Key Concepts and Formulas

• Vector between Two Points: Given two points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$, the vector from $P_1$ to $P_2$ is given by $\overrightarrow{P_1P_2} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$.
• Cross Product of Two Vectors: For two vectors $\vec{u} = u_x\hat{i} + u_y\hat{j} + u_z\hat{k}$ and $\vec{v} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}$, their cross product is given by the determinant: $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{vmatrix} = (u_y v_z - u_z v_y)\hat{i} - (u_x v_z - u_z v_x)\hat{j} + (u_x v_y - u_y v_x)\hat{k}$
• Magnitude of a Vector: For a vector $\vec{w} = w_x\hat{i} + w_y\hat{j} + w_z\hat{k}$, its magnitude is $|\vec{w}| = \sqrt{w_x^2 + w_y^2 + w_z^2}$.
• Area of a Quadrilateral in 3D Space: The area of a quadrilateral with diagonals represented by vectors $\vec{d_1}$ and $\vec{d_2}$ is given by: $\text{Area} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}|$ This formula is derived from the fact that the magnitude of the cross product of two vectors gives the area of the parallelogram formed by them. The area of any general quadrilateral is half the area of the parallelogram formed by its diagonals.

Step-by-Step Solution

We are given the vertices of quadrilateral $ABCD$ as $A(3,1,-1)$, $B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right)$, $C(2,2,1)$, and $D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$. Our goal is to find its area using the diagonal method.

Step 1: Identify the Diagonals of the Quadrilateral For a quadrilateral $ABCD$, the diagonals are $\overrightarrow{AC}$ and $\overrightarrow{BD}$. To use the area formula, we first need to express these diagonals as vectors.

Step 2: Calculate the Vectors Representing the Diagonals

We will use the formula for a vector between two points, $\overrightarrow{P_1P_2} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$.

• Calculate Diagonal Vector $\overrightarrow{AC}$: Points are $A(3,1,-1)$ and $C(2,2,1)$. $\overrightarrow{AC} = (2-3)\hat{i} + (2-1)\hat{j} + (1-(-1))\hat{k}$ $\overrightarrow{AC} = (-1)\hat{i} + (1)\hat{j} + (2)\hat{k}$ $\overrightarrow{AC} = -\hat{i} + \hat{j} + 2\hat{k}$

• Calculate Diagonal Vector $\overrightarrow{BD}$: Points are $B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right)$ and $D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$. $\overrightarrow{BD} = \left(\frac{10}{3} - \frac{5}{3}\right)\hat{i} + \left(\frac{2}{3} - \frac{7}{3}\right)\hat{j} + \left(\frac{-1}{3} - \frac{1}{3}\right)\hat{k}$ $\overrightarrow{BD} = \left(\frac{5}{3}\right)\hat{i} + \left(\frac{-5}{3}\right)\hat{j} + \left(\frac{-2}{3}\right)\hat{k}$ $\overrightarrow{BD} = \frac{5}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{2}{3}\hat{k}$

Step 3: Compute the Cross Product of the Diagonal Vectors

Now, we calculate the cross product $\overrightarrow{AC} \times \overrightarrow{BD}$. Let $\vec{d_1} = \overrightarrow{AC} = -\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{d_2} = \overrightarrow{BD} = \frac{5}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{2}{3}\hat{k}$. $\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ \frac{5}{3} & -\frac{5}{3} & -\frac{2}{3} \end{vmatrix}$ Expanding the determinant:

• $\hat{i}$ component: $(1)\left(-\frac{2}{3}\right) - (2)\left(-\frac{5}{3}\right) = -\frac{2}{3} + \frac{10}{3} = \frac{8}{3}$
• $\hat{j}$ component: (Remember the negative sign for the $\hat{j}$ term in the expansion) $- \left( (-1)\left(-\frac{2}{3}\right) - (2)\left(\frac{5}{3}\right) \right) = - \left( \frac{2}{3} - \frac{10}{3} \right) = - \left( -\frac{8}{3} \right) = \frac{8}{3}$
• $\hat{k}$ component: $(-1)\left(-\frac{5}{3}\right) - (1)\left(\frac{5}{3}\right) = \frac{5}{3} - \frac{5}{3} = 0$ Combining these components, we get the cross product vector: $\overrightarrow{AC} \times \overrightarrow{BD} = \frac{8}{3}\hat{i} + \frac{8}{3}\hat{j} + 0\hat{k} = \frac{8}{3}\hat{i} + \frac{8}{3}\hat{j}$

Step 4: Determine the Magnitude of the Cross Product

Next, we calculate the magnitude of the resulting cross product vector, $|\overrightarrow{AC} \times \overrightarrow{BD}|$. $|\overrightarrow{AC} \times \overrightarrow{BD}| = \sqrt{\left(\frac{8}{3}\right)^2 + \left(\frac{8}{3}\right)^2 + (0)^2}$ $|\overrightarrow{AC} \times \overrightarrow{BD}| = \sqrt{\frac{64}{9} + \frac{64}{9}}$ $|\overrightarrow{AC} \times \overrightarrow{BD}| = \sqrt{\frac{128}{9}}$ To simplify the radical: $|\overrightarrow{AC} \times \overrightarrow{BD}| = \frac{\sqrt{128}}{\sqrt{9}} = \frac{\sqrt{64 \times 2}}{3} = \frac{8\sqrt{2}}{3}$

Step 5: Calculate the Area of the Quadrilateral

Finally, we apply the area formula for a quadrilateral: $\text{Area} = \frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{BD}|$ Substitute the magnitude calculated in Step 4: $\text{Area} = \frac{1}{2} \times \frac{8\sqrt{2}}{3}$ $\text{Area} = \frac{4\sqrt{2}}{3}$

Common Mistakes & Tips

• Order of Subtraction for Vectors: Always subtract the coordinates of the initial point from the coordinates of the terminal point when forming a vector (e.g., $C-A$ for $\overrightarrow{AC}$). Reversing the order will result in a vector with opposite direction, which is fine for the magnitude, but consistency helps avoid errors.
• Fractional Arithmetic: Be meticulous with addition, subtraction, and multiplication involving fractions to prevent calculation errors.
• Cross Product Determinant Expansion: Pay close attention to the signs when expanding the determinant, especially the negative sign for the $\hat{j}$ component.
• Forgetting the $\frac{1}{2}$ Factor: The cross product magnitude gives the area of the parallelogram formed by the diagonals. The area of the quadrilateral is half of this value. Always include the $\frac{1}{2}$ in the final area calculation.

Summary

To find the area of the quadrilateral $ABCD$, we first determined the vectors representing its diagonals, $\overrightarrow{AC}$ and $\overrightarrow{BD}$. Then, we calculated the cross product of these diagonal vectors. The magnitude of this cross product vector was found to be $\frac{8\sqrt{2}}{3}$. Finally, by taking half of this magnitude, we obtained the area of the quadrilateral. The area is $\frac{4\sqrt{2}}{3}$ square units.

The final answer is $\boxed{\text{A}}$ which corresponds to option (A).