Question

If $\overrightarrow{e_1}=(1,1,1)$ and $\overrightarrow{e_2}=(1,1,-1)$ and $\overrightarrow{a}$ and $\overrightarrow{b}$ are two vectors such that $\overrightarrow{e_1}=2\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{e_2}=\overrightarrow{a}+2\overrightarrow{b}$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

• A. ${\cos }^{-1}\left(\frac{7}{9}\right)$
• B. ${\cos }^{-1}\left(\frac{7}{11}\right)$
• C. ${\cos }^{-1}(-\frac{7}{11})$
• D. ${\cos }^{-1}\left(\frac{6\sqrt{2}}{11}\right)$

Answer 1

Answer: (C)

${\cos }^{-1}(-\frac{7}{11})$

Solving $\overrightarrow{e_1}=2\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{e_2}=\overrightarrow{a}+2\overrightarrow{b}$ , we get $\overrightarrow{a}=\frac{2\overrightarrow{e_1}-\overrightarrow{e_2}}{3}$ , $\overrightarrow{b}=\frac{2\overrightarrow{e_2}-\overrightarrow{e_1}}{3}$

So, $\overrightarrow{a}=(\frac{1}{3},\frac{1}{3},1)$ and $\overrightarrow{b}=(\frac{1}{3},\frac{1}{3},-1)$

Now, $\overrightarrow{a}.\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|cos\theta =\frac{-7}{9}$

$\Rightarrow \frac{11}{9}cos\theta =\frac{-7}{9}$

$\Rightarrow cos\theta =\frac{-7}{11}$

So, Option (C)