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

The equation of the given line is
$\overrightarrow{r}=(2\hat{i}-\hat{j}+2\hat{k})+\lambda (3\hat{i}+4\hat{j}+2\hat{k}).....\left(1\right)$
The equation of the given plane is
$\overrightarrow{r}.(\hat{i}-\hat{j}+\hat{k})=\mathrm{5........}\left(2\right)$
Substituting the value of $\overrightarrow{r}$ in equation $\left(2\right)$, we get,

$\left[\right(2\hat{i}-\hat{j}+2\hat{k})+\lambda (3\hat{i}+4\hat{j}+2\hat{k})][\hat{i}-\hat{j}+\hat{k}]=5$

$⟹\left[\right(2+3\lambda )\hat{i}+(4\lambda -1)\hat{j}+(2+2\lambda \left)\hat{k}\right][\hat{i}-\hat{j}+\hat{k}]=5$
$⟹(2+3\lambda )-(4\lambda -1)+(2\lambda +2)=5$
$⟹\lambda =0$
Thus the point of intersection of the given line and the plane is,
$\overrightarrow{r}=2\hat{i}-\hat{j}+2\hat{k}=(2,-1,2)$
This means that the position vector of the point of intersection of the line and plane is given by the coordinates $(2,-1,2)$. The point is $(-1,-5,-10)$.
Hence, distance $d$ between the points, $(2,-1,2)$ and $(-1,-5,-10)$ is
$d=\sqrt{{(-1-2)}^2+{(-5+1)}^2+{(-10-2)}^2}=\sqrt{9+16+144}=\sqrt{169}=13$