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 given by:

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

where:

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

Calculate \( \overrightarrow{BD} \):

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

Simplify:

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

Calculate \( \overrightarrow{AC} \):

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

Simplify:

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

Now, compute \( \overrightarrow{BD} \times \overrightarrow{AC} \):

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

Expand:

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

Simplify:

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

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

The magnitude is:

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

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

The area is:

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

Answer 2

The problem asks for the area of the quadrilateral formed by the points \( A(3, 1, -1) \), \( B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right) \), \( C(2, 2, 1) \), and \( D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right) \). To solve this, we can calculate the area of the quadrilateral using the vector cross product method. We can treat this quadrilateral as composed of two triangles and sum their areas.

The vectors representing these triangles can be determined from the given points. For simplicity, let's find the area of triangles \( \triangle ABC \) and \( \triangle ACD \) and then sum these areas.

1. Calculate vectors for \( \triangle ABC \):
   • \( \overrightarrow{AB} = B - A = \left(\frac{5}{3} - 3, \frac{7}{3} - 1, \frac{1}{3} + 1\right) = \left(-\frac{4}{3}, \frac{4}{3}, \frac{4}{3}\right) \)
   • \( \overrightarrow{AC} = C - A = (2 - 3, 2 - 1, 1 + 1) = (-1, 1, 2) \)
2. Calculate the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \):
   • \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\frac{4}{3} & \frac{4}{3} & \frac{4}{3} \\ -1 & 1 & 2 \end{vmatrix} = \mathbf{i} \left(\frac{4}{3} \times 2 - \frac{4}{3} \times 1\right) - \mathbf{j} \left(-\frac{4}{3} \times 2 - \frac{4}{3} \times (-1)\right) + \mathbf{k} \left(-\frac{4}{3} \times 1 - (-1) \times \frac{4}{3}\right)\)
   • \( = \mathbf{i} \left(\frac{8}{3} - \frac{4}{3} \right) - \mathbf{j} \left(-\frac{8}{3} + \frac{4}{3}\right) + \mathbf{k} \left(-\frac{4}{3} + \frac{4}{3}\right) \)
   • \( = \mathbf{i} \left(\frac{4}{3}\right) + \mathbf{j} \left(\frac{4}{3}\right) + \mathbf{k} \left(0\right)\)
3. Magnitude of the cross product is:
   • \( \left|\overrightarrow{AB} \times \overrightarrow{AC}\right| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2 } = \sqrt{\frac{16}{9} + \frac{16}{9}} = \sqrt{\frac{32}{9}} = \frac{4\sqrt{2}}{3} \)
4. Area of \( \triangle ABC \) is:
   • \( \frac{1}{2} \times \left|\overrightarrow{AB} \times \overrightarrow{AC}\right| = \frac{1}{2} \times \frac{4\sqrt{2}}{3} = \frac{2\sqrt{2}}{3} \)

Repeat similar steps to find the area of \( \triangle ACD \):

1. Calculate vectors for \( \triangle ACD \):
   • \( \overrightarrow{AD} = D - A = \left(\frac{10}{3} - 3, \frac{2}{3} - 1, \frac{-1}{3} + 1\right) = \left(\frac{1}{3}, -\frac{1}{3}, \frac{2}{3}\right) \)
2. Calculate the cross product \( \overrightarrow{AC} \times \overrightarrow{AD} \):
   • \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & 2 \\ \frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \end{vmatrix} = \mathbf{i} \left(1\cdot\frac{2}{3} - 2\cdot(-\frac{1}{3})\right) - \mathbf{j} \left(-1\cdot\frac{2}{3} - 2\cdot\frac{1}{3}\right) + \mathbf{k} \left(-1 \cdot (-\frac{1}{3}) - 1\cdot\frac{1}{3}\right)\)
   • \( = \mathbf{i} \left(\frac{2}{3} + \frac{2}{3}\right) - \mathbf{j} \left(-\frac{2}{3} - \frac{2}{3}\right) + \mathbf{k} \left(\frac{1}{3} - \frac{1}{3}\right) \)
   • \( = \mathbf{i} \left(\frac{4}{3}\right) + \mathbf{j} \left(\frac{4}{3}\right) + \mathbf{k} \left(0\right)\)
3. Magnitude of the cross product is:
   • \( \left|\overrightarrow{AC} \times \overrightarrow{AD}\right| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2 } = \sqrt{\frac{16}{9} + \frac{16}{9}} = \sqrt{\frac{32}{9}} = \frac{4\sqrt{2}}{3} \)
4. Area of \( \triangle ACD \) is:
   • \( \frac{1}{2} \times \left|\overrightarrow{AC} \times \overrightarrow{AD}\right| = \frac{1}{2} \times \frac{4\sqrt{2}}{3} = \frac{2\sqrt{2}}{3} \)

The total area of quadrilateral \( ABCD \) is the sum of the areas of \( \triangle ABC \) and \( \triangle ACD \):

• Total Area = \( \frac{2\sqrt{2}}{3} + \frac{2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3} \)

Therefore, the area of the quadrilateral \( ABCD \) is \( \frac{4\sqrt{2}}{3} \).