Find the vector in w in W closest to v.

Question

Let $W$ be the subspace R_3 with basis $\{(1,1,0), (0,1,-1)\},$ and let $v = (2, 0, -4)$. Find the vector in $w$ in $W$ closest to v.

This is what I've tried:

I obtained an orthogonal basis, where $w_1 = (1, 1, 0)$ and $w_2 = (-1/2, 1/2, -1).$

I then used to definition of a projection = $(v, w_1)w_1 + (v, w_2)w_2$. However, there is no answer corresponding to the vector produced this way.

I know the answer is (0, 2, -2) but I have no idea how to get there.

Edit: I forgot a negative sign in the answer. Sorry about that folks

Answer 1

The mistake is forgetting to divide by the norm squared.

$$\frac{(v, w_1)}{\color{blue}{\|w_1\|^2}}w_1 + \frac{(v, w_2)}{\color{blue}{\|w_2\|^2}}w_2$$

$$\frac{2}{2}(1,1,0)+\frac{3}{\frac32}(-0.5,0.5,-1)=(1,1,0)+2(-0.5,0.5,-1)=(0,2,-2)$$

Note that $(0,2,2)$ is not inside $W$.

$$a(1,1,0)-2(0,1,-1)=(0,2,2)$$

that would mean $a=0$ .

Answer 2

No, the answer is not $(0,2,2)$ (it doesn't even belong to $W$); it's $(0,2,-2)$.

An orthonormal basis of $W$ is $(e_1,e_2)=\left(\frac1{\sqrt2}(1,1,0),\frac1{\sqrt6}(-1,1,-2)\right)$. So, the answer is$$\bigl\langle(2,0,-4),e_1\bigr\rangle e_1+\bigl\langle(2,0,-4),e_2\bigr\rangle e_2=(0,2,-2). $$