Question

Find the distance of the point $(-1,-5,-10)$ from the point of intersection of the line $\overrightarrow{r}=(2\hat{i}-\hat{j}+2\hat{k})+\lambda (3\hat{i}+4\hat{j}+2\hat{k})$ and the plane $\overrightarrow{r}(\hat{i}-\hat{j}+\hat{k})=5$

Answer 1

$\overrightarrow{r}(\hat{i}-\hat{j}+\hat{k})=5\Rightarrow (x\hat{i}+y\hat{j}+z\hat{k})\cdot (\hat{i}-\hat{j}+\hat{k})=5\Rightarrow x-y+z=5\cdots \left(1\right)$

$\overrightarrow{r}=(2\hat{i}-\hat{j}+2\hat{k})+\lambda (3\hat{i}+4\hat{j}+2\hat{k})\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda$
The required point will have coordinates -
$(3\lambda +2,4\lambda -1,2\lambda +2)\cdots \left(2\right)$
This point lies on the plane $\left(1\right)$
$\Rightarrow 3\lambda +2-4\lambda +1+2\lambda +2=5\Rightarrow \lambda =0$

Put the value of $\lambda$ in $\left(2\right)$
$\therefore$ The point of intersection of the given line and the plane is $(2,-1,2)$
Distance between $(2,-1,2)$ and $(-1,-5,-10)$ is $\sqrt{(2+1{)}^2+(-1+5{)}^2+(2+10{)}^2}=13$