$\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ are any 4 points and $|\overline{A B} \times \overline{C…
$\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ are any 4 points and
$|\overline{A B} \times \overline{C D}+\overline{B C} \times \overline{A D}+\overline{C A} \times \overline{B D}|=\lambda$
(Area of $\triangle A B C$ ) then $\lambda=$
$1$
$2$
$3$
$4$
Solution
Given four points A, B, C and D with position vectors $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{d}}$ respectively.
Now $|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{BC}} \times \overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{CA}} \times \overrightarrow{\mathrm{BD}}|=\lambda$ Area of $\varnothing$ $\mathrm{ABC}$.
$\begin{aligned} & \Rightarrow|(\vec{b}-\vec{a}) \times(\vec{a}-\vec{c})+(\vec{c}-\vec{b}) \times(\vec{d}-\vec{a})+(\vec{a}-\vec{c}) \times(\vec{d}-\vec{b})| \\ & \frac{\lambda}{2}|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}| \\ & 2|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|=\frac{\lambda}{2}|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|\end{aligned}$
Compare above equation,
$\begin{gathered}2=\frac{\lambda}{2} \\ \lambda=4\end{gathered}$