Question

If $\hat{a},\hat{b}$ and $\hat{c}$ are unit vectors satisfying $\hat{a}-\sqrt{3}\hat{c}+\hat{c}=\overline{0}$ then the angle between the vectors $\hat{a}$ and $\hat{c}$ is:

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

Answer 1

The correct option is B $\frac{\pi }{3}$

Given that

$\hat{a}-\sqrt{3}\hat{b}+\hat{c}=0$

$\Rightarrow \hat{a}+\hat{c}=\sqrt{3}\hat{c}$

Squaring on both sides

$\Rightarrow (\hat{a}+\hat{c}{)}^2=(\sqrt{3}\hat{c}{)}^2$

$\Rightarrow {\hat{a}}^2+{\hat{c}}^2+2(\hat{a}.\hat{c})=(\sqrt{3}\hat{c}{)}^2$

$\Rightarrow {\hat{a}}^2+{\hat{c}}^2+2(\hat{a}.\hat{c})=3{\hat{c}}^2$

Given that $\hat{a},\hat{b},\hat{c}$ are unit vectors

Therefore, $\Rightarrow 1^2+1^2+2(\hat{a}.\hat{c})=3\left(1^2\right)$

$\Rightarrow 2+2\left(|a||c|cos\right(\hat{a},\hat{c}\left)\right)=3$

$\Rightarrow 2(1\times 1cos(\hat{a},\hat{c}\left)\right)=1$

$\Rightarrow cos(\hat{a},\hat{c})=\frac{1}{2}$

$\Rightarrow cos(\hat{a},\hat{c})=cos\left({60}^0\right)$

Therefore angle between $\hat{a},\hat{c}={60}^0=\frac{\pi }{3}$