A, B, C and D are any four points in the space. If $|\vec{AB} \times \vec{CD}+\vec{BC} \times \vec{AD}+\vec{CA} \times \vec{BD}|=\lambda \Delta_{ABC}$, where $\Delta_{ABC}$ is the area of triangle ABC, then $\lambda$ is equal to:
Answer & explanation
Correct answer: option 3
Let P.V. of A, B, C and D be $\vec{a}, \vec{b}, \vec{c}$ and $\vec{0}$
$\Rightarrow \vec{A B} \times \vec{C D}=(\vec{b}-\vec{a}) \times-\vec{c}$,
$\vec{B C} \times \vec{A D}=(\vec{c}-\vec{b}) \times-\vec{a}$
and $\vec{C A} \times \vec{B D}=(\vec{a}-\vec{c}) \times-\vec{b}$
$\Rightarrow \vec{A B} \times \vec{C D}+\vec{B C} \times \vec{A D}+\vec{C A} \times \vec{B D}$
$=\vec{c} \times \vec{b}+\vec{a} \times \vec{c}+\vec{a} \times \vec{c}+\vec{b} \times \vec{a}-\vec{a} \times \vec{b}+\vec{c} \times \vec{b}$
$=2(\vec{c} \times \vec{b}+\vec{b} \times \vec{a}+\vec{a} \times \vec{c})$
$=2(\vec{c} \times(\vec{b}-\vec{a})-\vec{a} \times(\vec{b}-\vec{a}))$
$=2((\vec{c}-\vec{a}) \times(\vec{b}-\vec{a}))$
$=2(\vec{A C} \times \vec{A B})$
$\Rightarrow|\vec{A B} \times \vec{C D}+\vec{B C} \times \vec{A D}+\vec{C A} \times \vec{B D}|$
$=4\left|\frac{1}{2}(\vec{A C} \times \vec{A B})\right|$
$=4 \Delta_{A B C}$
Hence (3) is correct answer.