QJEEM24Mab5c1: JEE Main 2024 Physics - Class 11 JEEM24Mab5c1 Solution

Area of quadrilateral $A B C D=$ area of $\triangle A B C+$ area of $\triangle A D C$

In $\triangle A B C$

$\begin{aligned}& \overrightarrow{A B}=4,8,-8 \\& \overrightarrow{B C}=-2,-3,-4 \\& \overrightarrow{A B} \times \overrightarrow{B C}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\4 & 8 & -8 \\-2 & -3 & -4\end{array}\right| \\& =-56 \hat{i}+32 \hat{j}+4 \hat{k}\end{aligned}$

Area of $\triangle A B C=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B C}|$

$\begin{aligned}& =\frac{1}{2} \sqrt{56^2+32^2+4^2} \\& =\frac{1}{2} \sqrt{4176}=\frac{12 \sqrt{29}}{2}=6 \sqrt{29}\end{aligned}$

In $\triangle A D C=\overrightarrow{A D}=-2,-3,-4$

$\overrightarrow{D C}=4,8,-8$

$\overrightarrow{A D} \times \overrightarrow{D C}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -2 & -3 & -4 \\ 4 & 8 & -8\end{array}\right|$

$=56 \hat{i}-32 \hat{j}-4 k$

$\text { Area of } \frac{1}{2}|\overrightarrow{A D} \times \overrightarrow{D C}|$

$\begin{aligned}& =\frac{1}{2} \sqrt{4176} \\& =6 \sqrt{29}\end{aligned}$

Area of $A B C D=6 \sqrt{29}+6 \sqrt{29}$

$=12 \sqrt{29}$