Question

Find the distance between of the point (2,12,5) form the point of intersection of the line $\overrightarrow{r}=2\hat{i}-4\hat{j}+2\hat{k}+\lambda (3\hat{i}+4\hat{j}+2\hat{k})$ and the plane $\overrightarrow{r}.(\hat{i}-2\hat{j}+\hat{k})=0$

Answer 1

$\text{Equation of line is}$
$\overrightarrow{r}=2\hat{i}-4\hat{j}+2\hat{k}+\lambda (3\hat{i}+4\hat{j}+2\hat{k})$

$\text{Its cartesian form is}$
$\frac{x-2}{3}=\frac{y+4}{4}=\frac{z-2}{2}=\lambda ....\left(i\right)$

$\text{Equation of plane is}$ $\overrightarrow{r}.(\hat{i}-2\hat{j}+\hat{k})=0$

$\Rightarrow x-2y+z=0....\left(ii\right)$

$\text{To find point of intersection of (1) and (2) from (1),}$

$x=2+3\lambda ,y=-4+4\lambda ,z=2+2\lambda$

$\text{Put in (2),}$
$2+3\lambda -2(-4+4\lambda )+2+2\lambda =0$

$\Rightarrow 2+3\lambda +8-8\lambda +2+2\lambda =0$

$\Rightarrow 12-3\lambda =0\Rightarrow \lambda =4$

$\therefore \text{point of intersection of (1) and (2) is}$

$P(2+3\times 4,-4+4\times 4,2+2\times 4)$

$P(14,12,10)$

$\text{Distance of A(2,12,5) from P(14,12,10)}$

$|AP|=\sqrt{(14-2{)}^2+(12-12{)}^2+(10-5{)}^2}$

$=\sqrt{{12}^2+0^2+5^2}=13units$