Question

Given vectors $3\vec{a} - 5\vec{b}$ and $2\vec{a} + \vec{b}$ are mutually perpendicular, and so are $\vec{a} + 4\vec{b}$ and $-\vec{a} + \vec{b}$. Find the acute angle between $\vec{a}$ and $\vec{b}$:

• A. $\cos^{-1}\left(\dfrac{19}{5\sqrt{43}}\right)$
• B. $\cos^{-1}\left(\dfrac{9}{5\sqrt{43}}\right)$
• C. $\pi - \cos^{-1}\left(\dfrac{19}{5\sqrt{43}}\right)$
• D. $\pi - \cos^{-1}\left(\dfrac{9}{5\sqrt{43}}\right)$

Hint:

Dot product of perpendicular vectors is zero: apply this to set up equations.

Answer 1

The Correct Option is A

From the dot product condition: \[ (3\vec{a} - 5\vec{b}) \cdot (2\vec{a} + \vec{b}) = 0 \Rightarrow 6a^2 - 13(\vec{a} \cdot \vec{b}) - 5b^2 = 0 \] Use both equations and solve to find $\cos\theta = \dfrac{19}{5\sqrt{43}}$