Question

A line passing through the point A with position vector $\stackrel{\rightharpoondown }{a}=4\hat{i}+2\hat{j}+2\hat{k}$ is parallel to the vector $\stackrel{\rightharpoondown }{b}=2\hat{i}+3\hat{j}+6\hat{k}$. Find the length of the perpendicular drawn on this line from a point P with vector $\stackrel{\rightharpoondown }{r_1}=\hat{i}+2\hat{j}+3\hat{k}$.

Answer 1

Given position vector

$\overrightarrow{a}=4\hat{i}+2\hat{j}+2\hat{k}$

parallel vector

$\overrightarrow{b}=2\hat{i}+3\hat{j}+6\hat{k}$

eq of line

$\overrightarrow{r}=4\hat{i}+2\hat{j}+2\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$

Point P(1,2,3)

let a line from point $P(1,2,3)$ meets the vector at point $Q(x,y,z)$ on given line perpendicularly

cartesian form of given line

$\frac{x-4}{2}=\frac{y-2}{3}=\frac{z-2}{6}=\lambda$

$x=2\lambda +4$

$y=3\lambda +2$

$z=6\lambda +2$

normal vector of $PQ$

$a_2=2\lambda +4-1=2\lambda +3$

$b_2=3\lambda +2-2=3\lambda$

$c_2=6\lambda +2-3=6\lambda -1$

line $PQ$ is perpendicular to given line so

$a{a}_2+b{b}_2+c{c}_2=0$

$2(2\lambda +3)+3\left(3\lambda \right)+6(6\lambda -1)=0$

$4\lambda +6+9\lambda +36\lambda -6=0$

$49\lambda =0$

$\lambda =0$

$x=4$

$y=2$

$z=2$

distance from point $P$

$PQ=\sqrt{(4-1{)}^2+(2-2{)}^2+(2-3{)}^2}$

$PQ=\sqrt{3^2+(-1{)}^2}$

$PQ=\sqrt{10}$