Question

Given that $(\overrightarrow{x}-\hat{a}).(\overrightarrow{x}+\hat{a})=8$ and $\overrightarrow{x}.\hat{a}=2,$ then the angle between $(\overrightarrow{x}-\hat{a})$ and $(\overrightarrow{x}+\hat{a})$ is:

• A. ${\sec }^{-1}\left(\frac{\sqrt{21}}{4}\right)$
• B. ${\cos }^{-1}\left(\frac{3}{\sqrt{21}}\right)$
• C. ${\cos }^{-1}\left(\frac{5}{\sqrt{21}}\right)$
• D. None of these

Answer 1

The correct option is A ${\sec }^{-1}\left(\frac{\sqrt{21}}{4}\right)$

$(\overrightarrow{x}-\hat{a}).(\overrightarrow{x}+\hat{a})=8\Rightarrow {\left|\overrightarrow{x}\right|}^2=9\Rightarrow \left|\overrightarrow{x}\right|=3$

To determine $(\overrightarrow{x}-\hat{a}),$ we have $(\overrightarrow{x}-\hat{a}).(\overrightarrow{x}-\hat{a})$

$={\left|\overrightarrow{x}\right|}^2-2(\overrightarrow{x}.\hat{a})+\hat{a}.\hat{a}=9-4=6\Rightarrow {|\overrightarrow{x}-\hat{a}|}^2=6$

$\Rightarrow \left|(\overrightarrow{x}-\hat{a})\right|=\sqrt{6}$ and similarly, $|\overrightarrow{x}+\hat{a}|=\sqrt{14}$

Now let the angle between $(\overrightarrow{x}-\hat{a})$ and $(\overrightarrow{x}+\hat{a})$ be $\theta .$

Then $(\overrightarrow{x}-\hat{a}).(\overrightarrow{x}+\hat{a})=\sqrt{14}\times \sqrt{6}\cos \theta$

$\Rightarrow 8=\sqrt{14}\times \sqrt{6}\cos \theta \Rightarrow \cos \theta =\frac{4}{\sqrt{21}}$