The projection of the point (11,-1,6) onto the plane $3x+2y-7z-51=0$ is equal to

Question

The projection of the point $(11,-1,6)$ onto the plane $3x+2y-7z-51=0$ is equal to

(A) $(14,1,-1$

(B)$(4,2,-5)$

(C)$(18,2,1)$

(D) None of these

I'm exactly sure how to do this but if $(a,b,c)$ is the projection of the point then the line connecting $(a,b,c)$ and $(11,-1,6)$ must be orthgonal to the plane $3x+2y-7z-51=0$ and $(a,b,c)$ must be located on the this plane. But this does not take me anywhere .

I would like to have complete solution .

Answer 1

The point which is foot of perpendicular on the plane and the original point itself,when joined are parallel to the normal to the plane. So you get the Dr's of the line formed by the original point and foot of perpendicular same as Dr's of normal to the plane. I.e. $(3,2,-7)$. So you can take a parametric point :

$$(a,b,c)=(11,-1,6)+\lambda(3,2,-7)$$Now what you need to do is just satisfy this on the equation of plane to get the value of $\lambda$

Answer 2

The plane: $f(x) = 3x + 3 y + 7 z - 51 = 0$.

The normal $\vec{N}$ to this plane : $\nabla f = (3, 2, -7)$.

Vector joining original point to projection point is parallel to the normal $\vec{N}$:

A) $\vec{v_1} = (11, -1, 6) - (14, 1, -1) = (-3, -2, 7)$.

First time lucky, $( -3, -2, 7)$ is $(-1)(3, 2, -7)$, hence parallel.

B) $\vec{v_2} = (11, -1, 6) - (4, 2, -5) = (7, -3, 11)$.

C) $\vec{v_3} = (11, -1, 6) - (18, 2, 1) = (-7, -3, -5)$.

A is the answer.