Question

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 $x-y+z=5$ is

• A. 13.0
• B. 13
• C. 013
• D. 13.00

Answer 1

Answer: (A), (B), (C), (D)

Given plane equation is $x-y+$ $z=5$.
Any point on the line is $(2+3\lambda ,4\lambda -1,2+2\lambda )$

Now, for point of intersection, substitute equation of point in plane to obtain parameter $\lambda$.

$\therefore 2+3\lambda -(4\lambda -1)+2+2\lambda =5\therefore 5+\lambda =5\therefore \lambda =0$

Hence, point of intersection of line and plane is $(2,-1,2)$.

Distance between $(-1,-5,-10)$ and

$(2,-1,2)=\sqrt{(2+1{)}^2+(-1+5{)}^2+(2+10{)}^2}=\sqrt{169}=13$