Question

Angle between diagonals of a parallelogram whose side are represented by $\overrightarrow{a}=2\hat{i}+\hat{j}+\hat{k}$ and $\hat{b}=\hat{i}-\hat{j}-\hat{k}$

• A. ${\cos }^{-1}\left(\frac{1}{3}\right)$
• B. ${\cos }^{-1}\left(\frac{3}{5}\right)$
• C. ${\sin }^{-1}\sqrt{\frac{8}{9}}$
• D. ${\tan }^{-1}\left(\frac{4}{3}\right)$

Answer 1

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

Two diagonals of parallelogram are respectively $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$

Angle between them, is

$\cos \theta =\frac{(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})}{|\overrightarrow{a}+\overrightarrow{b}||\overrightarrow{a}-\overrightarrow{b}|}$

As from the given data, we get

$\overrightarrow{a}+\overrightarrow{b}=3\hat{i}+0\hat{j}+0\hat{k},then|\overrightarrow{a}+\overrightarrow{b}|=3$

$\overrightarrow{a}-\overrightarrow{b}=\hat{i}+2\hat{j}+2\hat{k},then|\overrightarrow{a}-\overrightarrow{b}|=3$

and $(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=3$

On putting that value in above formula, we get

$\cos \theta =\frac{3}{3\times 3}=\frac{1}{3}$

So the angle between two diagonals $\theta ={\cos }^{-1}\left(\frac{1}{3}\right)$