If A(3,1,-1), B(5/3, 7/3, 1/3), C(2,2,1) and D(10/3, 2/3, -1/3) are the verticle of a quadrilateral ABCD, then its area is

Question

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

(1) \(\frac{4 \sqrt{2}}{3}\)

(2) \(\frac{5 \sqrt{2}}{3}\)

(3) \(2 \sqrt{2}\)

(4) \(\frac{2 \sqrt{2}}{3}\)

Answer 1

Correct option is (1) \(\frac{4 \sqrt{2}}{3}\)

 \(\overrightarrow{AC}=-\hat i+\hat j+2\hat k\)

\(\overrightarrow{BD} = \frac{5}{3}\hat i-\frac{5}{3}\hat j-\frac{2}{3}\hat k\)

area of quadrilateral = \(\frac{1}{2}|\overrightarrow{AC}\times\overrightarrow{BD}|\)

\(=\frac{1}{2} \begin{bmatrix} \hat i & \hat j & \hat k \\[0.3em] -1 & 1 & 2 \\[0.3em] \frac{5}{3} & \frac{-5}{3} & \frac{-2}{3} \end{bmatrix}\)

\(=\frac{8}{6}(\hat i+\hat j)=\frac{4}{3}\sqrt2\)