Question

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two non-zero vectors perpendicular to each other and $|\overrightarrow{a}|=|\overrightarrow{b}|.$ If $|\overrightarrow{a}\times \overrightarrow{b}|=|\overrightarrow{a}|$, then the angle between the vectors $(\overrightarrow{a}+\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b}))$ and $\overrightarrow{a}$ is equal to:

• A. ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• B. ${\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• C. ${\sin }^{-1}\left(\frac{1}{\sqrt{6}}\right)$
• D. ${\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Given : $|\overrightarrow{a}\times \overrightarrow{b}|=|\overrightarrow{a}|=|\overrightarrow{b}|$
Let $|\overrightarrow{a}|=a$
Now,
$\overrightarrow{a}\cdot (\overrightarrow{a}+\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b}))=a^2(\because \overrightarrow{a}\cdot \overrightarrow{b}=0,\overrightarrow{a}\cdot (\overrightarrow{a}\times \overrightarrow{b})=0)\Rightarrow a|\overrightarrow{a}+\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b})|\cos \theta =a^2\Rightarrow \cos \theta =\frac{a}{|\overrightarrow{a}+\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b})|}\cdots \left(1\right)$

Now,
${|\overrightarrow{a}+\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b})|}^2=(\overrightarrow{a}+\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b}))\cdot (\overrightarrow{a}+\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b}))=(a^2+0+0)+(0+a^2+0)+(0+0+a^2)=3a^2$
Therefore,
$\cos \theta =\frac{a}{a\sqrt{3}}\therefore \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$