The area of the quadrilateral ABCD with vertices A(2,1,1),B(1,2,5),C(-2,-3,5) and D(1,-6,-…

Question

The area of the quadrilateral $ABCD$ with vertices $A(2,1,1),B(1,2,5),C(-2,-3,5)$ and $D(1,-6,-7)$ is equal to :

Accepted Answer

$\begin{gathered} \overset{	o \leftarrow }{AC}=(-2-2)\hat{i}+(-3-1)\hat{j}+(5-1)\hat{k} \ =-4\hat{i}-4\hat{j}+4\hat{k} \ \overset{	o \leftarrow }{BD}=(1-1)\hat{i}+(-6-2)\hat{j}+(-7-5)\hat{k} \ =-8\hat{j}-12\hat{k} \end{gathered}$

So, area of quadrilateral $=\frac{1}{2}|\overset{	o \leftarrow }{AC}	imes \overset{	o \leftarrow }{BD}|$

$\begin{gathered} =\frac{1}{2}\parallel \begin{gathered} \hat{i} \quad \hat{j} \quad \hat{k} \ -4 \quad -4 \quad 4 \ 0 \quad -8 \quad -12 \end{gathered}\parallel \ =\frac{1}{2}|(48+32)\hat{i}-(48-0)\hat{j}+(32-0)\hat{k}| \ =\frac{1}{2}|80\hat{i}-48\hat{j}+32\hat{k}| \ =\frac{1}{2}16|15\hat{i}-3\hat{j}+2\hat{k}| \ =8\sqrt{25+9+4}=8\sqrt{38} \end{gathered}$

Solutions · 1