How to find the norm of this vector and the angles between him and other?

Question

I am stuck on this exercise, so I would appreciate some help. Here is the information given : $$ \|\vec{a}\|= \sqrt2$$ $$ \|\vec{b}\|=\|\vec{d}\|=1$$ $$ \vec{a}\|\vec{d}$$ $$ \vec{c}= \vec{a}+\vec{b}$$ $$\alpha=\angle( \vec{a}, \vec{b})=\frac\pi4 $$

What is the norm of the vector $\vec{c}$ ? And what is the angle $\theta = \angle(\vec{c},\vec{d})$?

Thank you already !

What did I do: $$cos(\frac\pi4)=ab/||a|| ||b||$$ $$ab=\sqrt2$$

$$ ||c||^2 = ||a||^2 + 2*ab + ||b||^2$$ $$ ||c||^2 = 4 + 2*\sqrt2 + 1^2$$ $$ ||c||^2 = 5 + 2*\sqrt2$$ $$ ||c|| = \sqrt{ 5 + 2*\sqrt2}$$

Answer 1

$\vec a\cdot\vec b$ is in fact equal to $\cos \frac{\pi}{4}\|a\|\|b\|$, that is $\frac{1}{\sqrt{2}}\sqrt{2}=1$.

Then $\|a\|^2$ is $\sqrt{2}^2$, that is $2$. Therefore you get $\|c\|=\sqrt{2+2+1}$ as expected.