Question

The perpendicular distance of a corner of unit cube from a diagonal not passing through it is

• A. $\sqrt{\frac{2}{3}}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $1$

Answer 1

The correct option is A $\sqrt{\frac{2}{3}}$

Let P.V. of
$\overline{A}=\overline{0}$
$\overline{B}=\overline{b}$
$\overline{C}=\overline{c}+\overline{b}$
$\overline{D}=\overline{c}$
$\overline{G}=\overline{d}$
$\overline{E}=\overline{c}+\overline{b}+\overline{a}$
$\overline{AE}$ is the main diagonal$=\overline{c}+\overline{b}+\overline{d}$
& $\overline{c}$ be a Corner$=\overline{c}+\overline{b}$
So line equation of $AC=\overline{o}+t(\overline{c}+\overline{b}+\overline{d})$
So, $\perp$ distance $=\frac{|(\overline{c}+\overline{b})\times (\overline{c}+\overline{b})+\overline{d}|}{|\overline{c}+\overline{b}+\overline{d}|}$
$=\frac{|(\overline{c}+\overline{b})\times \overline{d}|}{\sqrt{3}}$
$=\frac{|\overline{c}+\overline{b}||\overline{d}|\sin \theta }{\sqrt{3}}$
$=\frac{\sqrt{2}}{3}$