Question

Find the distance of the point P$(3,4,4)$ from the point, where the line joining the point A$(3,-4,-5)$ and B$(2,-3,1)$ intersects the plane $2x+y+z=7$.

Answer 1

The line joining the points A and B is parallel to the vector pointing from A to B, hence its equation is:

$\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda$.

Any point on this line is $(3-\lambda ,-4+\lambda ,-5+6\lambda )$

If this point lies on the given plane, then:

$2x+y+z=7\Rightarrow 2(3-\lambda )+(-4+\lambda )+(-5+6\lambda )=7\Rightarrow -3+5\lambda =7\Rightarrow \lambda =2$

So, the point on the plane is $(1,-2,7)$.

So, the distance of this point from point P is $\sqrt{2^2+6^2+3^2}=7.$