Given the vectors a and b such that a+b and 2a-b are perpendicular and a-b and 4a+b are perpendicular, find the angle between a and b

Question

$(a+b)\cdot(2a-b)=0$

$(a-b)\cdot(4a+b)=0 $

$2a\cdot a + a \cdot b - b \cdot b = 0$

$4a\cdot a -3 a \cdot b - b \cdot b = 0$

$b \cdot b=2a\cdot a+a \cdot b$ then I found

$a\cdot a=2a\cdot b$

$b \cdot b=5a\cdot b$

Then I dont know how to continue to find an angle.

Can someone help. Appreciate that

Answer 1

$0 = (\vec a + \vec b)\cdot(2\vec a - \vec b) = 2\vec a\cdot\vec a-\vec b\cdot\vec b +\vec a\cdot\vec b = 2a^2 - b^2 + ab\cos(\theta)$

$0 = (\vec a - \vec b)\cdot(4\vec a + \vec b) = 4\vec a\cdot\vec a-\vec b\cdot\vec b -3\vec a\cdot\vec b = 4a^2 - b^2 - 3ab\cos(\theta)$

Multiply the first equation by $3$, then add the second equation. It follows that $10 a^2 = 4b^2$, hence $b = \frac {1}{2}\sqrt{10} \space a$. Substitute this result into one of the two equations and you get $\cos(\theta) = \frac {1}{10}\sqrt {10}$.