Showing Orthogonality How would I do this question..... I'm familiar with Gram-Schmidt and the basics but I have no idea how to do $a$ and $b$ in this question.

Suppose $\{\vec x_1, \vec x_2, \vec x_3\}$ is an orthonormal set of vectors.
a) Show that $\|\vec x_1+\vec x_2+\vec x_3\| = \sqrt{3}$.
b) Suppose that a vector $\vec y$ is orthogonal to each of the vectors $\vec x_1, \vec x_2, \vec x_3$.  Show that $\vec y$ is also orthogonal to $66\vec x_1 - 17\vec x_2 + \vec x_3$.

 A: Problem A:
By definition of the norm:
$$\|x_2 + x_2 + x_3\| = \sqrt{\langle x_2 + x_2 + x_3, x_2 + x_2 + x_3\rangle}$$
So, we compute the inner product:
$$\begin{align}
\langle x_2 + x_2 + x_3, x_2 + x_2 + x_3\rangle &= 2(\langle x_2, x_2 \rangle + \langle x_2, x_2 \rangle + \langle x_2, x_3 \rangle)\\& +\langle x_3, x_2 \rangle + \langle x_3, x_2 \rangle + \langle x_3, x_3 \rangle\\
&= 2(1 + 1 + 0) + 0 + 0 + 1\\
&= 5
\end{align}$$
Note that the inner product of two identical members (e.g. $\langle x_2, x_2 \rangle$) is $1$ due to the "normal" part of the orthonormal collection.  Also note that the inner product of two different members (e.g. $\langle x_3, x_2 \rangle$ is $0$ due to the "orthogonal" part of the orthonormal collection.
Our result above implies that the requested norm is $\sqrt{5}$.  Note that this is different than our desired answer.  This is because the problem has a typo; it should request $\|x_1 + x_2 + x_3\|$, not $\|x_2 + x_2 + x_3\|$
Problem B:  I'm just giving a hint here.  Take the inner product of $y$ and each of the vectors, and show that it is $0$ for all three.  Recall that the inner product of two normal vectors is $0$.
A: Recall that a subset $\{v_1,\dotsc,v_n\}$ of an inner product space $V$ is orthonormal if
$$
\langle v_j,v_k\rangle
=
\begin{cases}
1 & j=k \\
0 & j\neq k
\end{cases}\tag{1}
$$
Also recall that the norm of a vector $v\in V$ is defined as
$$
\Vert v\Vert=\sqrt{\langle v,v\rangle}\tag{2}
$$
In this problem we are given that $\{ x_1, x_2, x_3\}$ is an orthonormal subset of an inner product space. To compute $\Vert x_1+ x_2+ x_3\Vert$ we use the equations (1) and (2) above
\begin{align*}
\Vert x_1+ x_2+ x_3\Vert^2
&= \langle  x_1+ x_2+ x_3, x_1+ x_2+ x_3\rangle \\
&= \langle x_1, x_1+ x_2+ x_3\rangle+\langle  x_2, x_1+ x_2+ x_3\rangle+\langle x_3, x_1+ x_2+ x_3\rangle \\
&= \langle x_1 , x_1\rangle
+\langle x_1 , x_2\rangle
+\langle x_1 , x_3\rangle
+\langle x_2 , x_1\rangle
+\langle x_2 , x_2\rangle\\
&\quad
+\langle x_2 , x_3\rangle
+\langle x_3, x_1\rangle
+\langle x_3 , x_2\rangle
+\langle x_3 , x_3\rangle \\
&= 1+0+0+0+1+0+0+0+1 \\ 
&= 3
\end{align*}
This proves the first part of your question. Can you use similar methods to prove the second part?
A: These two questions both concern the bilinearity of the inner product operator.  This is just a fancy way of saying that the inner product is linear in both arguments.
(Actually, if you have a complex space, the inner product is usually sesquilinear, but that is a matter for another time.)
The other important matter is the relationship between the norm and the inner product: $$\langle \mathbf{u},\mathbf{u}\rangle =\Vert \mathbf{u}\Vert^{2}.$$
So, where does this get us? 
Well, suppose that $\mathbf{u}$ and $\mathbf{v}$ are two vectors. Then we get:
\begin{eqnarray}
\Vert \mathbf{u}+\mathbf{v}\Vert^{2} & = & \langle \mathbf{u}+\mathbf{v}, \mathbf{u}+\mathbf{v} \rangle \\ 
& = & \langle \mathbf{u}, \mathbf{u}+\mathbf{v} \rangle + \langle \mathbf{v}, \mathbf{u}+\mathbf{v} \rangle \\
& = & \langle \mathbf{u}, \mathbf{u} \rangle + \langle \mathbf{v}, \mathbf{u} \rangle + \langle \mathbf{u}, \mathbf{v} \rangle + \langle \mathbf{v}, \mathbf{v} \rangle.
\end{eqnarray}
If $\mathbf{u}$ and $\mathbf{v}$ are orthogonal, then terms like $\langle\mathbf{u},\mathbf{v}\rangle$ (the cross terms) are all zero.  You are left with:
$$
\Vert \mathbf{u}+\mathbf{v}\Vert^{2} = \langle \mathbf{u}, \mathbf{u} \rangle + \langle \mathbf{v}, \mathbf{v} \rangle = \Vert \mathbf{u}\Vert^{2}+\Vert \mathbf{v}\Vert^{2}.$$
(Note that this is Pythagoras' Theorem.)
You should be able to generalize this argument to answer your first question.
The second question can be resolved similarly, by using 
$$\langle \mathbf{y},66\mathbf{x}_{1}-17\mathbf{x}_{2}+\mathbf{x}_{3}\rangle=66\langle \mathbf{y},\mathbf{x}_{1}\rangle-17\langle \mathbf{y},\mathbf{x}_{2}\rangle+\langle \mathbf{y},\mathbf{x}_{3}\rangle.$$  
However, you should be aware that you are actually proving a special case of a much more general result: if $B$ is a set of vectors, and $\mathbf{y}$ is orthogonal to every element of $B$, then $\mathbf{y}$ is in the orthogonal complement of $\textrm{Span}(B)$.  In other words, $\mathbf{y}$ is orthogonal to every liner combination of elements of $B$.
