If \(\overrightarrow{AC}=2\hat{i}+\hat{j}+\hat{k}\) and \(\overrightarrow{BD}=-\hat{i}+3\hat{j}+2\hat{k}\) then the area of the quadrilateral ABCD is

Concept:

The area of the quadrilateral ABCD = \(\rm \dfrac 12 |\vec {AC} \times \vec {BD}|\)  , where \(\rm \vec {AC} \;\; \text {and }\;\; \vec {BD}\) are diagonals.

Calculations:

Let \(\overrightarrow{AC}=2\hat{i}+\hat{j}+\hat{k}\)    and  \(\overrightarrow{BD}=-\hat{i}+3\hat{j}+2\hat{k}\) are the diagonal of the quadrilateral ABCD.

The area of the quadrilateral ABCD = \(\rm \dfrac 12 |\vec {AC} \times \vec {BD}|\) ....(1) 

⇒\(\rm (\vec {AC} \times \vec {BD}) =\)\(\begin{vmatrix} \vec i&\vec j & \vec k \\ 2&1 &1 \\ -1&3 &2 \end{vmatrix}\)

⇒\(\rm (\vec {AC} \times \vec {BD}) =\)\(\rm -\vec i - 5\vec j + 7 \vec k\)

⇒\(\rm |\vec {AC} \times \vec {BD}| = \sqrt {(-1)^2+(-5)^2+(7)^2}\)

⇒\(\rm |\vec {AC} \times \vec {BD}| = \sqrt {75} = 5 \sqrt 3\)

From equation (1), we have

The area of the quadrilateral ABCD = \(\rm \dfrac{5\sqrt3}{2}\)