Question

The shortest distance of diagonal of unit cube and edge skew to it is

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

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}$

A$\equiv (1,1,1)$

$A\equiv (0,0,0)$

$B\equiv (1,0,1)$

$C\equiv (1,0,0)$

Mid-points of $OA\equiv D$

$\frac{(\frac{0+1}{2},\frac{0+1}{2},\frac{0+1}{2})}{(\frac{1}{2},\frac{1}{2},-\frac{1}{2})}$

Mid- points of $BC\equiv E$

$\equiv (\frac{1+1}{2},\frac{0+0}{2},\frac{1+0}{2})$

$=(1,0,\frac{1}{2})$

$\begin{array}{cc}\text{Hence,}& \\ \text{Distance}& =\sqrt{{(1-\frac{1}{2})}^2+{(0-\frac{1}{2})}^2+{(\frac{1}{2}-\frac{1}{2})}^2}\\ & =\frac{1}{\sqrt{2}}\end{array}$