Practicing Success
Given figure is cuboid then the acute angle between the diagonal OA and BE is |
$\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ $\cos ^{-1}\left(\frac{1}{6}\right)$ $\cos ^{-1}\left(\frac{2}{3}\right)$ $\cos ^{-1}\left(\frac{1}{3}\right)$ |
$\cos ^{-1}\left(\frac{1}{3}\right)$ |
$\overrightarrow{O A}=a \hat{i}+a \hat{j}+a \hat{k} ~~~~|\overrightarrow{OA}|=\sqrt{a^2+a^2+a^2}=\sqrt{3} a$ $\overrightarrow{B E}=a \hat{i}-a \hat{j}-a \hat{A} ~~~~|\overrightarrow{B E}|=\sqrt{3}a$ So $|\overrightarrow{O A} . \overrightarrow{B E}|=|\overrightarrow{O A}||\overrightarrow{B E}| \cos \theta$ $\Rightarrow \cos \theta=\left|\frac{a^2-a^2-a^2}{a \sqrt{3} \times a \sqrt{3}}\right| =\frac{1}{3}$ so $\theta=\cos ^{-1} \frac{1}{3}$ Option: D |