$Find$ $||v||$ given that: $\widehat uv$ = π/4, $||u||$ = 3, $(u-v)$ $\bot$ $u$
Here, I know that $<uv>$ = $||u|| * ||v|| * cos$ $\widehat uv$
So I know that $<uv>$ = $3 * ||v|| * cos π/4$
and thus that $<uv>$ = $\left(\frac{\sqrt 2}{2}\right) ||v||$
But I'm having trouble using my last bit of data to finish the problem.
Similarly, I have another exercise:
$Find$ $||v||$ given that: $\widehat uv$ = π/4, $||u||$ = 3, $\widehat (u+v)u = π/6$ (having a formatting issue with the angle)
So I already have some information from the previous exercise, namely that $<uv>$ = $\left(\frac{\sqrt 2}{2}\right) ||v||$
I also know that $<(u+v)u>$ = $||u+v|| * 3 * cos π/6$
Thus I know that $<uu> + <vu>$ = $||u+v||$ * $\left(\frac{3\sqrt 3}{2}\right)$
And that $||u||^2 + <vu>$ = $||u+v||$ * $\left(\frac{3\sqrt 3}{2}\right)$
9 $+<vu>$ = $||u+v||$ * $\left(\frac{3\sqrt 3}{2}\right)$
I don't really know how to proceed from here. Someone who was trying to help me showed me that $\widehat (u+v)v = π/4 - π/6 $
But I don't know how to use that information to find an answer.
Thanks, SE!
Guide:
if you have
$$\|u\|^2+\langle v, u \rangle =\|u + v \| \cdot \left( \frac{3\sqrt3}2\right)\tag{1}$$
and
$$\langle u, v\rangle=\left(\frac{\sqrt2}2 \right)\|v\|\tag{2}.$$
We can square $(1)$,
$$\|u\|^4+2\|u\|^2\langle u, v\rangle+\langle u, v \rangle^2= \left(\|u \|^2+2\langle u,v \rangle + \|v\|^2 \right)\cdot \left( \frac{3\sqrt3}2\right)\tag{3}$$
You can substituing $(2)$ into $(3)$, and note that since we know the value of $\|u \|$, it become a quadratic equation in $\|v\|$.