JEE Main 2024 — 30 January, Shift 1 — Question 18

Let the vertices of the parallelogram be $A, B, C, D$ in counterclockwise order. We are given the coordinates of opposite vertices $A(2, 3, 5)$ and $C(-3, 4, -2)$. We are also given the diagonal vector $\vec{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$.

In a parallelogram, the diagonals bisect each other. Let $M$ be the midpoint of both diagonals $AC$ and $BD$.

First, find the vector $\vec{AC}$: $\vec{AC} = C - A = (-3-2)\hat{i} + (4-3)\hat{j} + (-2-5)\hat{k}$ $\vec{AC} = -5\hat{i} + \hat{j} - 7\hat{k}$

The area of a parallelogram can be found using the cross product of its two diagonals. If the diagonals are $\vec{d_1}$ and $\vec{d_2}$, the area of the parallelogram is $\frac{1}{2} |\vec{d_1} \times \vec{d_2}|$. Here, the two diagonals are $\vec{AC}$ and $\vec{BD}$.

Calculate the cross product $\vec{AC} \times \vec{BD}$: $\vec{AC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -7 \\ 1 & 2 & 3 \end{vmatrix}$ $= \hat{i}((1)(3) - (-7)(2)) - \hat{j}((-5)(3) - (-7)(1)) + \hat{k}((-5)(2) - (1)(1))$ $= \hat{i}(3 + 14) - \hat{j}(-15 + 7) + \hat{k}(-10 - 1)$ $= 17\hat{i} - (-8)\hat{j} - 11\hat{k}$ $= 17\hat{i} + 8\hat{j} - 11\hat{k}$

Now, find the magnitude of the cross product: $|\vec{AC} \times \vec{BD}| = \sqrt{(17)^2 + (8)^2 + (-11)^2}$ $= \sqrt{289 + 64 + 121}$ $= \sqrt{474}$

The area of the parallelogram is $\frac{1}{2} |\vec{AC} \times \vec{BD}|$: Area $= \frac{1}{2} \sqrt{474}$

The final answer is $\boxed{\frac{1}{2}\sqrt{474}}$.