Let A(2,3,5) and C(-3,4,-2) be opposite vertices of a parallelogram A B C D . If the diagonal \overrightarrowBD=\hati+2 \hatj+3 \hatk , then the area of the parallelogram is equal to :

3D Geometry PYQ — JEE Main 2024 · 30 Jan · Shift 1 · Q44 with Solution

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Correct answer: D

Use the diagonals of the parallelogram

For parallelogram $ABCD$ , the opposite vertices are $A$ and $C$ , so one diagonal is

\overrightarrow{AC}=C-A=(-3-2)\hat i+(4-3)\hat j+(-2-5)\hat k.

Thus,

\overrightarrow{AC}=-5\hat i+\hat j-7\hat k.

The other diagonal is given:

\overrightarrow{BD}=\hat i+2\hat j+3\hat k.

Area of a parallelogram in terms of diagonals

If the diagonals are $\vec d_1$ and $\vec d_2$ , then area of the parallelogram is

\text{Area}=\frac12\left|\vec d_1\times \vec d_2\right|.

So here,

\text{Area}=\frac12\left|\overrightarrow{AC}\times\overrightarrow{BD}\right|.

Compute the cross product

\overrightarrow{AC}\times\overrightarrow{BD} = \begin{vmatrix} \hat i & \hat j & \hat k\\ -5 & 1 & -7\\ 1 & 2 & 3 \end{vmatrix}.

Expanding:

=\hat i\,(1\cdot 3-(-7)\cdot 2) -\hat j\,((-5)\cdot 3-(-7)\cdot 1) +\hat k\,((-5)\cdot 2-1\cdot 1).

=\hat i(3+14)-\hat j(-15+7)+\hat k(-10-1).

=17\hat i+8\hat j-11\hat k.

Find its magnitude

\left|\overrightarrow{AC}\times\overrightarrow{BD}\right| =\sqrt{17^2+8^2+(-11)^2} =\sqrt{289+64+121} =\sqrt{474}.

Hence the area

\text{Area}=\frac12\sqrt{474}.

Match with the options

This corresponds to Option D.

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