Question

If A, B, C, D are four points in space satisfying
$\overrightarrow{AB}.\overrightarrow{CD}=k\left[{\left|\overrightarrow{AD}\right|}^2+{\left|\overrightarrow{BC}\right|}^2-{\left|\overrightarrow{AC}\right|}^2-{\left|\overrightarrow{BD}\right|}^2\right],$
then the value of k is

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

Answer 1

Answer: (C)

Let $\vec{b},\vec{c},\vec{d}$ be the position vectors of B, C, D w.r.t.
A as origin. So, $\overrightarrow{AB}=\vec{b},\overrightarrow{CD}=\vec{d}-\vec{c},\overrightarrow{AD}=\vec{d},$
$\overrightarrow{BC}=\vec{c}-\vec{b},\overrightarrow{AC}=\vec{c}and\overrightarrow{BD}=\vec{d}-\vec{b}$
$Now,L.H.S.=\vec{b}.\left(\vec{d}-\vec{c}\right)and$
$RHS=k\left[{\left|\vec{d}\right|}^2+{\left|\vec{c}-\vec{b}\right|}^2-{\left|\vec{c}\right|}^2-{\left|\vec{d}-\vec{b}\right|}^2\right]$
$=k\left[\overrightarrow{d.}\vec{d}+\vec{c}.\vec{c}+\vec{b}.-2\vec{c}.\vec{b}-\vec{c}.\vec{c}-\vec{d}.\vec{d}-\vec{b}.\vec{b}+2\vec{d}.\vec{b}\right]$
$=2k\left[\vec{b}.\left(\vec{d}-\vec{c}\right)\right]\Rightarrow k=\frac{1}{2}\cdot$