Verified Solution • 2025–26 Session moderate

QJEEM23Ma865e: JEE Main 2023 Physics - Class 11 JEEM23Ma865e Solution: The area of the quadrilateral with vertices and is equal to :

The area of the quadrilateral ABCD\mathrm{ABCD} with vertices A(2,1,1),B(1,2,5),C(−2,−3,5)\mathrm{A}(2,1,1), \mathrm{B}(1,2,5), \mathrm{C}(-2,-3,5) and D(1,−6,−7)\mathrm{D}(1,-6,-7) is equal to :

Multiple Choice Options

  • A 48
  • B 8388 \sqrt{38}
  • C 54
  • D 9389 \sqrt{38}

Step-by-Step Verified Solution

amp; Here AC→=(−2−2)i^+(−3−1)j^+(5−1)k^amp;=−4i^−4j^+4k^amp;BD→=(1−1)i^+(−6−2)j^+(−7−5)k^amp;=−8j^−12k^\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 =12∣AC→×BD→∣=\frac{1}{2}| \overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}} \mid

amp;=12∥i^amp;j^amp;k^−4amp;−4amp;40amp;−8amp;−12∥amp;=12∣(48+32)i^−(48−0)j^+(32−0)k^∣amp;=12∣80i^−48j^+32k^∣amp;=1216∣15i^−3j^+2k^∣amp;=825+9+4=838 sq units. \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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