Determine the foot of the perpendicular from the points $Q(1,2,3,4)$ on the plane generated by $A(1,1,0,0), B(1,0,1,0), C(0,1,1,1)$
Let me define $W=span\{(1,1,0,0), (1,0,1,0), (0,1,1,1)\}$
Then I want to calculate Projection of $Q$ on $W$ $P_W(Q)$, which will give me the foot, where
$P_W(Q) = \frac{\langle Q,A \rangle}{||A||^2}A+\frac{\langle Q,B \rangle}{||B||^2}B+\frac{\langle Q,C \rangle}{||C||^2}C$
Is this approach fine?
CALCULATIONS :
$||A|| = \sqrt{2}, ||B||=\sqrt{3}, ||C||=\sqrt{3}$
$\langle Q,A \rangle= 3, \langle Q,B \rangle= 4, \langle Q,C \rangle=9$
So my final answer is $(\frac72,\frac92,5,3)$ but this answer doesn't match the one given in my notes.
So can someone please very the answer with me?
Your approach was a good idea, but you overlooked the fact that in order to compute the orthogonal projection onto a span of vectors by adding up the individual projections, those vectors must be orthogonal (see this answer for an illustration of how this fails in $\mathbb R^2$). So, as you noted in the comments, if you first apply Gram-Schmidt to the spanning vectors to get an orthonormal basis, you can then use this method.
There’s another way to attack this problem that’s less work, however. Observe that the orthogonal projection $\mathbf\pi_W\mathbf v$ of $\mathbf v$ onto $W$ is whatever’s left over after removing the part of $\mathbf v$ that’s in $W^\perp$ (its orthogonal rejection from $W$). Here, $W^\perp$ is only one-dimensional, so if you can come up with a basis for it, you only need to compute the projection onto a single vector and subtract instead of going through the whole G-S process and then computing three projections. In this case, it’s not too hard to work out that $(1,-1,-1,2)$ is orthogonal to all three vectors that generate $W$, so spans $W^\perp$. Even if you can’t spot this by inspection, I think that row-reducing a $3\times4$ matrix that consists of zeros and ones in order to find its null space is still less work than the other approach.