Determine the orthogonal projection of the vectors $(1,0,0)^t$ to $W$ let  W  be the  subspace of $\mathbb{R}^3$  spanned  by the vectors   $(1,1,0)^t$ and $(0,1,1)^t$  .Determine the  orthogonal projection  of the vectors  $(1,0,0)^t$  to $W$ 
My attempt  : i know  one  formula that Projection of $f(t)$ along $g(t)$ = $\frac{\langle f,g \rangle }{\langle g,g \rangle}.g$
But here  im confused that how  can  i applied this formula on  the given  question
Any hints/solution will be aprrecaited
thanks u
 A: Geometrically, $W=\langle (1,1,0), (0,1,1) \rangle$ is a two-dimensional subspace of $\mathbb{R}^3$, i.e. it's a plane in our physical space. If we find an orthogonal line to this plane, every vector in our space can then be written as a linear combination of some vector in $W$ and some vector along the orthogonal line.
Let's find a vector that is orthogonal to both of these vectors (i.e. orthogonal to the plane spanned by $W$):
$$(a,b,c) \cdot (1,1,0)^t = a+b = 0$$
$$(a,b,c) \cdot (0,1,1)^t = b+c = 0$$
The solution to the system above is $(a,-a,a)$. So, the orthogonal complement of $W$ is spanned by $(1,-1,1)$, i.e. $$W^{\perp}=\langle(1,-1,1)\rangle$$
Now that you have $\mathbb{R}^3 = W \oplus W^{\perp}$, you can write any vector as a linear combination of the vectors $(1,1,0),(0,1,1),(1,-1,1)$. Now you can write $(1,0,0)$ as such a linear combination and then ignore the coefficient of the third basis vector $(1,-1,1)$.
Edit
Since what we have found is a basis, we can find $\alpha, \beta, \gamma$ such that
$$(1,0,0) = \alpha (1,1,0) + \beta (0,1,1) + \gamma (1,-1,1)$$
This gives a linear system of equations
$$\alpha + \gamma = 1$$
$$\alpha + \beta - \gamma = 0$$
$$\beta + \gamma = 0$$
This system has only one solution, obviously. I won't solve it here and I'll leave that to you, but the final solution to the system is $$(\alpha,\beta,\gamma)=(\frac{2}{3},-\frac{1}{3},\frac{1}{3})$$
Now, the orthogonal projection on $W$ should have no component that's orthogonal to it. Therefore $$\vec{p}=\frac{2}{3}(1,1,0) - \frac{1}{3}(0,1,1)=(\frac{2}{3},\frac{1}{3}, -\frac{1}{3})$$
Alternatively, you could use your own projection formulas to find $\alpha$ and $\beta$ without solving the linear system. But I believe thinking about the problem geometrically is useful.
A: There are a couple of ways to apply the formula that you know for orthogonal projection onto a single vector to this problem.  
You can compute the projections onto each of the vectors that generate $W$ and add them up, but that will only work if those vectors are an orthogonal basis of $W$. (This is essentially what you’re doing when you compute the coordinates of a vector.) The two given vectors are not orthogonal, so you’ll have to do something to convert them into an orthogonal basis. The usual procedure for doing this is the Gram-Schmidt process, which you’ve probably learned already. In this case, we can leave $(1,1,0)^T$ alone and apply the process to the second vector, giving $$(0,1,1)^T-{\langle(0,1,1)^T,(1,1,0)^T\rangle \over \langle(1,1,0)^T,(1,1,0)^T\rangle}(1,1,0)^T=\left(-\frac12,\frac12,1\right)^T.$$ Since you’re working in $\mathbb R^3$, you can instead use cross products to generate a vector in $W$ that’s orthogonal to $(1,1,0)^T$: $((1,1,0)^T\times(0,1,1)^T)\times(1,1,0)^T = (-1,1,2)$, which is a scalar multiple of the vector produced by G-S above, as it should be.  
There’s a way that’s potentially much less work, though. For any vector, if you subtract its orthogonal projection onto $W$ from it, what’s left is orthogonal to $W$. So as a general principle, you can compute an orthogonal projection onto a subspace $W$ by subtracting the projection onto the orthogonal complement $W^\perp$ from the vector. In this case, $W$ is two-dimensional, so $W^\perp$ is one-dimensional: you can reduce this problem to a single application of the projection formula if you can find any nonzero vector that’s orthogonal to $W$. As before, since you’re working in $\mathbb R^3$, you can use a cross product to do this: $(1,1,0)^T\times(0,1,1)^T = (1,-1,1)^T$. Project $(1,0,0)^T$ onto this vector and subtract the result from $(1,0,0)^T$ to find its projection onto $W$.
