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3D Geometry question

2024 · 6 Apr · Shift 1 · Q35
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3D Geometry question

2024 · 6 Apr · Shift 1 · Q35

JEE MainMathematics3D GeometryMCQ+4 / −1
If A(3,1,−1),B(53,73,13),C(2,2,1)A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1)A(3,1,−1),B(35​,37​,31​),C(2,2,1) and D(103,23,−13)D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)D(310​,32​,3−1​) are the vertices of a quadrilateral ABCDA B C DABCD, then its area is
  1. A
    423\frac{4 \sqrt{2}}{3}342​​
  2. B
    223\frac{2 \sqrt{2}}{3}322​​
  3. C
    523\frac{5 \sqrt{2}}{3}352​​
  4. D
    222 \sqrt{2}22​
View written solutionFree

Correct answer: A

  1. Find the side vectors

Given A(3,1,−1),B(53,73,13),C(2,2,1),D(103,23,−13).A(3,1,-1),\quad B\left(\frac53,\frac73,\frac13\right),\quad C(2,2,1),\quad D\left(\frac{10}3,\frac23,-\frac13\right).A(3,1,−1),B(35​,37​,31​),C(2,2,1),D(310​,32​,−31​).

Let us compute the consecutive side vectors:

AB→=B−A=(53−3,73−1,13−(−1))=(−43,43,43).\overrightarrow{AB}=B-A=\left(\frac53-3,\frac73-1,\frac13-(-1)\right)=\left(-\frac43,\frac43,\frac43\right).AB=B−A=(35​−3,37​−1,31​−(−1))=(−34​,34​,34​).

BC→=C−B=(2−53,2−73,1−13)=(13,−13,23).\overrightarrow{BC}=C-B=\left(2-\frac53,2-\frac73,1-\frac13\right)=\left(\frac13,-\frac13,\frac23\right).BC=C−B=(2−35​,2−37​,1−31​)=(31​,−31​,32​).

CD→=D−C=(103−2,23−2,−13−1)=(43,−43,−43)=−AB→.\overrightarrow{CD}=D-C=\left(\frac{10}3-2,\frac23-2,-\frac13-1\right)=\left(\frac43,-\frac43,-\frac43\right)=-\overrightarrow{AB}.CD=D−C=(310​−2,32​−2,−31​−1)=(34​,−34​,−34​)=−AB.

DA→=A−D=(3−103,1−23,−1−(−13))=(−13,13,−23)=−BC→.\overrightarrow{DA}=A-D=\left(3-\frac{10}3,1-\frac23,-1-\left(-\frac13\right)\right)=\left(-\frac13,\frac13,-\frac23\right)=-\overrightarrow{BC}.DA=A−D=(3−310​,1−32​,−1−(−31​))=(−31​,31​,−32​)=−BC.

So opposite sides are equal and parallel. Hence ABCDABCDABCD is a parallelogram.


  1. Area of parallelogram

Area of parallelogram ABCDABCDABCD is Area=∣AB→×BC→∣.\text{Area}=\left|\overrightarrow{AB}\times\overrightarrow{BC}\right|.Area=​AB×BC​.

Now,

\qquad \overrightarrow{BC}=\left(\frac13,-\frac13,\frac23\right).$$ Compute the cross product: $$\overrightarrow{AB}\times\overrightarrow{BC} =\begin{vmatrix} \hat i & \hat j & \hat k \\ -\frac43 & \frac43 & \frac43 \\ \frac13 & -\frac13 & \frac23 \end{vmatrix}.$$ $$=\hat i\left(\frac43\cdot\frac23-\frac43\cdot\left(-\frac13\right)\right) -\hat j\left(-\frac43\cdot\frac23-\frac43\cdot\frac13\right) +\hat k\left(-\frac43\cdot\left(-\frac13\right)-\frac43\cdot\frac13\right).$$ $$=\hat i\left(\frac89+\frac49\right)-\hat j\left(-\frac89-\frac49\right)+\hat k\left(\frac49-\frac49\right).$$ $$=\left(\frac43,\frac43,0\right).$$ Therefore, $$\text{Area}=\sqrt{\left(\frac43\right)^2+\left(\frac43\right)^2} =\frac43\sqrt{2} =\frac{4\sqrt{2}}{3}.$$ --- 3. **Match with options** Thus the area of quadrilateral $ABCD$ is $$\boxed{\frac{4\sqrt{2}}{3}}.$$ So the correct option is **A**. --- 4. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** Hence, they agree.
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