Question

Let $A,B,C,D$ are any four points in space. If $|\overrightarrow{AB}\times \overrightarrow{CD}+\overrightarrow{BC}\times \overrightarrow{AD}+\overrightarrow{CA}\times \overrightarrow{BD}|=\lambda \times$(Area of $△ABC$), then the value of $\lambda$ is:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be the position vectors of $A,B,C.$
area of $\Delta ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$
$\therefore \text{area of}\Delta ABC=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}|\cdots \left(i\right)$

Now,$|\overrightarrow{AB}\times \overrightarrow{CD}+\overrightarrow{BC}\times \overrightarrow{AD}+\overrightarrow{CA}\times \overrightarrow{BD}|$
Let $\overrightarrow{D}$ be the origin of reference.
So, $\overrightarrow{DA}=\overrightarrow{a},\overrightarrow{DB}=\overrightarrow{b},\overrightarrow{DC}=\overrightarrow{a}$
$=\left|\right(\overrightarrow{b}-\overrightarrow{a})\times (-\overrightarrow{c})+(\overrightarrow{c}-\overrightarrow{b})\times (-\overrightarrow{a})+(\overrightarrow{a}-\overrightarrow{c})\times (-\overrightarrow{b}\left)\right|$
$=2|\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}|$
$=2\left(2\text{area of}\Delta ABC\right)$
$=4$ (Area of triangle $ABC$ )