JEE Main 2024 — 6 April, Shift 1 — Question 2

Given the vertices of the quadrilateral $ABCD$:

$$A(3, 1, -1), \quad B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), \quad C(2, 2, 1), \quad D\left(\frac{10}{3}, \frac{2}{3}, -\frac{1}{3}\right)$$

The area of any quadrilateral can be computed using half the magnitude of the vector cross product of its diagonals:

$$\text{Area} = \frac{1}{2} |\vec{AC} \times \vec{BD}|$$

Determine the Diagonal Vectors}

Let us compute the components of the diagonal vectors $\vec{AC}$ and $\vec{BD}$:

$$\vec{AC} = (2 - 3)\hat{i} + (2 - 1)\hat{j} + (1 - (-1))\hat{k} = -\hat{i} + \hat{j} + 2\hat{k}$$ $$\vec{BD} = \left(\frac{10}{3} - \frac{5}{3}\right)\hat{i} + \left(\frac{2}{3} - \frac{7}{3}\right)\hat{j} + \left(-\frac{1}{3} - \frac{1}{3}\right)\hat{k}$$ $$\vec{BD} = \frac{5}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{2}{3}\hat{k} = \frac{1}{3}(5\hat{i} - 5\hat{j} - 2\hat{k})$$

Calculate the Cross Product $\vec{AC} \times \vec{BD}$

Using the matrix determinant method for the vector cross product:

$$\vec{AC} \times \vec{BD} = \frac{1}{3} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ 5 & -5 & -2 \end{vmatrix}$$

Expanding along the first row:

$$\vec{AC} \times \vec{BD} = \frac{1}{3} \left[ \hat{i}\big(1(-2) - 2(-5)\big) - \hat{j}\big((-1)(-2) - 2(5)\big) + \hat{k}\big((-1)(-5) - 1(5)\big) \right]$$ $$\vec{AC} \times \vec{BD} = \frac{1}{3} \left[ \hat{i}(-2 + 10) - \hat{j}(2 - 10) + \hat{k}(5 - 5) \right]$$ $$\vec{AC} \times \vec{BD} = \frac{1}{3} (8\hat{i} + 8\hat{j} + 0\hat{k}) = \frac{8}{3}\hat{i} + \frac{8}{3}\hat{j}$$

Compute the Magnitude and Final Area

Now, find the magnitude of the resulting vector:

$$|\vec{AC} \times \vec{BD}| = \sqrt{\left(\frac{8}{3}\right)^2 + \left(\frac{8}{3}\right)^2 + 0^2}$$ $$|\vec{AC} \times \vec{BD}| = \sqrt{\frac{64}{9} + \frac{64}{9}} = \sqrt{\frac{128}{9}} = \frac{8\sqrt{2}}{3}$$

Substituting this value back into our structural area formula:

$$\text{Area} = \frac{1}{2} \cdot \frac{8\sqrt{2}}{3} = \frac{4\sqrt{2}}{3}$$

Final Answer The area of the quadrilateral $ABCD$ is $\frac{4\sqrt{2}}{3}$ square units.