Question

Find the distance from a point $(-1,-5,-10)$ to the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=5$.

Answer 1

Let $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=r$

So the point $M(3r+2,4r-1,12r+2)$ lie on the given line.

If this intersect line the plane $x-y+z=5$ at the point M.

So put the value of $x,y,z$ in $x-y+z=5$

$\Rightarrow (3x+2)-(4r-1)+(12r+2)=5$

$\Rightarrow 11r+5=5$

$\Rightarrow 11r=0$

So the point of intersection of line and the plane is $(2,-1,2)$

Now the distance from $(2,-1,2)$ and the point $(-1,-5,-10)$ is

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

$=\sqrt{3^2+4^2+{12}^2}$

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

$=\sqrt{169}=13$.