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, $\overrightarrow{AB}= \vec{b}, \, \overrightarrow{CD} =\vec{d} -\vec{c}, \overrightarrow{AD}=\vec{d},$
$\overrightarrow{BC}=\vec{c}-\vec{b}, \overrightarrow{AC}=\vec{c} \, and \, \overrightarrow{BD}=\vec{d} -\vec{b}$
$Now, L.H.S. =\vec{b}. \left(\vec{d}-\vec{c}\right)\, and$
$RHS=k\left[\left|\vec{d}\right|^{2}+\left|\vec{c}-\vec{b}\right|^{2}-\left|\vec{c}\right|^{2}-\left|\vec{d}-\vec{b}\right|^{2}\right]$
$=k \left[\vec{d.}\,\vec{d}+\vec{c}.\, \vec{c}+\vec{b}.-2 \vec{c}.\vec{b}-\vec{c}. \vec{c}-\vec{d}. \vec{d}-\vec{b}.\vec{b}+2 \vec{d}. \vec{b}\right]$
$=2 k\left[\vec{b}. \left(\vec{d}-\vec{c}\right)\right] \Rightarrow \, k=\frac{1}{2}\cdot$