2\widehat{j}+\widehat{k}
\displaystyle \text{If }\theta\text{ is the angle between the line and the plane, then}
\displaystyle \sin\theta=\frac{|\overrightarrow{d}\cdot\overrightarrow{n}|}{|\overrightarrow{d}||\overrightarrow{n}|}
\displaystyle =\frac{|3(2)+4(-2)+5(1)|}{\sqrt{3^{2}+4^{2}+5^{2}}\sqrt{2^{2}+(-2)^{2}+1^{2}}}
\displaystyle =\frac{|6-8+5|}{\sqrt{50}\sqrt{9}}
\displaystyle =\frac{3}{3\sqrt{50}}
\displaystyle =\frac{1}{\sqrt{50}}
\displaystyle =\frac{\sqrt{2}}{10}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 6. }\text{The direction cosines of the line joining the points }(4,3,-5)\text{ and }(-2,1,-8)\text{ are}
\displaystyle \text{(a) }\left(\frac{6}{7},\frac{2}{7},\frac{3}{7}\right) \qquad \text{(b) }\left(\frac{2}{7},\frac{3}{7},-\frac{6}{7}\right)
\displaystyle \text{(c) }\left(\frac{6}{7},\frac{3}{7},\frac{2}{7}\right) \qquad \text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(4,3,-5)\text{ and }B(-2,1,-8).
\displaystyle \overrightarrow{AB}=(-2-4)\widehat{i}+(1-3)\widehat{j}+(-8+5)\widehat{k}
\displaystyle =-6\widehat{i}-2\widehat{j}-3\widehat{k}
\displaystyle |\overrightarrow{AB}|=\sqrt{(-6)^2+(-2)^2+(-3)^2}=\sqrt{49}=7
\displaystyle \therefore \text{Direction cosines of the line are}
\displaystyle \left(\frac{-6}{7},\frac{-2}{7},\frac{-3}{7}\right)
\displaystyle \text{or equivalently }\left(\frac{6}{7},\frac{2}{7},\frac{3}{7}\right)
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 7. }\text{The point of intersection of lines }\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1} \\ \text{and }\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}\text{ is}
\displaystyle \text{(a) }(5,7,-2) \qquad \text{(b) }(-3,3,6)
\displaystyle \text{(c) }(2,10,4) \qquad \text{(d) }\left(\frac{21}{5},\frac{5}{3},\frac{10}{3}\right)
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}=\lambda
\displaystyle \therefore x=5+3\lambda,\ y=7-\lambda,\ z=\lambda-2
\displaystyle \text{Let }\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}=\mu
\displaystyle \therefore x=-3-36\mu,\ y=3+2\mu,\ z=6+4\mu
\displaystyle \text{At the point of intersection,}
\displaystyle 5+3\lambda=-3-36\mu \qquad (1)
\displaystyle 7-\lambda=3+2\mu \qquad (2)
\displaystyle \lambda-2=6+4\mu \qquad (3)
\displaystyle \text{From (2), }\lambda=4-2\mu
\displaystyle \text{From (3), }\lambda=8+4\mu
\displaystyle \therefore 4-2\mu=8+4\mu
\displaystyle \therefore -6\mu=4
\displaystyle \therefore \mu=-\frac{2}{3}
\displaystyle \therefore \lambda=4-2\left(-\frac{2}{3}\right)=\frac{16}{3}
\displaystyle \text{Hence, }x=5+3\left(\frac{16}{3}\right)=21
\displaystyle y=7-\frac{16}{3}=\frac{5}{3}
\displaystyle z=\frac{16}{3}-2=\frac{10}{3}
\displaystyle \therefore \text{Point of intersection is }\left(21,\frac{5}{3},\frac{10}{3}\right)
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 8. }\text{Equation of the line passing through }(2,-1,1)\text{ and parallel to the line } \\ \frac{x-5}{4}=\frac{y+2}{-3}=\frac{z}{5}\text{ is}
\displaystyle \text{(a) }\frac{x-2}{4}=\frac{y+1}{-3}=\frac{z-1}{5} \qquad \text{(b) }\frac{x-2}{4}=\frac{y+1}{3}=\frac{z-1}{5}
\displaystyle \text{(c) }\frac{x-2}{-4}=\frac{y+1}{-3}=\frac{z-1}{5} \qquad \text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{Direction ratios of the given line are }4,-3,5.
\displaystyle \text{Required line passes through }(2,-1,1)\text{ and is parallel to the given line.}
\displaystyle \therefore \text{its direction ratios are also }4,-3,5.
\displaystyle \therefore \text{Equation of the required line is}
\displaystyle \frac{x-2}{4}=\frac{y+1}{-3}=\frac{z-1}{5}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 9. }\text{The angle between the lines }x=1,\ y=2\text{ and } \\ y=-1,\ z=0\text{ is}
\displaystyle \text{(a) }\frac{\pi}{6} \qquad \text{(b) }\frac{\pi}{3} \qquad \text{(c) }\frac{\pi}{2} \qquad \text{(d) }0^\circ
\displaystyle \text{Answer:}
\displaystyle x=1,\ y=2\text{ represents a line parallel to the }Z\text{-axis.}
\displaystyle \therefore \text{direction ratios of the first line are }0,0,1.
\displaystyle y=-1,\ z=0\text{ represents a line parallel to the }X\text{-axis.}
\displaystyle \therefore \text{direction ratios of the second line are }1,0,0.
\displaystyle \text{Let }\theta\text{ be the angle between the two lines.}
\displaystyle \cos\theta=\frac{0(1)+0(0)+1(0)}{\sqrt{0^{2}+0^{2}+1^{2}}\sqrt{1^{2}+0^{2}+0^{2}}}
\displaystyle =0
\displaystyle \therefore \theta=\frac{\pi}{2}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 10. }\text{If a line passes through a point }(-2,-7,3),\text{ then find the} \\ \text{equations of a line parallel to axes.}
\displaystyle \text{Answer:}
\displaystyle \text{A line parallel to the }X\text{-axis has direction ratios }(1,0,0).
\displaystyle \text{Hence, through }(-2,-7,3),\text{ its equation is}
\displaystyle \frac{x+2}{1}=\frac{y+7}{0}=\frac{z-3}{0}
\displaystyle \text{or }y=-7,\ z=3.
\displaystyle \text{A line parallel to the }Y\text{-axis has direction ratios }(0,1,0).
\displaystyle \text{Hence, through }(-2,-7,3),\text{ its equation is}
\displaystyle \frac{x+2}{0}=\frac{y+7}{1}=\frac{z-3}{0}
\displaystyle \text{or }x=-2,\ z=3.
\displaystyle \text{A line parallel to the }Z\text{-axis has direction ratios }(0,0,1).
\displaystyle \text{Hence, through }(-2,-7,3),\text{ its equation is}
\displaystyle \frac{x+2}{0}=\frac{y+7}{0}=\frac{z-3}{1}
\displaystyle \text{or }x=-2,\ y=-7.
\\

\displaystyle \textbf{Question 11. }\text{Find }k,\text{ so that the lines }\frac{1-x}{3}=\frac{y-2}{2k}=\frac{z-3}{2}\text{ and } \\ \frac{x-1}{3k}=\frac{y-5}{1}=\frac{6-z}{5}\text{ are at right angles.}
\displaystyle \text{Answer:}
\displaystyle \text{Direction ratios of the first line are }-3,\ 2k,\ 2
\displaystyle \text{Direction ratios of the second line are }3k,\ 1,\ -5
\displaystyle \text{Since the lines are at right angles,}
\displaystyle (-3)(3k)+(2k)(1)+(2)(-5)=0
\displaystyle -9k+2k-10=0
\displaystyle -7k-10=0
\displaystyle \therefore k=-\frac{10}{7}
\\

\displaystyle \textbf{Question 12. }\text{A plane meets the coordinate axes in }A,B\text{ and }C\text{ such that} \\ \text{the centroid of the }\triangle ABC\text{ is the point }(p,q,r),\text{ show that the equation} \\ \text{of the plane is }\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=3.
\displaystyle \text{Answer:}
\displaystyle \text{Let the plane meet the coordinate axes at }A(a,0,0),\ B(0,b,0)\text{ and }C(0,0,c).
\displaystyle \text{Centroid of }\triangle ABC=\left(\frac{a+0+0}{3},\frac{0+b+0}{3},\frac{0+0+c}{3}\right)
\displaystyle =\left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right)
\displaystyle \text{Given, the centroid is }(p,q,r).
\displaystyle \therefore \frac{a}{3}=p,\ \frac{b}{3}=q,\ \frac{c}{3}=r
\displaystyle \therefore a=3p,\ b=3q,\ c=3r
\displaystyle \text{The intercept form of the equation of a plane is}
\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
\displaystyle \therefore \frac{x}{3p}+\frac{y}{3q}+\frac{z}{3r}=1
\displaystyle \therefore \frac{x}{p}+\frac{y}{q}+\frac{z}{r}=3
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 13. }\text{Find the coordinates of the point, where the line passing through }(5,1,6) \\ \text{and }(3,4,1)\text{ crosses }YZ\text{-plane.}
\displaystyle \text{Answer:}
\displaystyle \text{Direction ratios of the line joining }(5,1,6)\text{ and }(3,4,1)\text{ are}
\displaystyle (3-5,\ 4-1,\ 1-6)=(-2,3,-5)
\displaystyle \therefore \text{Equation of the line is}
\displaystyle \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda
\displaystyle \therefore x=5-2\lambda,\ y=1+3\lambda,\ z=6-5\lambda
\displaystyle \text{Since the point lies on the }YZ\text{-plane, }x=0
\displaystyle \therefore 5-2\lambda=0
\displaystyle \therefore \lambda=\frac{5}{2}
\displaystyle \therefore y=1+3\left(\frac{5}{2}\right)=\frac{17}{2}
\displaystyle \therefore z=6-5\left(\frac{5}{2}\right)=-\frac{13}{2}
\displaystyle \therefore \text{The required point is }\left(0,\frac{17}{2},-\frac{13}{2}\right).
\\

\displaystyle \textbf{Question 14. }\text{Write the direction cosines of the normal to plane } \\ 3x+4y+12z=52.
\displaystyle \text{Answer:}
\displaystyle \text{The normal vector to the plane }3x+4y+12z=52\text{ is}
\displaystyle \overrightarrow{n}=3\widehat{i}+4\widehat{j}+12\widehat{k}
\displaystyle |\overrightarrow{n}|=\sqrt{3^{2}+4^{2}+12^{2}}=\sqrt{169}=13
\displaystyle \therefore \text{Direction cosines of the normal are}
\displaystyle \left(\frac{3}{13},\frac{4}{13},\frac{12}{13}\right).
\\

\displaystyle \textbf{Question 15. }\text{Given that }P(3,2,-4),\ Q(5,4,-6)\text{ and }R(9,8,-10) \\ \text{ are collinear. Find the ratio in which }Q\text{ divides }PR.
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{PQ}=(5-3,\ 4-2,\ -6+4)=(2,2,-2)
\displaystyle \overrightarrow{QR}=(9-5,\ 8-4,\ -10+6)=(4,4,-4)
\displaystyle \therefore \overrightarrow{QR}=2\overrightarrow{PQ}
\displaystyle \therefore PQ:QR=1:2
\displaystyle \therefore Q\text{ divides }PR\text{ internally in the ratio }1:2.
\displaystyle \therefore \text{Required ratio }=1:2
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\displaystyle \textbf{Question 16. }\text{Find the vector equation of the line which is parallel to the vector } \\ 3\widehat{i}-2\widehat{j}+6\widehat{k}\text{ and which passes through the point }(1,-2,3).
\displaystyle \text{Answer:}
\displaystyle \text{Position vector of the point }(1,-2,3)\text{ is }
\displaystyle \widehat{i}-2\widehat{j}+3\widehat{k}
\displaystyle \text{Given direction vector is }3\widehat{i}-2\widehat{j}+6\widehat{k}
\displaystyle \text{Hence, the vector equation of the line is}
\displaystyle \overrightarrow{r}=(\widehat{i}-2\widehat{j}+3\widehat{k})+\lambda(3\widehat{i}-2\widehat{j}+6\widehat{k})
\displaystyle \text{where }\lambda\in R.
\\

\displaystyle \textbf{Question 17. }\text{Find the equation of a line in cartesian form which is parallel to } \\ 2\widehat{i}-\widehat{j}+3\widehat{k}\text{ and which passes through the point }(5,-2,4).
\displaystyle \text{Answer:}
\displaystyle \text{The given vector }2\widehat{i}-\widehat{j}+3\widehat{k}\text{ gives the direction ratios }2,-1,3.
\displaystyle \text{The line passes through the point }(5,-2,4).
\displaystyle \therefore \text{The cartesian equation of the line is}
\displaystyle \frac{x-5}{2}=\frac{y+2}{-1}=\frac{z-4}{3}.
\\

\displaystyle \textbf{Question 18. }\text{Find the direction ratios and direction cosines of the line passing} \\ \text{through two points }(2,-4,5)\text{ and }(0,1,-1).
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(2,-4,5)\text{ and }B(0,1,-1).
\displaystyle \overrightarrow{AB}=(0-2,\ 1+4,\ -1-5)=(-2,5,-6)
\displaystyle \therefore \text{Direction ratios of the line are }-2,\ 5,\ -6.
\displaystyle |\overrightarrow{AB}|=\sqrt{(-2)^2+5^2+(-6)^2}
\displaystyle =\sqrt{4+25+36}=\sqrt{65}
\displaystyle \therefore \text{Direction cosines of the line are}
\displaystyle \left(\frac{-2}{\sqrt{65}},\frac{5}{\sqrt{65}},\frac{-6}{\sqrt{65}}\right).
\\

\displaystyle \textbf{Question 19. }\text{Prove that the lines }x=ay+b,\ z=cy+d\text{ and } \\ x=a'y+b',\ z=c'y+d'\text{ are perpendicular, if }aa'+cc'+1=0.
\displaystyle \text{Answer:}
\displaystyle \text{For the line }x=ay+b,\ z=cy+d,
\displaystyle \text{let }y=\lambda.
\displaystyle \therefore x=a\lambda+b,\ y=\lambda,\ z=c\lambda+d
\displaystyle \therefore \text{direction ratios of the first line are }a,\ 1,\ c.
\displaystyle \text{For the line }x=a'y+b',\ z=c'y+d',
\displaystyle \text{let }y=\mu.
\displaystyle \therefore x=a'\mu+b',\ y=\mu,\ z=c'\mu+d'
\displaystyle \therefore \text{direction ratios of the second line are }a',\ 1,\ c'.
\displaystyle \text{Since }aa'+cc'+1=0,
\displaystyle aa'+1+cc'=0
\displaystyle \therefore (a)(a')+(1)(1)+(c)(c')=0
\displaystyle \therefore \text{the scalar product of their direction vectors is }0.
\displaystyle \therefore \text{the given lines are perpendicular.}
\\

\displaystyle \textbf{Question 20. }\text{Find the angle between the line } \\ \overrightarrow{r}=(2\widehat{i}+2\widehat{j}+\widehat{k})+\lambda(2\widehat{i}-3\widehat{j}+2\widehat{k})\text{ and the plane }\overrightarrow{r}\cdot(3\widehat{i}-2\widehat{j}+5\widehat{k})=4.
\displaystyle \text{Answer:}
\displaystyle \text{Direction vector of the line is}
\displaystyle \overrightarrow{d}=2\widehat{i}-3\widehat{j}+2\widehat{k}
\displaystyle \text{Normal vector to the plane is}
\displaystyle \overrightarrow{n}=3\widehat{i}-2\widehat{j}+5\widehat{k}
\displaystyle \text{Let }\theta\text{ be the angle between the line and the plane.}
\displaystyle \therefore \sin\theta=\frac{|\overrightarrow{d}\cdot\overrightarrow{n}|}{|\overrightarrow{d}||\overrightarrow{n}|}
\displaystyle =\frac{|(2)(3)+(-3)(-2)+(2)(5)|}{\sqrt{2^{2}+(-3)^{2}+2^{2}}\sqrt{3^{2}+(-2)^{2}+5^{2}}}
\displaystyle =\frac{|6+6+10|}{\sqrt{17}\sqrt{38}}
\displaystyle =\frac{22}{\sqrt{646}}
\displaystyle \therefore \theta=\sin^{-1}\left(\frac{22}{\sqrt{646}}\right)
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\displaystyle \textbf{Question 21. }\text{Find the shortest distance between the lines } \\ \overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+\lambda(\widehat{i}-\widehat{j}+\widehat{k})\text{ and }\overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\mu(2\widehat{i}+\widehat{j}+2\widehat{k}).
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a_1}=\widehat{i}+2\widehat{j}+\widehat{k},\ \overrightarrow{b_1}=\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{a_2}=2\widehat{i}-\widehat{j}-\widehat{k},\ \overrightarrow{b_2}=2\widehat{i}+\widehat{j}+2\widehat{k}
\displaystyle \text{Shortest distance}=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})|}{|\overrightarrow{b_1}\times\overrightarrow{b_2}|}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=\widehat{i}-3\widehat{j}-2\widehat{k}
\displaystyle \overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-1&1\\2&1&2\end{vmatrix}=-3\widehat{i}+3\widehat{k}
\displaystyle |(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})|=|(\widehat{i}-3\widehat{j}-2\widehat{k})\cdot(-3\widehat{i}+3\widehat{k})|
\displaystyle =|-3-6|=9
\displaystyle |\overrightarrow{b_1}\times\overrightarrow{b_2}|=\sqrt{(-3)^2+3^2}=3\sqrt{2}
\displaystyle \therefore \text{Shortest distance}=\frac{9}{3\sqrt{2}}=\frac{3}{\sqrt{2}}
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\displaystyle \textbf{Question 22. }\text{Show that the points whose position vectors are } \\ (-2\widehat{i}+3\widehat{j}+5\widehat{k}),\ (\widehat{i}+2\widehat{j}+3\widehat{k})\text{ and }(7\widehat{i}-\widehat{k})\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=(-2,3,5),\ B=(1,2,3),\ C=(7,0,-1)
\displaystyle \overrightarrow{AB}=(1+2)\widehat{i}+(2-3)\widehat{j}+(3-5)\widehat{k}
\displaystyle =3\widehat{i}-\widehat{j}-2\widehat{k}
\displaystyle \overrightarrow{BC}=(7-1)\widehat{i}+(0-2)\widehat{j}+(-1-3)\widehat{k}
\displaystyle =6\widehat{i}-2\widehat{j}-4\widehat{k}
\displaystyle =2(3\widehat{i}-\widehat{j}-2\widehat{k})
\displaystyle \therefore \overrightarrow{BC}=2\overrightarrow{AB}
\displaystyle \therefore \overrightarrow{AB}\parallel\overrightarrow{BC}
\displaystyle \therefore A,B\text{ and }C\text{ are collinear.}
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\displaystyle \textbf{Question 23. }\text{Find the angle between the pair of lines given by } \\ \overrightarrow{r}=3\widehat{i}+2\widehat{j}-4\widehat{k}+\lambda(\widehat{i}+2\widehat{j}+2\widehat{k})\text{ and }\overrightarrow{r}=5\widehat{i}-2\widehat{j}+\mu(3\widehat{i}+2\widehat{j}+6\widehat{k}).
\displaystyle \text{Answer:}
\displaystyle \text{Direction vector of the first line is }\overrightarrow{b_1}=\widehat{i}+2\widehat{j}+2\widehat{k}
\displaystyle \text{Direction vector of the second line is }\overrightarrow{b_2}=3\widehat{i}+2\widehat{j}+6\widehat{k}
\displaystyle \text{Let }\theta\text{ be the angle between the lines.}
\displaystyle \cos\theta=\frac{|\overrightarrow{b_1}\cdot\overrightarrow{b_2}|}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}
\displaystyle =\frac{|1(3)+2(2)+2(6)|}{\sqrt{1^{2}+2^{2}+2^{2}}\sqrt{3^{2}+2^{2}+6^{2}}}
\displaystyle =\frac{|3+4+12|}{3\cdot7}
\displaystyle =\frac{19}{21}
\displaystyle \therefore \theta=\cos^{-1}\left(\frac{19}{21}\right)
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\displaystyle \textbf{Question 24. }\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 each other.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}=\lambda
\displaystyle \therefore x=1+3\lambda,\ y=-1+2\lambda,\ z=1+5\lambda
\displaystyle \text{Let }\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}=\mu
\displaystyle \therefore x=-2+4\mu,\ y=1+3\mu,\ z=-1-2\mu
\displaystyle \text{If the lines intersect, then}
\displaystyle 1+3\lambda=-2+4\mu \qquad (1)
\displaystyle -1+2\lambda=1+3\mu \qquad (2)
\displaystyle 1+5\lambda=-1-2\mu \qquad (3)
\displaystyle \text{From (1), }3\lambda-4\mu=-3
\displaystyle \text{From (2), }2\lambda-3\mu=2
\displaystyle \text{Solving these two equations,}
\displaystyle \lambda=-17,\ \mu=-12
\displaystyle \text{Substituting in (3),}
\displaystyle 1+5(-17)=-84
\displaystyle -1-2(-12)=23
\displaystyle \therefore -84\neq23
\displaystyle \therefore \text{there is no common point on the two lines.}
\displaystyle \therefore \text{the given lines do not intersect each other.}
\\

\displaystyle \textbf{Question 25. }\text{Find the shortest distance between the lines } \\ \overrightarrow{r}=(1-t)\widehat{i}+(t-2)\widehat{j}+(3-2t)\widehat{k}\text{ and }\overrightarrow{r}=(s+1)\widehat{i}+(2s-1)\widehat{j}-(2s+1)\widehat{k}.
\displaystyle \text{Answer:}
\displaystyle \text{First line can be written as}
\displaystyle \overrightarrow{r}=(\widehat{i}-2\widehat{j}+3\widehat{k})+t(-\widehat{i}+\widehat{j}-2\widehat{k})
\displaystyle \therefore \overrightarrow{a_1}=\widehat{i}-2\widehat{j}+3\widehat{k},\ \overrightarrow{b_1}=-\widehat{i}+\widehat{j}-2\widehat{k}
\displaystyle \text{Second line can be written as}
\displaystyle \overrightarrow{r}=(\widehat{i}-\widehat{j}-\widehat{k})+s(\widehat{i}+2\widehat{j}-2\widehat{k})
\displaystyle \therefore \overrightarrow{a_2}=\widehat{i}-\widehat{j}-\widehat{k},\ \overrightarrow{b_2}=\widehat{i}+2\widehat{j}-2\widehat{k}
\displaystyle \text{Shortest distance}=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})|}{|\overrightarrow{b_1}\times\overrightarrow{b_2}|}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=\widehat{j}-4\widehat{k}
\displaystyle \overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-1&1&-2\\1&2&-2\end{vmatrix}
\displaystyle =2\widehat{i}-4\widehat{j}-3\widehat{k}
\displaystyle |(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})|=|(\widehat{j}-4\widehat{k})\cdot(2\widehat{i}-4\widehat{j}-3\widehat{k})|
\displaystyle =|-4+12|=8
\displaystyle |\overrightarrow{b_1}\times\overrightarrow{b_2}|=\sqrt{2^{2}+(-4)^{2}+(-3)^{2}}=\sqrt{29}
\displaystyle \therefore \text{Shortest distance}=\frac{8}{\sqrt{29}}
\\

\displaystyle \textbf{Question 26. }\text{Find the coordinates of foot of perpendicular drawn from the} \\ \text{point }(0,2,3)\text{ on the line }\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(0,2,3)\text{ and }H\text{ be the foot of the perpendicular on the line.}
\displaystyle \text{Equation of the line is } \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda
\displaystyle \therefore H(-3+5\lambda,\ 1+2\lambda,\ -4+3\lambda)
\displaystyle \overrightarrow{PH}=(-3+5\lambda)\widehat{i}+(-1+2\lambda)\widehat{j}+(-7+3\lambda)\widehat{k}
\displaystyle \text{Direction vector of the line is }5\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle \text{Since }PH\perp\text{ line,}
\displaystyle \overrightarrow{PH}\cdot(5\widehat{i}+2\widehat{j}+3\widehat{k})=0
\displaystyle 5(-3+5\lambda)+2(-1+2\lambda)+3(-7+3\lambda)=0
\displaystyle -15+25\lambda-2+4\lambda-21+9\lambda=0
\displaystyle 38\lambda-38=0
\displaystyle \therefore \lambda=1
\displaystyle \therefore H=(2,3,-1)
\displaystyle \therefore \text{The foot of the perpendicular is }(2,3,-1).
\\

\displaystyle \textbf{Question 27. }\text{Find the shortest distance between the lines } \\ l_1:\frac{x-1}{1}=\frac{y+2}{-1}=\frac{z-1}{1}\text{ and }l_2:\frac{x-2}{2}=\frac{y+1}{1}=\frac{z-1}{2}.
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a_1}=\widehat{i}-2\widehat{j}+\widehat{k},\ \overrightarrow{b_1}=\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{a_2}=2\widehat{i}-\widehat{j}+\widehat{k},\ \overrightarrow{b_2}=2\widehat{i}+\widehat{j}+2\widehat{k}
\displaystyle \text{Shortest distance}=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})|}{|\overrightarrow{b_1}\times\overrightarrow{b_2}|}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=\widehat{i}+\widehat{j}
\displaystyle \overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-1&1\\2&1&2\end{vmatrix}=-3\widehat{i}+3\widehat{k}
\displaystyle |(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})|=|(\widehat{i}+\widehat{j})\cdot(-3\widehat{i}+3\widehat{k})|
\displaystyle =|-3|=3
\displaystyle |\overrightarrow{b_1}\times\overrightarrow{b_2}|=\sqrt{(-3)^2+3^2}=3\sqrt{2}
\displaystyle \therefore \text{Shortest distance}=\frac{3}{3\sqrt{2}}=\frac{1}{\sqrt{2}}
\\

\displaystyle \textbf{Question 28. }\text{Prove that the lines }\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}\text{ and } \\ \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}\text{ are coplanar. Also, find the plane containing these two lines.}
\displaystyle \text{Answer:}
\displaystyle \text{For the first line, }\overrightarrow{a_1}=-\widehat{i}-3\widehat{j}-5\widehat{k},\ \overrightarrow{b_1}=3\widehat{i}+5\widehat{j}+7\widehat{k}
\displaystyle \text{For the second line, }\overrightarrow{a_2}=2\widehat{i}+4\widehat{j}+6\widehat{k},\ \overrightarrow{b_2}=\widehat{i}+4\widehat{j}+7\widehat{k}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=3\widehat{i}+7\widehat{j}+11\widehat{k}
\displaystyle \overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&5&7\\1&4&7\end{vmatrix}
\displaystyle =7\widehat{i}-14\widehat{j}+7\widehat{k}
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})
\displaystyle =(3\widehat{i}+7\widehat{j}+11\widehat{k})\cdot(7\widehat{i}-14\widehat{j}+7\widehat{k})
\displaystyle =21-98+77=0
\displaystyle \therefore \text{the given lines are coplanar.}
\displaystyle \text{Normal to the required plane is }7\widehat{i}-14\widehat{j}+7\widehat{k}
\displaystyle \therefore \text{normal can be taken as }\widehat{i}-2\widehat{j}+\widehat{k}
\displaystyle \text{The plane passes through }(-1,-3,-5).
\displaystyle \therefore 1(x+1)-2(y+3)+1(z+5)=0
\displaystyle \therefore x+1-2y-6+z+5=0
\displaystyle \therefore x-2y+z=0
\displaystyle \therefore \text{Required plane is }x-2y+z=0.
\\

\displaystyle \textbf{Question 29. }\text{Find the coordinates of foot of perpendicular drawn from the point } \\ (0,2,3)\text{ on line }\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}. \text{ Also, find the length of perpendicular.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(0,2,3)\text{ and let }H\text{ be the foot of the perpendicular on the line.}
\displaystyle \text{Equation of the line is } \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda
\displaystyle \therefore H(-3+5\lambda,\ 1+2\lambda,\ -4+3\lambda)
\displaystyle \overrightarrow{PH}=(-3+5\lambda)\widehat{i}+(-1+2\lambda)\widehat{j}+(-7+3\lambda)\widehat{k}
\displaystyle \text{Direction vector of the line is }5\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle \text{Since }PH\perp\text{ line,}
\displaystyle \overrightarrow{PH}\cdot(5\widehat{i}+2\widehat{j}+3\widehat{k})=0
\displaystyle 5(-3+5\lambda)+2(-1+2\lambda)+3(-7+3\lambda)=0
\displaystyle -15+25\lambda-2+4\lambda-21+9\lambda=0
\displaystyle 38\lambda-38=0
\displaystyle \therefore \lambda=1
\displaystyle \therefore H=(2,3,-1)
\displaystyle \text{Hence, the foot of the perpendicular is }(2,3,-1).
\displaystyle \text{Length of perpendicular}=|PH|
\displaystyle =\sqrt{(2-0)^2+(3-2)^2+(-1-3)^2}
\displaystyle =\sqrt{4+1+16}
\displaystyle =\sqrt{21}
\displaystyle \therefore \text{Length of the perpendicular }=\sqrt{21}.
\\

\displaystyle \textbf{Question 30. }\text{Show that lines }\overrightarrow{r}=(\widehat{i}+\widehat{j}-\widehat{k})+\lambda(3\widehat{i}-\widehat{j})\text{ and } \\ \overrightarrow{r}=(4\widehat{i}-\widehat{k})+\mu(2\widehat{i}+3\widehat{k})\text{ intersect each other. Find their point of intersection.}
\displaystyle \text{Answer:}
\displaystyle \text{For the first line,}
\displaystyle x=1+3\lambda,\quad y=1-\lambda,\quad z=-1
\displaystyle \text{For the second line,}
\displaystyle x=4+2\mu,\quad y=0,\quad z=-1+3\mu
\displaystyle \text{At the point of intersection, corresponding coordinates must be equal.}
\displaystyle 1-\lambda=0
\displaystyle \therefore \lambda=1
\displaystyle -1=-1+3\mu
\displaystyle \therefore \mu=0
\displaystyle \text{Now, }x=1+3(1)=4
\displaystyle \text{and }x=4+2(0)=4
\displaystyle \therefore \text{all three coordinates satisfy simultaneously.}
\displaystyle \text{Hence, the two lines intersect.}
\displaystyle \text{The point of intersection is }(4,0,-1).
\\

\displaystyle \textbf{Question 31. }\text{By computing shortest distance, determine whether the following} \\ \text{pair of lines intersect or not }\overrightarrow{r}=(4\widehat{i}+5\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}).
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a_1}=4\widehat{i}+5\widehat{j},\ \overrightarrow{b_1}=\widehat{i}+2\widehat{j}-3\widehat{k}
\displaystyle \overrightarrow{a_2}=\widehat{i}-\widehat{j}+2\widehat{k},\ \overrightarrow{b_2}=2\widehat{i}+4\widehat{j}-5\widehat{k}
\displaystyle \text{Shortest distance}=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})|}{|\overrightarrow{b_1}\times\overrightarrow{b_2}|}
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=-3\widehat{i}-6\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&2&-3\\2&4&-5\end{vmatrix}
\displaystyle =2\widehat{i}-\widehat{j}
\displaystyle |(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})|=|(-3\widehat{i}-6\widehat{j}+2\widehat{k})\cdot(2\widehat{i}-\widehat{j})|
\displaystyle =|-6+6|=0
\displaystyle \therefore \text{Shortest distance}=0
\displaystyle \therefore \text{the given lines intersect each other.}
\\

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