Given X=(4,8,4,-8)\in R^4 and S = span\{(1,2,1,0),(-1,1,-1,1), (1,0,1,2)\} is a subspace,
Find the orthogonal projections of X onto both S and S^\perp.

Now, I've already found S^\perp = span\{(-6,-3,4,1)\} so I can decompose X into Y\in S and Z\in S^\perp s.t.
X = Y + Z.

Let Z = a(-6,-3,4,1), a in R.

So, (4,8,4,-8)=Y+a(-6,-3,4,1)

(I chose to use Z instead of Y since dim(Z)=1, but dim(Y)=3, so fewer inner-products.)

Dotting both sides with (-6,-3,4,1), we get:

(4,8,4,-8)\cdot(-6,-3,4,1)=a\|(-6,-3,4,1)\|^2 since Y\cdot(-6,-3,4,1) = 0.

So a = \frac{-24}{31} and Z = \frac{-24}{31}(-6,-3,4,1).

Y = X - Z so
Y = (-14,-1,16,-5)

Hope that's right!