If ( a+3 b)(7 a-5 b) and ( a-4 b)(7 a-2 b), then the angle between a and b (in degrees) is…

Question

If $(\vec a+3\vec b)\perp(7\vec a-5\vec b)$ and $(\vec a-4\vec b)\perp(7\vec a-2\vec b)$, then the angle between $\vec a$ and $\vec b$ (in degrees) is \_\_\_\_\_\_\_\_.

Answer: 60

Accepted Answer

$(\vec a+3\vec b)\cdot(7\vec a-5\vec b)=0$: $7|\vec a|^2-15|\vec b|^2+16\vec a\cdot\vec b=0$ \ldots  (1)

$(\vec a-4\vec b)\cdot(7\vec a-2\vec b)=0$: $7|\vec a|^2+8|\vec b|^2-30\vec a\cdot\vec b=0$ \ldots  (2)

Solving (1) and (2) simultaneously gives $|\vec a|=|\vec b|$ and $\cos	heta = \dfrac{\vec a\cdot\vec b}{|\vec a|^2}=\dfrac12$, so $	heta=60^{\circ}$.

Solutions · 1