Inner Product Space Puzzle

Question

Question: Let $V$ be a real inner product space, and let $\mathbf{u}, \mathbf{v} \in V$ be such that: $$\left\lVert\mathbf{u}\right\rVert = 3, \qquad \left\lVert\mathbf{u}+\mathbf{v}\right\rVert =4, \qquad \left\lVert\mathbf{u}-\mathbf{v}\right\rVert = 6.$$ Find $\left\lVert\mathbf{v}\right\rVert$.

I don't know exactly why, but I am kinda stuck on this. My main attempts have involved using $\left\lVert\mathbf{u}\right\rVert = \sqrt{\left\langle\mathbf{u},\mathbf{u}\right\rangle}$, followed by a bunch of algebraic manipulation and the application of vector space rules in an attempt to isolate $\left\langle\mathbf{v}, \mathbf{v}\right\rangle$.

Any hints/tips would be appreciated.

Answer 1

We know $$ \langle u + v, u + v \rangle = 16\quad\text{and}\quad \langle u - v, u - v\rangle = 36 $$ Therefore, $$ \langle u + v, u + v \rangle + \langle u - v, u - v \rangle = 2\langle u, u \rangle + 2 \langle v, v\rangle = 52 $$ Now since $\langle u, u \rangle = 9$, we have $$ \langle v, v\rangle = 17 $$

Answer 2

In every inner product space we have:

$||u+v||^2+||u-v||^2=2||u||^2+2||v||^2$

From

$\left\lVert\mathbf{u}\right\rVert = 3, \qquad \left\lVert\mathbf{u}+\mathbf{v}\right\rVert =4, \qquad \left\lVert\mathbf{u}-\mathbf{v}\right\rVert = 6$

we get $||v||^2=17$