Question

The vectors $\overline{AB}=3\hat{i}-2\hat{j}+2\hat{k}$ and $\overline{BC}=-\hat{i}+2\hat{k}$ are the adjacent sides of a parallelogram ABCD then the angle between the diagonals is

• A. ${\cos }^{-1}\left(\sqrt{\frac{1}{85}}\right)$
• B. $\pi -{\cos }^{-1}\left(\sqrt{\frac{49}{85}}\right)$
• C. ${\cos }^{-1}\left(\frac{1}{2\sqrt{2}}\right)$
• D. ${\cos }^{-1}\left(\sqrt{\frac{3}{10}}\right)$

Answer 1

The correct option is D ${\cos }^{-1}\left(\sqrt{\frac{3}{10}}\right)$

$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=2\hat{i}-2\hat{j}+4\hat{k}$
$\overrightarrow{BD}=\overrightarrow{-AB}+\overrightarrow{BC}=-4\hat{i}+2\hat{j}$
Let Angle between $\overrightarrow{AC}$ & $\overrightarrow{BC}$ is $\theta \therefore \frac{\overrightarrow{AC}.\overrightarrow{BC}}{\left|\overrightarrow{AC}\right|\left|\overrightarrow{BD}\right|}=\cos \theta \Rightarrow \cos \theta =\frac{-12}{4\sqrt{6}\sqrt{5}}=-\sqrt{\frac{3}{10}}.\Rightarrow$
Acute angle between diagonals = ${\cos }^{-1}\sqrt{\frac{3}{10}}$