Question

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

Answer 1

Equation of given line is

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

Equation of plane is

$\overrightarrow{r}\cdot (i-2j+k)=0$

For line to intersect with plane

$[(2+3\lambda )i+(-4+4\lambda )j+(2+2\lambda )k].[i-2j+k]=0$

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

$\Rightarrow \lambda =4$

$\overrightarrow{r}=14i+12j+10k$

$\therefore$ Points of intersection is $(14,12,10)$

$\therefore$ distance $=\sqrt{{(14-2)}^2+{(12-12)}^2+{(10-5)}^2}=\sqrt{144+25}=13$ units.