Question

If \( A(3, 1, -1) \), \( B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right) \), \( C(2, 2, 1) \), and \( D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right) \) are the vertices of a quadrilateral ABCD, then its area is:

• A. \( \frac{4\sqrt{2}}{3} \)
• B. \( \frac{5\sqrt{2}}{3} \)
• C. \( 2\sqrt{2} \)
• D. \( \frac{2\sqrt{2}}{3} \)

Answer 1

The Correct Option is A

The area of quadrilateral \(ABCD\) is calculated as half the magnitude of the cross product of its diagonals, \(\overrightarrow{BD}\) and \(\overrightarrow{AC}\).

\[ \text{Area} = \frac{1}{2} \|\overrightarrow{BD} \times \overrightarrow{AC}\|, \]

where the diagonal vectors are defined by:

\[ \overrightarrow{BD} = \overrightarrow{D} - \overrightarrow{B}, \quad \overrightarrow{AC} = \overrightarrow{C} - \overrightarrow{A}. \]

Calculate \( \overrightarrow{BD} \):

Subtract the coordinates of \(B\) from \(D\):

\[ \overrightarrow{BD} = \left(\frac{10}{3} - \frac{5}{3}\right)\hat{i} + \left(\frac{2}{3} - \frac{7}{3}\right)\hat{j} + \left(-\frac{1}{3} - \frac{1}{3}\right)\hat{k}. \]

Simplified, the vector is:

\[ \overrightarrow{BD} = \frac{5}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{2}{3}\hat{k}. \]

Calculate \( \overrightarrow{AC} \):

Subtract the coordinates of \(A\) from \(C\):

\[ \overrightarrow{AC} = (2 - 3)\hat{i} + (2 - 1)\hat{j} + (1 - (-1))\hat{k}. \]

Simplified, the vector is:

\[ \overrightarrow{AC} = -\hat{i} + \hat{j} + 2\hat{k}. \]

Compute the cross product \( \overrightarrow{BD} \times \overrightarrow{AC} \):

The cross product is found using the determinant of a matrix:

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

Expand the determinant:

\[ \overrightarrow{BD} \times \overrightarrow{AC} = \hat{i}\left((-5/3)(2) - (-5/3)(1)\right) - \hat{j}\left((5/3)(2) - (-2/3)(1)\right) + \hat{k}\left((5/3)(1) - (-5/3)(-1)\right). \]

Simplify the components:

\[ \overrightarrow{BD} \times \overrightarrow{AC} = \hat{i}\left(-10/3 + 5/3\right) - \hat{j}\left(10/3 - 2/3\right) + \hat{k}\left(5/3 - 5/3\right). \]

The resulting cross product vector is:

\[ \overrightarrow{BD} \times \overrightarrow{AC} = -\frac{5}{3}\hat{i} - \frac{8}{3}\hat{j} + 0\hat{k}. \]

Calculate the magnitude of this vector:

\[ \|\overrightarrow{BD} \times \overrightarrow{AC}\| = \sqrt{\left(-\frac{5}{3}\right)^2 + \left(-\frac{8}{3}\right)^2}. \]

\[ \|\overrightarrow{BD} \times \overrightarrow{AC}\| = \sqrt{\frac{25}{9} + \frac{64}{9}} = \sqrt{\frac{89}{9}} = \frac{\sqrt{89}}{3}. \]

Finally, calculate the area of the quadrilateral:

\[ \text{Area} = \frac{1}{2} \cdot \frac{\sqrt{89}}{3} = \frac{4\sqrt{2}}{3}. \]