Question

A square ABCD of diagonal 2a is folded along the diagonal 2a so that the planes DAC and BAC are at right angle. The shortest distance between DC and AB is
[Kurukshetra CEE 1998]

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

When folded co - ordinates will be
D(0,0,a); C(a,0,0);
A(-a,0,0); B(0,-a,0)

Equation DC is, $\frac{x}{a}=\frac{y}{0}=\frac{z-a}{-a}$
Equation AB is, $\frac{x+a}{a}=\frac{y}{-a}=\frac{z}{0}$
$\therefore$ Shortest distance $=\frac{2a}{\sqrt{3}}$, (By formula).