Question

ABCD is a square of unit side. It is folded along the diagonal AC, so that the plane ABC is perpendicular to the plane ACD. The shortest distance between the lines AB and CD is

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

Answer 1

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

Let A(1,0,0), B(1,1,0), C(0,1,0) D(0,0,0) be the vertices of square. In new position, $B=(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}})$
Equ of CD is $\overline{r}=s\left(\hat{j}\right)$
Equ of AB is $\overline{r}=\hat{I}+t\left(\overline{AB}\right)$
Apply S.D. $=\frac{(\overrightarrow{a}-\overrightarrow{c}).(\overrightarrow{b}\times \overrightarrow{d})}{|\overrightarrow{b}\times \overrightarrow{d}|}$