Question

If $\bar{a}$ and $\bar{b}$ are two vectors such that $|\bar{a}|=|\bar{b}|=\sqrt{2}$ with $\bar{a}.\bar{b}=-1$, then the angle between $\bar{a}$ and $\bar{b}$ is

• A. $\frac{2\pi}{3}$
• B. $\frac{5\pi}{6}$
• C. $\frac{5\pi}{9}$
• D. $\frac{3\pi}{4}$

Answer 1

Answer: (A)

$\frac{2\pi}{3}$

Answer 2

The angle $\theta$ between two vectors $\bar{a}$ and $\bar{b}$ is determined by the formula for the dot product: $\bar{a} \cdot \bar{b} = |\bar{a}| |\bar{b}| \cos \theta$.

Given $|\bar{a}|=\sqrt{2}$, $|\bar{b}|=\sqrt{2}$, and $\bar{a}.\bar{b}=-1$.

Substitute the given values into the formula:

$-1 = (\sqrt{2})(\sqrt{2}) \cos \theta$

$-1 = 2 \cos \theta$

$\cos \theta = -\frac{1}{2}$

The angle $\theta$ between two vectors is conventionally taken to be in the range $[0, \pi]$. In this range, the value of $\theta$ for which $\cos \theta = -\frac{1}{2}$ is $\theta = \frac{2\pi}{3}$.