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Mathematics Question on Quadrilaterals

If A(3,1,−1)A(3, 1, -1), B(53,73,13)B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1)C(2, 2, 1), and D(103,23,−13)D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right) are the vertices of a quadrilateral ABCD, then its area is:

A

423\frac{4\sqrt{2}}{3}

B

523\frac{5\sqrt{2}}{3}

C

222\sqrt{2}

D

223\frac{2\sqrt{2}}{3}

Answer

423\frac{4\sqrt{2}}{3}

Explanation

Solution

The area of quadrilateral ABCDABCD is given by:

Area=12∥BD→×AC→∥,\text{Area} = \frac{1}{2} \|\overrightarrow{BD} \times \overrightarrow{AC}\|,

where:

BD→=D→−B→,AC→=C→−A→.\overrightarrow{BD} = \overrightarrow{D} - \overrightarrow{B}, \quad \overrightarrow{AC} = \overrightarrow{C} - \overrightarrow{A}.

Calculate BD→\overrightarrow{BD}:

BD→=(103−53)i^+(23−73)j^+(−13−13)k^.\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:

BD→=53i^−53j^−23k^.\overrightarrow{BD} = \frac{5}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{2}{3}\hat{k}.

Calculate AC→\overrightarrow{AC}:

AC→=(2−3)i^+(2−1)j^+(1−(−1))k^.\overrightarrow{AC} = (2 - 3)\hat{i} + (2 - 1)\hat{j} + (1 - (-1))\hat{k}.

Simplify:

AC→=−i^+j^+2k^.\overrightarrow{AC} = -\hat{i} + \hat{j} + 2\hat{k}.

Now, compute BD→×AC→\overrightarrow{BD} \times \overrightarrow{AC}:

BD→×AC→=∣i^j^k^ 53−53−23 −112∣.\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:

BD→×AC→=i^((−5/3)(2)−(−5/3)(1))−j^((5/3)(2)−(−2/3)(1))+k^((5/3)(1)−(−5/3)(−1)).\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:

BD→×AC→=i^(−10/3+5/3)−j^(10/3−2/3)+k^(5/3−5/3).\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).

BD→×AC→=−53i^−83j^+0k^.\overrightarrow{BD} \times \overrightarrow{AC} = -\frac{5}{3}\hat{i} - \frac{8}{3}\hat{j} + 0\hat{k}.

The magnitude is:

∥BD→×AC→∥=(−53)2+(−83)2.\|\overrightarrow{BD} \times \overrightarrow{AC}\| = \sqrt{\left(-\frac{5}{3}\right)^2 + \left(-\frac{8}{3}\right)^2}.

∥BD→×AC→∥=259+649=899=893.\|\overrightarrow{BD} \times \overrightarrow{AC}\| = \sqrt{\frac{25}{9} + \frac{64}{9}} = \sqrt{\frac{89}{9}} = \frac{\sqrt{89}}{3}.

The area is:

Area=12⋅893=423.\text{Area} = \frac{1}{2} \cdot \frac{\sqrt{89}}{3} = \frac{4\sqrt{2}}{3}.