\displaystyle \textbf{Question 1: }~\text{Find the perpendicular distance of the point }(3,-1,11) \\ \text{ from the line }\frac{x}{2}=\frac{y-2}{-3}=\frac{z-3}{4}.
\displaystyle \text{Answer:}
\displaystyle \text{Given equation of line }AB\text{ is}
\displaystyle \frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda\ \text{[say]}
\displaystyle \text{or }\frac{x}{2}=\lambda,\ \frac{y-2}{3}=\lambda
\displaystyle \text{and }\frac{z-3}{4}=\lambda
\displaystyle \text{or }x=2\lambda,\ y=3\lambda+2
\displaystyle \text{and }z=4\lambda+3
\displaystyle \therefore\ \text{Any point }P\text{ on the given line}
\displaystyle =(2\lambda,3\lambda+2,4\lambda+3)
\displaystyle Q(3,-1,11)
\displaystyle \text{Let }P\text{ be the foot of perpendicular drawn from point}
\displaystyle Q(3,-1,11)\text{ on line }AB.
\displaystyle \text{Now, DR's of line }QP=(2\lambda-3,3\lambda+2+1,4\lambda+3-11)
\displaystyle =(2\lambda-3,3\lambda+3,4\lambda-8)
\displaystyle \text{Here, }a_1=2\lambda-3,\ b_1=3\lambda+3,\ c_1=4\lambda-8
\displaystyle \text{and }a_2=2,\ b_2=3,\ c_2=4
\displaystyle \text{Since, }QP\perp AB
\displaystyle \therefore\ a_1a_2+b_1b_2+c_1c_2=0
\displaystyle \text{or }2(2\lambda-3)+3(3\lambda+3)+4(4\lambda-8)=0
\displaystyle \text{or }4\lambda-6+9\lambda+9+16\lambda-32=0
\displaystyle \text{or }29\lambda-29=0
\displaystyle \text{or }\lambda=1
\displaystyle \therefore\ \text{Foot of perpendicular}
\displaystyle P=(2,3+2,4+3)
\displaystyle =(2,5,7)
\displaystyle \text{Now, equation of perpendicular }QP,\text{ where }Q(3,-1,11)\text{ and }P(2,5,7)\text{ is}
\displaystyle \frac{x-3}{2-3}=\frac{y+1}{5+1}=\frac{z-11}{7-11}
\displaystyle \text{Using two points form of equation of line}
\displaystyle \text{i.e. }\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}
\displaystyle \text{or }\frac{x-3}{-1}=\frac{y+1}{6}=\frac{z-11}{-4}
\displaystyle \text{Now, length of perpendicular }QP=\text{ distance between points }Q(3,-1,11)\text{ and }P(2,5,7)
\displaystyle =\sqrt{(2-3)^2+(5+1)^2+(7-11)^2}
\displaystyle =\sqrt{1+36+16}
\displaystyle =\sqrt{53}
\displaystyle \text{Hence, length of perpendicular is }\sqrt{53}.

\displaystyle \textbf{Question 2: }~\text{Find the perpendicular distance of the point }(1,0,0)\text{ from the} \\ \text{line }\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}.\text{ Also, find the coordinates of the foot of} \\ \text{the perpendicular and the equation of the perpendicular. \ [CBSE\ 2005,\ 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given: Point }P(1,0,0)\text{ and equation of line }\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}
\displaystyle \text{Let }PQ\text{ be the perpendicular drawn from }P\text{ to given line whose endpoint/foot is }Q.
\displaystyle \text{Thus, to find distance }PQ\text{ we have to first find coordinates of }Q.
\displaystyle \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}=\lambda\ (\text{let})
\displaystyle \Rightarrow x=2\lambda+1,\ y=-3\lambda-1,\ z=8\lambda-10
\displaystyle \text{Therefore, coordinates of }Q(2\lambda+1,-3\lambda-1,8\lambda-10)
\displaystyle \text{Now as we know (TIP): if two points }A(x_1,y_1,z_1)\text{ and }B(x_2,y_2,z_2)\text{ are on a line, then its direction ratios are proportional to }(x_2-x_1,y_2-y_1,z_2-z_1)
\displaystyle \text{Hence, direction ratios of }PQ\text{ is}
\displaystyle =(2\lambda+1-1,-3\lambda-1-0,8\lambda-10-0)
\displaystyle =(2\lambda,-3\lambda-1,8\lambda-10)
\displaystyle \text{and by comparing with given line equation, direction ratios of the given line are}
\displaystyle =(2,-3,8)
\displaystyle \text{Since }PQ\text{ is perpendicular to given line, therefore by ``condition of perpendicularity''}
\displaystyle a_1a_2+b_1b_2+c_1c_2=0\text{, where }a\text{ and }b\text{ terms are direction ratios of perpendicular lines}
\displaystyle \Rightarrow 2(2\lambda)+(-3)(-3\lambda-1)+8(8\lambda-10)=0
\displaystyle \Rightarrow 4\lambda+9\lambda+3+64\lambda-80=0
\displaystyle \Rightarrow 77\lambda-77=0
\displaystyle \Rightarrow \lambda=1
\displaystyle \text{Therefore coordinates of }Q
\displaystyle \text{i.e. foot of perpendicular}
\displaystyle =Q(2(1)+1,-3(1)-1,8(1)-10)
\displaystyle =Q(3,-4,-2)
\displaystyle \text{Now,}
\displaystyle \text{Distance between }PQ
\displaystyle \text{Distance between two points }A(x_1,y_1,z_1)\text{ and }B(x_2,y_2,z_2)\text{ is given by}
\displaystyle =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
\displaystyle =\sqrt{(1-3)^2+(0+4)^2+(0+2)^2}
\displaystyle =\sqrt{(-2)^2+(4)^2+(2)^2}
\displaystyle =\sqrt{4+16+4}
\displaystyle =\sqrt{24}
\displaystyle =2\sqrt{6}\text{ unit}

\displaystyle \textbf{Question 3: }~\text{Find the foot of the perpendicular drawn from the point }A(1,0,3) \\ \text{ to the line} \text{joining the points }B(4,7,1)\text{ and }C(3,5,3).
\displaystyle \text{Answer:}
\displaystyle \text{Let }D\text{ be the foot of the perpendicular drawn from the point }A(1,0,3)\text{ to the line }BC.
\displaystyle \text{The coordinates of a general point on the line }BC\text{ are given by}
\displaystyle \frac{x-4}{4-3}=\frac{y-7}{7-5}=\frac{z-1}{1-3}=\lambda
\displaystyle \Rightarrow x=\lambda+4
\displaystyle y=2\lambda+7
\displaystyle z=-2\lambda+1
\displaystyle \text{Let the coordinates of }D\text{ be}
\displaystyle (\lambda+4,2\lambda+7,-2\lambda+1)
\displaystyle \text{The direction ratios of }AD\text{ are proportional to}
\displaystyle (\lambda+4-1,2\lambda+7-0,-2\lambda+1-3)
\displaystyle (\lambda+3,2\lambda+7,-2\lambda-2)
\displaystyle \text{The direction ratios of the line }BC\text{ are proportional to }(1,2,-2)\text{ but }AD\perp BC.
\displaystyle \therefore\ 1(\lambda+3)+2(2\lambda+7)-2(-2\lambda-2)=0
\displaystyle \Rightarrow \lambda+3+4\lambda+14+4\lambda+4=0
\displaystyle \Rightarrow 9\lambda+21=0
\displaystyle \Rightarrow \lambda=-\frac{7}{3}
\displaystyle \text{Substituting}
\displaystyle \Rightarrow \lambda=-\frac{7}{3}\text{ in }(\lambda+4,2\lambda+7,-2\lambda+1)
\displaystyle \text{we get the coordinates of }D
\displaystyle =\left(\frac{5}{3},\frac{7}{3},\frac{17}{3}\right)

\displaystyle \textbf{Question 4: }~\text{A(1,0,4),\ B(0,-11,3),\ C(2,-3,1)} \text{ are three points and } D \\ \text{ is the foot of }\text{perpendicular } \text{ from } A\text{ on }BC.\text{ Find the coordinates of }D.
\displaystyle \text{Answer:}
\displaystyle \text{Point }D\text{ is the foot of the perpendicular drawn from the point }A(1,0,4)\text{ to the line }BC.
\displaystyle \text{The coordinates of a general point on the line }BC\text{ are given by}
\displaystyle \frac{x-0}{2-0}=\frac{y+11}{-3+11}=\frac{z-3}{1-3}=\lambda
\displaystyle \Rightarrow x=2\lambda
\displaystyle y=8\lambda-11
\displaystyle z=-2\lambda+3
\displaystyle \text{Let the coordinates of }D\text{ be}
\displaystyle (2\lambda,8\lambda-11,-2\lambda+3)
\displaystyle \text{The direction ratios of }AD\text{ are proportional to}
\displaystyle (2\lambda-1,8\lambda-11-0,-2\lambda+3-4)
\displaystyle \text{i.e. }(2\lambda-1,8\lambda-11,-2\lambda-1)
\displaystyle \text{The direction ratios of the line }BC\text{ are proportional to }(2,8,-2)\text{ but }AD\perp BC.
\displaystyle \therefore\ 2(2\lambda-1)+8(8\lambda-11)-2(-2\lambda-1)=0
\displaystyle \Rightarrow 4\lambda-2+64\lambda-88+4\lambda+2=0
\displaystyle \Rightarrow 72\lambda-88=0
\displaystyle \Rightarrow \lambda=\frac{11}{9}
\displaystyle \text{Substituting}
\displaystyle \Rightarrow \lambda=\frac{11}{9}\text{ in }(2\lambda,8\lambda-11,-2\lambda+3)
\displaystyle \text{we get the coordinates of }D\text{ as}
\displaystyle =\left(\frac{22}{9},-\frac{11}{9},\frac{5}{9}\right)

\displaystyle \textbf{Question 5: }~\text{Find the foot of perpendicular from the point }(2,3,4)\text{ to the line } \\ \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}.\text{ Also, find the perpendicular distance from the given point} \\ \text{to the line.  }
\displaystyle \text{Answer:}
\displaystyle \text{Let }L\text{ be the foot of the perpendicular drawn from the point }P(2,3,4)\text{ to the given line.}
\displaystyle \text{The coordinates of a general point on the line}
\displaystyle \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}
\displaystyle \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}=\lambda
\displaystyle \text{They can be re-written as}
\displaystyle \frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}=\lambda
\displaystyle \Rightarrow x=-2\lambda+4
\displaystyle y=6\lambda
\displaystyle z=-3\lambda+1
\displaystyle \text{Let the coordinates of }L\text{ be}
\displaystyle (-2\lambda+4,6\lambda,-3\lambda+1)
\displaystyle \text{The direction ratios of }PL\text{ are proportional to}
\displaystyle (-2\lambda+4-2,6\lambda-3,-3\lambda+1-4)
\displaystyle \text{i.e. }(-2\lambda+2,6\lambda-3,-3\lambda-3)
\displaystyle \text{The direction ratios of the given line are proportional to }(-2,6,-3)\text{, but }PL\perp\text{ the given line.}
\displaystyle \therefore -2(-2\lambda+2)+6(6\lambda-3)-3(-3\lambda-3)=0
\displaystyle \Rightarrow 4\lambda-4+36\lambda-18+9\lambda+9=0
\displaystyle \Rightarrow 49\lambda-13=0
\displaystyle \Rightarrow \lambda=\frac{13}{49}
\displaystyle \text{Substituting}
\displaystyle \Rightarrow \lambda=\frac{13}{49}\text{ in }(-2\lambda+4,6\lambda,-3\lambda+1)
\displaystyle \text{we get the coordinates of }L\text{ as}
\displaystyle =\left(\frac{170}{49},\frac{78}{49},\frac{10}{49}\right)
\displaystyle \therefore PL=\sqrt{\left(\frac{170}{49}-2\right)^2+\left(\frac{78}{49}-3\right)^2+\left(\frac{10}{49}-4\right)^2}
\displaystyle =\sqrt{\frac{44521}{2401}}
\displaystyle =\sqrt{\frac{909}{49}}
\displaystyle =\frac{3}{7}\sqrt{101}
\displaystyle \text{Hence, the length of the perpendicular from }P\text{ on }PL\text{ is }\frac{3}{7}\sqrt{101}\text{ units.}

\displaystyle \textbf{Question 6: }~\text{Find the equation of the perpendicular drawn from the point }P(2,4,-1) \\ \text{ to the line } \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}.\text{ Also, write down the coordinates of the foot of the} \\ \text{perpendicular from }P.
\displaystyle \text{Answer:}
\displaystyle \text{Let }L\text{ be the foot of the perpendicular drawn from the point }P(2,4,-1)\text{ to the given line.}
\displaystyle \text{The coordinates of a general point on the line}
\displaystyle \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}\text{ are given by}
\displaystyle \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\lambda
\displaystyle \Rightarrow x=\lambda-5
\displaystyle y=4\lambda-3
\displaystyle z=-9\lambda+6
\displaystyle \text{Let the coordinates of }L\text{ be}
\displaystyle (\lambda-5,4\lambda-3,-9\lambda+6)
\displaystyle \text{The direction ratios of }PL\text{ are proportional to}
\displaystyle (\lambda-5-2,4\lambda-3-4,-9\lambda+6+1)
\displaystyle \text{i.e. }(\lambda-7,4\lambda-7,-9\lambda+7)
\displaystyle \text{The direction ratios of the given line are proportional to }(1,4,-9)\text{, but }PL\perp\text{ the given line.}
\displaystyle \therefore 1(\lambda-7)+4(4\lambda-7)-9(-9\lambda+7)=0
\displaystyle \Rightarrow \lambda-7+16\lambda-28+81\lambda-63=0
\displaystyle \Rightarrow 98\lambda-98=0
\displaystyle \Rightarrow \lambda=1
\displaystyle \text{Substituting}
\displaystyle \Rightarrow \lambda=1\text{ in }(\lambda-5,4\lambda-3,-9\lambda+6)
\displaystyle \text{we get the coordinates of }L\text{ as }(-4,1,-3)
\displaystyle \text{Equation of the line }PL\text{ is}
\displaystyle \frac{x-2}{-4-2}=\frac{y-4}{1-4}=\frac{z+1}{-3+1}
\displaystyle \text{or }\frac{x-2}{-6}=\frac{y-4}{-3}=\frac{z+1}{-2}

\displaystyle \textbf{Question 7: }~\text{Find the length of the perpendicular drawn from the point }(5,4,-1) \\ \text{ to the line }\overrightarrow{r}=\widehat{i}+\lambda(2\widehat{i}+9\widehat{j}+5\widehat{k}).
\displaystyle \text{Answer:}
\displaystyle \text{Let the point }(5,4,-1)\text{ be }P\text{ and the point through which the line passes be }Q(1,0,0).
\displaystyle \text{The line is parallel to the vector }\overrightarrow{b}=2\widehat{i}+9\widehat{j}+5\widehat{k}
\displaystyle \text{Now, }\overrightarrow{PQ}=-4\widehat{i}-4\widehat{j}+\widehat{k}
\displaystyle \therefore\ \overrightarrow{b}\times\overrightarrow{PQ}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&9&5\\-4&-4&1\end{vmatrix}
\displaystyle =29\widehat{i}-22\widehat{j}+28\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b}\times\overrightarrow{PQ}\right|=\sqrt{29^2+(-22)^2+28^2}
\displaystyle =\sqrt{841+484+784}
\displaystyle =\sqrt{2109}
\displaystyle \left|\overrightarrow{b}\right|=\sqrt{2^2+9^2+5^2}
\displaystyle =\sqrt{4+81+25}
\displaystyle =\sqrt{110}
\displaystyle d=\frac{\left|\overrightarrow{b}\times\overrightarrow{PQ}\right|}{\left|\overrightarrow{b}\right|}
\displaystyle =\frac{\sqrt{2109}}{\sqrt{110}}
\displaystyle =\sqrt{\frac{2109}{110}}

\displaystyle \textbf{Question 8: }~\text{Find the foot of the perpendicular drawn from the point } \\ \widehat{i}+6\widehat{j}+3\widehat{k}\text{ to the line }\overrightarrow{r}=\widehat{j}+2\widehat{k}+\lambda(\widehat{i}+2\widehat{j}+3\widehat{k}).\text{ Also, find the length of the} \\ \text{perpendicular.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }L\text{ be the foot of the perpendicular drawn from the point }P(\widehat{i}+6\widehat{j}+3\widehat{k})\text{ to the line }\overrightarrow{r}=\widehat{j}+2\widehat{k}+\lambda(\widehat{i}+2\widehat{j}+3\widehat{k}).
\displaystyle \text{Let the position vector of }L\text{ be }\overrightarrow{r}=\widehat{j}+2\widehat{k}+\lambda(\widehat{i}+2\widehat{j}+3\widehat{k})=\lambda\widehat{i}+(1+2\lambda)\widehat{j}+(2+3\lambda)\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{PL}=\text{Position vector of }L-\text{Position vector of }P
\displaystyle \Rightarrow \overrightarrow{PL}=\{\lambda\widehat{i}+(1+2\lambda)\widehat{j}+(2+3\lambda)\widehat{k}\}-(\widehat{i}+6\widehat{j}+3\widehat{k})
\displaystyle \Rightarrow \overrightarrow{PL}=(\lambda-1)\widehat{i}+(2\lambda-5)\widehat{j}+(3\lambda-1)\widehat{k}\quad\text{...(2)}
\displaystyle \text{Since }\overrightarrow{PL}\text{ is perpendicular to the given line, which is parallel to }\overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle \text{we have }
\displaystyle \overrightarrow{PL}\cdot\overrightarrow{b}=0
\displaystyle \Rightarrow \{(\lambda-1)\widehat{i}+(2\lambda-5)\widehat{j}+(3\lambda-1)\widehat{k}\}\cdot(\widehat{i}+2\widehat{j}+3\widehat{k})=0
\displaystyle \Rightarrow 1(\lambda-1)+2(2\lambda-5)+3(3\lambda-1)=0
\displaystyle \Rightarrow \lambda-1+4\lambda-10+9\lambda-3=0
\displaystyle \Rightarrow 14\lambda-14=0
\displaystyle \Rightarrow \lambda=1
\displaystyle \text{Substituting }\lambda=1\text{ in (1),}
\displaystyle \text{we get the position vector of }L\text{ as }\widehat{i}+3\widehat{j}+5\widehat{k}
\displaystyle \text{Substituting }\lambda=1\text{ in (2),}
\displaystyle \text{we get }\overrightarrow{PL}=-3\widehat{j}+2\widehat{k}
\displaystyle =\sqrt{(-3)^2+2^2}
\displaystyle =\sqrt{13}
\displaystyle \text{Hence, the length of the perpendicular from point }P\text{ on }PL\text{ is }\sqrt{13}\text{ units.}

\displaystyle \textbf{Question 9: }~\text{Find the equation of the perpendicular drawn from the point }P(-1,3,2) \\ \text{ to the line }\overrightarrow{r}=(2\widehat{j}+3\widehat{k})+\lambda(2\widehat{i}+\widehat{j}+3\widehat{k}).\text{ Also, find the coordinates of the foot of the} \\ \text{perpendicular from }P.
\displaystyle \text{Answer:}


\displaystyle \text{Let }L\text{ be the foot of the perpendicular drawn from the point }P(-1,3,2)\text{ to the line }\overrightarrow{r}=(2\widehat{j}+3\widehat{k})+\lambda(2\widehat{i}+\widehat{j}+3\widehat{k}).
\displaystyle \text{Let the position vector of }L\text{ be }\overrightarrow{r}=(2\widehat{j}+3\widehat{k})+\lambda(2\widehat{i}+\widehat{j}+3\widehat{k})=2\lambda\widehat{i}+(2+\lambda)\widehat{j}+(3+3\lambda)\widehat{k}\quad\text{...(1)}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{PL}=\text{Position vector of }L-\text{Position vector of }P
\displaystyle \Rightarrow \overrightarrow{PL}=\{2\lambda\widehat{i}+(2+\lambda)\widehat{j}+(3+3\lambda)\widehat{k}\}-(-\widehat{i}+3\widehat{j}+2\widehat{k})
\displaystyle \Rightarrow \overrightarrow{PL}=(2\lambda+1)\widehat{i}+(\lambda-1)\widehat{j}+(3\lambda+1)\widehat{k}\quad\text{...(2)}
\displaystyle \text{Since }\overrightarrow{PL}\text{ is perpendicular to the given line, which is parallel to }\overrightarrow{b}=2\widehat{i}+\widehat{j}+3\widehat{k}
\displaystyle \text{we have }
\displaystyle \overrightarrow{PL}\cdot\overrightarrow{b}=0
\displaystyle \Rightarrow \{(2\lambda+1)\widehat{i}+(\lambda-1)\widehat{j}+(3\lambda+1)\widehat{k}\}\cdot(2\widehat{i}+\widehat{j}+3\widehat{k})=0
\displaystyle \Rightarrow 2(2\lambda+1)+1(\lambda-1)+3(3\lambda+1)=0
\displaystyle \Rightarrow 4\lambda+2+\lambda-1+9\lambda+3=0
\displaystyle \Rightarrow 14\lambda+4=0
\displaystyle \Rightarrow \lambda=-\frac{2}{7}
\displaystyle \text{Substituting }\lambda=-\frac{2}{7}\text{ in (1),}
\displaystyle \text{we get the position vector of }L\text{ as }-\frac{4}{7}\widehat{i}+\frac{12}{7}\widehat{j}+\frac{15}{7}\widehat{k}
\displaystyle \text{So, the coordinates of the foot of the perpendicular from }P\text{ to the given line is }L\left(-\frac{4}{7},\frac{12}{7},\frac{15}{7}\right)
\displaystyle \text{Substituting }\lambda=-\frac{2}{7}\text{ in (2), we get }\overrightarrow{PL}=\frac{3}{7}\widehat{i}-\frac{9}{7}\widehat{j}+\frac{1}{7}\widehat{k}
\displaystyle \text{Equation of the perpendicular drawn from }P\text{ to the given line is}
\displaystyle \overrightarrow{r}=\text{Position vector of }P+\lambda(\overrightarrow{PL})
\displaystyle =(-\widehat{i}+3\widehat{j}+2\widehat{k})+\lambda\left(\frac{3}{7}\widehat{i}-\frac{9}{7}\widehat{j}+\frac{1}{7}\widehat{k}\right)

\displaystyle \textbf{Question 10: }~\text{Find the foot of the perpendicular from }(0,2,7)\text{ on the line } \\ \frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }L\text{ be the foot of the perpendicular drawn from the point }P(0,2,7)\text{ to the given line.}
\displaystyle \text{The coordinates of a general point on the line }\frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}\text{ are given by}
\displaystyle \frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}=\lambda
\displaystyle \Rightarrow x=-\lambda-2
\displaystyle y=3\lambda+1
\displaystyle z=-2\lambda+3
\displaystyle \text{Let the coordinates of }L\text{ be }(-\lambda-2,3\lambda+1,-2\lambda+3)
\displaystyle \text{The direction ratios of }PL\text{ are proportional to }
\displaystyle (-\lambda-2-0,3\lambda+1-2,-2\lambda+3-7)
\displaystyle \text{i.e. }(-\lambda-2,3\lambda-1,-2\lambda-4)
\displaystyle \text{The direction ratios of the given line are proportional to }(-1,3,-2)\text{, but }PL\perp\text{ the given line.}
\displaystyle \therefore -1(-\lambda-2)+3(3\lambda-1)-2(-2\lambda-4)=0
\displaystyle \Rightarrow \lambda+2+9\lambda-3+4\lambda+8=0
\displaystyle \Rightarrow 14\lambda+7=0
\displaystyle \Rightarrow \lambda=-\frac{1}{2}
\displaystyle \text{Substituting }\lambda=-\frac{1}{2}\text{ in }(-\lambda-2,3\lambda+1,-2\lambda+3)
\displaystyle \text{we get the coordinates of }L\text{ as }\left(-\frac{3}{2},-\frac{1}{2},4\right)

\displaystyle \textbf{Question 11: }~\text{Find the foot of the perpendicular from }(1,2,-3)\text{ to the line } \\ \frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }L\text{ be the foot of the perpendicular drawn from the point }P(1,2,-3)\text{ to the given line.}
\displaystyle \text{The coordinates of a general point on the line }\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}\text{ are given by}
\displaystyle \frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}=\lambda
\displaystyle \Rightarrow x=2\lambda-1
\displaystyle y=-2\lambda+3
\displaystyle z=-\lambda
\displaystyle \text{Let the coordinates of }L\text{ be }(2\lambda-1,-2\lambda+3,-\lambda)
\displaystyle \text{The direction ratios of }PL\text{ are proportional to}
\displaystyle (2\lambda-1-1,-2\lambda+3-2,-\lambda+3)
\displaystyle \text{i.e. }(2\lambda-2,-2\lambda+1,-\lambda+3)
\displaystyle \text{The direction ratios of the given line are proportional to }(2,-2,-1)\text{, but }PL\perp\text{ the given line.}
\displaystyle \therefore 2(2\lambda-2)-2(-2\lambda+1)-1(-\lambda+3)=0
\displaystyle \Rightarrow 4\lambda-4+4\lambda-2+\lambda-3=0
\displaystyle \Rightarrow 9\lambda-9=0
\displaystyle \Rightarrow \lambda=1
\displaystyle \text{Substituting }\lambda=1\text{ in }(2\lambda-1,-2\lambda+3,-\lambda)
\displaystyle \text{we get the coordinates of }L\text{ as }(1,1,-1)

\displaystyle \textbf{Question 12: }~\text{Find the equation of line passing through the points }A(0,6,-9) \\ \text{ and }B(-3,-6,3). \text{ If }D\text{ is the foot of perpendicular drawn from a point }C(7,4,-1) \\ \text{ on the line }AB, \text{ then find the coordinates of the point }D\text{ and the} \\ \text{equation of line }CD.\text{ [CBSE\ 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Equation of line }AB\text{ passing through the points }A(0,6,-9)\text{ and }B(-3,-6,3)\text{ is}
\displaystyle \frac{x-0}{-3-0}=\frac{y-6}{-6-6}=\frac{z+9}{3+9}
\displaystyle \text{or }\frac{x}{1}=\frac{y-6}{4}=\frac{z+9}{-4}
\displaystyle \text{Here, }D\text{ is the foot of the perpendicular drawn from }C(7,4,-1)\text{ on }AB.
\displaystyle \text{The coordinates of a general point on }AB\text{ are given by}
\displaystyle \frac{x}{1}=\frac{y-6}{4}=\frac{z+9}{-4}=\lambda
\displaystyle \Rightarrow x=\lambda
\displaystyle y=4\lambda+6
\displaystyle z=-4\lambda-9
\displaystyle \text{Let the coordinates of }D\text{ be }(\lambda,4\lambda+6,-4\lambda-9)
\displaystyle \text{The direction ratios of }CD\text{ are proportional to}
\displaystyle (\lambda-7,4\lambda+6-4,-4\lambda-9+1)
\displaystyle \text{i.e. }(\lambda-7,4\lambda+2,-4\lambda-8)
\displaystyle \text{The direction ratios of }AB\text{ are proportional to }(1,4,-4)\text{, but }CD\perp AB.
\displaystyle \therefore 1(\lambda-7)+4(4\lambda+2)-4(-4\lambda-8)=0
\displaystyle \Rightarrow \lambda-7+16\lambda+8+16\lambda+32=0
\displaystyle \Rightarrow 33\lambda+33=0
\displaystyle \Rightarrow \lambda=-1
\displaystyle \text{Substituting }\lambda=-1\text{ in }(\lambda,4\lambda+6,-4\lambda-9)
\displaystyle \text{we get the coordinates of }D\text{ as }(-1,2,-5)
\displaystyle \text{Equation of }CD\text{ is,}
\displaystyle \frac{x-7}{-1-7}=\frac{y-4}{2-4}=\frac{z+1}{-5+1}
\displaystyle \text{or }\frac{x-7}{4}=\frac{y-4}{1}=\frac{z+1}{2}

\displaystyle \textbf{Question 13: }~\text{Find the distance of the point }(2,4,-1)\text{ from the line } \\ \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}.
\displaystyle \text{Answer:}
\displaystyle \text{We know that the distance }d\text{ from point }P\text{ to the line }l\text{ having equation }\overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}\text{ is given by }d=\frac{\left|\overrightarrow{b}\times\overrightarrow{PQ}\right|}{\left|\overrightarrow{b}\right|}
\displaystyle \text{where }Q\text{ is any point on the line }l.
\displaystyle \text{The equation of the given line is }\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}.
\displaystyle \text{Let }P(2,4,-1)\text{ be the given point.}
\displaystyle \text{Now, in order to find the required distance we need to find a point }Q\text{ on the given line.}
\displaystyle \text{The given line passes through the point }(-5,-3,6).
\displaystyle \text{So, take this point as the required point }Q(-5,-3,6).
\displaystyle \text{Also, the line is parallel to the vector }\overrightarrow{b}=\widehat{i}+4\widehat{j}-9\widehat{k}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{PQ}=(-5\widehat{i}-3\widehat{j}+6\widehat{k})-(2\widehat{i}+4\widehat{j}-\widehat{k})
\displaystyle =-7\widehat{i}-7\widehat{j}+7\widehat{k}
\displaystyle \overrightarrow{b}\times\overrightarrow{PQ}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&4&-9\\-7&-7&7\end{vmatrix}
\displaystyle =-35\widehat{i}+56\widehat{j}+21\widehat{k}
\displaystyle \Rightarrow \left|\overrightarrow{b}\times\overrightarrow{PQ}\right|=\sqrt{(-35)^2+56^2+21^2}
\displaystyle =\sqrt{1225+3136+441}
\displaystyle =\sqrt{4802}=49\sqrt{2}
\displaystyle \text{Let }d\text{ be the required distance.}
\displaystyle d=\frac{\left|\overrightarrow{b}\times\overrightarrow{PQ}\right|}{\left|\overrightarrow{b}\right|}
\displaystyle =\frac{49\sqrt{2}}{\sqrt{1^2+4^2+(-9)^2}}
\displaystyle =\frac{49\sqrt{2}}{\sqrt{98}}
\displaystyle =\frac{49\sqrt{2}}{7\sqrt{2}}=7
\displaystyle \text{Thus, the distance of the given point from the given line is }7\text{ units.}

\displaystyle \textbf{Question 14: }~\text{Find the coordinates of the foot of perpendicular drawn from} \\ \text{the point} A(1,8,4)\text{ to the line joining the points }B(0,-1,3)\text{ and }C(2,-3,-1). 
\displaystyle \text{Answer:}
\displaystyle \text{The Cartesian equation of the line joining points }B(0,-1,3)\text{ and }C(2,-3,-1)\text{ is }\frac{x-0}{2-0}=\frac{y-(-1)}{-3-(-1)}=\frac{z-3}{-1-3}
\displaystyle \text{Or }\frac{x}{2}=\frac{y+1}{-2}=\frac{z-3}{-4}
\displaystyle \text{Let }L\text{ be the foot of the perpendicular drawn from the point }A(1,8,4)\text{ to the line}
\displaystyle \frac{x}{2}=\frac{y+1}{-2}=\frac{z-3}{-4}
\displaystyle \text{The coordinates of a general point on the line }\frac{x}{2}=\frac{y+1}{-2}=\frac{z-3}{-4}\text{ are given by }\frac{x}{2}=\frac{y+1}{-2}=\frac{z-3}{-4}=\lambda
\displaystyle \text{Or }x=2\lambda,\ y=-2\lambda-1,\ z=-4\lambda+3
\displaystyle \text{Let the coordinates of }L\text{ be }(2\lambda,-2\lambda-1,-4\lambda+3)
\displaystyle \text{Therefore, the direction ratios of }AL\text{ are proportional to }(2\lambda-1,-2\lambda-1-8,-4\lambda+3-4)
\displaystyle \text{or }(2\lambda-1,-2\lambda-9,-4\lambda-1)
\displaystyle \text{Direction ratios of the given line are proportional to }(2,-2,-4)\text{. But }AL\perp\text{ the given line.}
\displaystyle \therefore 2(2\lambda-1)+(-2)(-2\lambda-9)+(-4)(-4\lambda-1)=0
\displaystyle \Rightarrow 4\lambda-2+4\lambda+18+16\lambda+4=0
\displaystyle \Rightarrow 24\lambda+20=0
\displaystyle \Rightarrow \lambda=-\frac{5}{6}
\displaystyle \text{Putting,}
\displaystyle \lambda=-\frac{5}{6}\text{ in }(2\lambda,-2\lambda-1,-4\lambda+3)
\displaystyle \text{we get,}
\displaystyle \left(2\left(-\frac{5}{6}\right),-2\left(-\frac{5}{6}\right)-1,-4\left(-\frac{5}{6}\right)+3\right)=\left(-\frac{5}{3},\frac{2}{3},\frac{19}{3}\right)
\displaystyle \text{Thus, the required coordinates of the foot of the perpendicular are }\left(-\frac{5}{3},\frac{2}{3},\frac{19}{3}\right)

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.