If $\vec{a}$ and $\vec{b}$ are two vectors, such that $|\vec{a}| = 1$, $|\vec{b}| = 2$ and $\vec{a} \cdot \vec{b} = \sqrt{3}$, then the angle between $2\vec{a}$ and $-\vec{b}$ is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{5\pi}{6}$ ##
$|2\vec{a}| = |2||\vec{a}| = 2 \times 1 = 2$
$|-\vec{b}| = |-1||\vec{b}| = 1 \times 2 = 2$
$(2\vec{a}) \cdot (-\vec{b}) = 2(-1)(\vec{a} \cdot \vec{b})$
$= -2(\vec{a} \cdot \vec{b}) = -2\sqrt{3}$
$(2\vec{a}) \cdot (-\vec{b}) = |2\vec{a}| |-\vec{b}| \cos \theta$
$-2\sqrt{3} = (2)(2) \cos \theta$
$-2\sqrt{3} = 4 \cos \theta$
$\cos \theta = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}$
Since, $\cos 150^\circ = -\frac{\sqrt{3}}{2}$.
Therefore, $\theta = 150^\circ$ or $\theta = \frac{5\pi}{6}$ radians.