# Find and angle between two vectors

$$\def\vec#1{{\overrightarrow #1}}$$ Well I have an practice problem, though a little bit more complicated that it seems from the title. Here it is

If $$(2\vec{a} -\vec{b}) \perp (\vec{a}+\vec{b})$$ and $$(\vec{a} - 3\vec{b}) \perp (2\vec{a}+ \vec{b})$$, then what is the angle between $$\vec{a}$$ and $$\vec{b}$$?

I've used the fact that if $$\vec{x} \perp \vec{y} \Rightarrow \vec{x}\cdot\vec{y} = 0$$ to derive the relation $$\cos\angle(\vec{a},\vec{b}) = -\frac{1}{3}\frac{\left|\vec{b}\right|}{\left|\vec{a}\right|}$$. Basically I calculated the scalar product for both pairs of vectors given above, then solved for the angle. I am pretty sure that this is correct, though I have no idea if it is the complete answer. Nothing else is given as a known value, and I can't think of any way to get anything else from the given data. Can someone help with this? Is my solution complete? If not, what else is there to do?

• You have three unknowns: $|\vec a|$, $|\vec b|$, and $\vec a\cdot\vec b$. At most you have two equations involving the three unknowns, so at best you have a parametric solution. Aug 19, 2021 at 13:50

If you expand the brackets you can eliminate $$\underline{a}\cdot\underline{b}$$ and get $$a^2=\frac23b^2$$
Then $$\cos\theta=\frac{\underline{a}\cdot\underline{b}}{|a||b|}=\frac{b^2-2a^2}{|a||b|}=\frac{b^2-\frac43b^2}{\sqrt{\frac23}b^2}=-\frac{1}{\sqrt{6}}$$