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

Since 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 $\overrightarrow{r}.(\hat{i}-\hat{j}+\hat{k})=5$ intersects each other so,

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

That is , $2+1+2+\lambda (3-4+2)=5$

$\Rightarrow \lambda =0$

$\therefore$ the point of intersection is $\overrightarrow{r}=2\hat{i}-\hat{j}+2\hat{k}+0(3\hat{i}+4\hat{j}+2\hat{k})=2\hat{i}-\hat{j}+2\hat{k}$ i.e., Q (2,-1,2)

Also let P(-1,-5, -10)

So, the required distance is PQ =$\sqrt{(2+1{)}^2+(-1+5{)}^2+(2+10{)}^2}=\sqrt{9+16+144}$ = 13 units.