JEE MAIN - Mathematics (2023 - 8th April Evening Shift - No. 8)
The area of the quadrilateral $$\mathrm{ABCD}$$ with vertices $$\mathrm{A}(2,1,1), \mathrm{B}(1,2,5), \mathrm{C}(-2,-3,5)$$ and $$\mathrm{D}(1,-6,-7)$$ is equal to :
48
$$8 \sqrt{38}$$
54
$$9 \sqrt{38}$$
Explanation
$$
\begin{aligned}
& \text { Here } \overrightarrow{\mathrm{AC}}=(-2-2) \hat{i}+(-3-1) \hat{j}+(5-1) \hat{k} \\\\
& =-4 \hat{i}-4 \hat{j}+4 \hat{k} \\\\
& \overrightarrow{\mathrm{BD}}=(1-1) \hat{i}+(-6-2) \hat{j}+(-7-5) \hat{k} \\\\
& =-8 \hat{j}-12 \hat{k}
\end{aligned}
$$
So, area of quadrilateral $=\frac{1}{2}| \overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}} \mid$
$$ \begin{aligned} & =\frac{1}{2}\left\|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -4 & -4 & 4 \\ 0 & -8 & -12 \end{array}\right\| \\\\ & =\frac{1}{2}|(48+32) \hat{i}-(48-0) \hat{j}+(32-0) \hat{k}| \\\\ & =\frac{1}{2}|80 \hat{i}-48 \hat{j}+32 \hat{k}| \\\\ & =\frac{1}{2} 16|15 \hat{i}-3 \hat{j}+2 \hat{k}| \\\\ & =8 \sqrt{25+9+4}=8 \sqrt{38} \text { sq units. } \end{aligned} $$
So, area of quadrilateral $=\frac{1}{2}| \overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}} \mid$
$$ \begin{aligned} & =\frac{1}{2}\left\|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -4 & -4 & 4 \\ 0 & -8 & -12 \end{array}\right\| \\\\ & =\frac{1}{2}|(48+32) \hat{i}-(48-0) \hat{j}+(32-0) \hat{k}| \\\\ & =\frac{1}{2}|80 \hat{i}-48 \hat{j}+32 \hat{k}| \\\\ & =\frac{1}{2} 16|15 \hat{i}-3 \hat{j}+2 \hat{k}| \\\\ & =8 \sqrt{25+9+4}=8 \sqrt{38} \text { sq units. } \end{aligned} $$
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