Question

If $A,B,C,D$ be $4$ distinct points in space such that $\overrightarrow{AB}$ is not perpendicular to $\overrightarrow{CD}$ and satisfies $\overrightarrow{AB}.\overrightarrow{CD}=\frac{1}{\lambda }[{\overline{AD}}^2+{\overline{BC}}^2-{\overline{AC}}^2-{\overline{BD}}^2],$ then the value of $\lambda$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Let $O$ be the origin of reference $\overline{a},\overline{b},\overline{c},\overline{d}$ be the position vector of $A,B,C,D$ respectively relative to $O$ .
$\overrightarrow{AB}.\overrightarrow{CD}=(\overline{b}-\overline{a}).(\overline{d}-\overline{c})$
$=\overline{b}.\overline{d}+\overline{c}.\overline{a}-\overline{b}.\overline{c}-\overline{a}.\overline{d}......\left(1\right)$
${\overline{AD}}^2+{\overline{BC}}^2-{\overline{AC}}^2-{\overline{BD}}^2$
$={\overline{d}}^2+{\overline{a}}^2-2\overline{d}.\overline{a}+{\overline{c}}^2+{\overline{b}}^2-2\overline{c}.\overline{b}$
$-{\overline{c}}^2-{\overline{a}}^2+2\overline{c}.\overline{a}-{\overline{d}}^2-{\overline{b}}^2+2\overline{d}.\overline{b}$
$=2[\overline{b}.\overline{d}+\overline{c}.\overline{a}-\overline{b}.\overline{c}-\overline{a}.\overline{d}]$
$=2[\overline{AB}.\overline{CD}]$

Thus, $[\overline{AB}.\overline{CD}]$ $=\frac{1}{2}[{\overline{AD}}^2+{\overline{BC}}^2-{\overline{AC}}^2-{\overline{BD}}^2]$

Hence, $\lambda =2$.