A) The vectors $\vec{a}$ and $\vec{b}$ makes a $45°$ angle, and the module of $\vec{a}$ is 3. Determine the module of $\vec{b}$ so that the vector $\vec{a} - \vec{b}$ is perpendicular to $\vec{a}$.
Well, I tried using the dot product, because $$\langle \vec{a}; \vec{b} \rangle = ||\vec{a}||.||\vec{b}||.cos(45°)$$ $$\langle \vec{a}; \vec{b} \rangle = 3.\frac{\sqrt {2}}{2} ||\vec{b}||$$
And then I also know that: $$\langle \vec{a-b}; \vec{a} \rangle =0$$
But then I don't know what to do with that... I mean, I don't know what to do with the information given but the things that I've written.
All you need to do is evaluate the scalar product further: $$ \langle \vec{a} - \vec{b}, \vec{a} \rangle = \langle \vec{a} , \vec{a} \rangle - \langle \vec{b}, \vec{a} \rangle = ||\vec{a}||^2 - \frac{3 \sqrt{2}}{2} ||\vec{b}|| = 0$$ and solve for $||\vec{b}||$.