If $\vec a,\vec b$ are unit vectors such that the vector $\vec a + 3\vec b$ is perpendicular to $7\vec a-5\vec b$ and $\vec a-4\vec b$ is perpendicular to $7\vec a-2\vec b$, then the angle between $\vec a$ and $\vec b$ is
Answer & explanation
Correct answer: option 3
Let θ be the angle between $\vec a$ and $\vec b$.
We have,
$(\vec a+3\vec b) ⊥ (7\vec a-5\vec b)$
$⇒(\vec a+3\vec b).(7\vec a-5\vec b)=0$
$⇒7|\vec a|^2+16 (\vec a.\vec b)-|\vec b|^2=0$
$⇒7+16\cos θ-15=0⇒\cos θ=\frac{1}{2}⇒θ=\frac{π}{3}$
And,
$(\vec a-4\vec b) ⊥ (7\vec a-2\vec b)$
$⇒(\vec a-4\vec b). (7\vec a-2\vec b)=0$
$⇒7|\vec a|^2+8|\vec b|^2-30 (\vec a.\vec b)=0$
$⇒15-30\cos θ=0⇒\cos θ=\frac{1}{2}⇒θ=\frac{π}{3}$