Let $a = <3, -4>$. Find all vectors $b=<b_{1},b_{2}>$ such that the angle between a and b is $\arccos(\frac{2}{\sqrt5})$ and $||\mathrm{proj}_{a}b||=2$.
My Try:
First of all we know that: $$\theta = \arccos\left(\dfrac{2}{\sqrt5}\right)$$
And also from the formula:
$$\cos(\theta)=\dfrac{a \cdot b}{||a|| ||b||}$$
So we have the inequality:
$$\dfrac{a\cdot b}{||a||||b||}=\dfrac{2}{sqrt5}$$
And if we substitute the a and b we get:
$$\dfrac{3b_{1}-4b_{2}}{5\sqrt{b_{1}^2+b_{2}^2}} = \dfrac{2}{\sqrt5}$$
And also from the magnitude of projection we will get:
$$\dfrac{3b_{1}-4b_{2}}{\sqrt{b_{1}^2+b_{2}^2}} = 2$$
But I couldn't go further. Thanks!