Question

Angle between vectors $a$ and $b$ where $a,b\text{and}c$ are unit vectors satisfying $a+b+\sqrt{3}c=0$ is

Answer 1

Step 1: Calculate the value of $\cos \theta$.

Understand that here vectors e $a,b\text{and}c$ are unit vectors satisfying $a+b+\sqrt{3}c=0$.

Therefore,

$\begin{array}{c}a+b+\sqrt{3}c=0\\ a+b=\sqrt{3}c\end{array}$

Square both sides.

$\begin{array}{c}{\left|a\right|}^2+{\left|b\right|}^2+2\left|a\right|\left|b\right|\cos \theta ={\left(\sqrt{3}\right)}^2{\left|c\right|}^2\\ 1+1+2\times 1\times 1\times \cos \theta =3\times 1\\ 1+1+2\cos \theta =3\\ 2+2\cos \theta =3\\ 2\cos \theta =3-2\\ 2\cos \theta =1\\ \cos \theta =\frac{1}{2}\end{array}$

Step 2:Calculate the value of $\theta$.

Take cos inverse on both sides.

Therefore,

$\begin{array}{c}{\cos }^{-1}\left(\cos \theta \right)={\cos }^{-1}\left(\frac{1}{2}\right)\\ \theta ={\cos }^{-1}\left(\cos \frac{\pi }{3}\right)\\ \theta =\frac{\pi }{3}\end{array}$

Hence,angle between vectors $a$ and $b$ where $a,b\text{and}c$ are unit vectors satisfying $a+b+\sqrt{3}c=0$ is $\frac{\pi }{3}$.