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

Any point on the given line

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

is : $2+3\lambda ,-1+4\lambda ,2+2\lambda$

Since this point lies on the plane $\overrightarrow{r}.(\hat{i}-\hat{j}+\hat{k})=5$

$\therefore (2+3\lambda )-(-1+4\lambda )+(2+2\lambda )=5$ $2+3\lambda +1-4\lambda +2+2\lambda =5$
$\lambda =0$
$\therefore$ Point of intersection of the given line and the given plane is $(2,-1,2)$

Now, distance between $(-1,-5,-10)$ and $(2,-1,2)$

= $\sqrt{(2+1{)}^2+(-1+5{)}^2+(2+10{)}^2}$

= $\sqrt{9+16+144}=\sqrt{169}$

= $13\text{units}$