What is the projection of the vector $v = [-3, 2, 3]^T$ on the plane with equation $1x + 1y - 2z = 0$?

Question

What is the projection of the vector $v = [-3, 2, 3]^T$ on the plane with equation $1x + 1y - 2z = 0$?

So my solution is first i use the component vector formula of ${a \cdot b}\over\bigl|a\bigr| $

= $-3 + 2 -6\over\sqrt{9+4+36}$ = $-7\over\sqrt49$

Now the projection formula ${a \cdot b}\over\bigl|a^2\bigr|$$\cdot a$

=$-7\over\sqrt49$$ \cdot <-3,2,3>$

=$-1\over\sqrt7$$ \cdot <-3,2,3>$

=

But the answers are either

a) $1\over6$$ \cdot <-2,12,11>$

b) $1\over3$$ \cdot <-29,56,18>$

c) $1\over3$$ \cdot <-11,4,7>$

d) $1\over6$$ \cdot <-11,19,4>$

I do not really see how i went wrong, and why my answer would differ from the correct one

Answer 1

it is much easier to find the projection of $(-3, 2, 3)^\top$ onto $(1,1,-2)^\top$ first and then subtract that from $(-3,2,3)^\top.$

$(-3,2,3)^\top = k(1,1,-2)^\top + \epsilon$ so that $\epsilon$ is orthogonal to $(1,1,-2)^\top.$ taking the dot product gives you $-3 + 2 -6=k(1+1+4) \to k = -\frac 76$ and $$\begin{align}\epsilon &= (-3,2,3)^\top +\frac 76 (1,1,-2)^\top\\ &=(-11, 19, 4)^\top/6\end{align}$$ is the required projection.

Answer 2

A normal vector of the plane is $(1,1,-2)^t$, it has length $\sqrt{6}$. Thus the part of $v$ in direction of the above as unit normal vector $n$ is: $$ n\cdot v=(1/\sqrt{6})(1,1,-2)^t\cdot(-3,2,3)^t = -7/\sqrt{6} $$ We subtract this from $v$ to have the rest within the plane: $$ (-3,2,3)^t-(-7/\sqrt{6})(1/\sqrt{6})(1,1,-2)^t = (-3,2,3)^t + (7/6,7/6,-7/3)^t = (-11/6,19/6,2/3)^t =(1/6)(-11,19,4)^t $$ Test: $$ (1,1,-2)\cdot(-11,19,4) = 0 $$ That vector is orthogonal to the normal vector, thus within the plane.

In fact that test succeeds only for one answer.

Answer 3

First, the $9+4+36$ in the square root is wrong, you should have used $1+1+4$ instead if you look at the formula more carefully.

Secondly, the formula is $\Large{{a \cdot b}\over\bigl|a\bigr|^2}$ and you forgot a square in the bottom.

Thirdly, the formula gives you the projection to the normal vector and you need to subtract this from $a$ to get the projection to the plane.