# The area of the quadilateral $ABCD$ where $A(0,4,1),B(2,3,-1),C(4,5,0)$ and $D(2,6,2)$ is equal to

$(A)\;9\;sq. units\qquad(B)\;18\;sq. units\qquad(C)\;27\;sq. units\qquad(D)\;81\;sq. units$

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• Area of a quadilateral $ABCD$ is $= \large\frac{1}{2} $$| \overrightarrow{AC} x \overrightarrow{BD}| Step 1 Let the given points be A(0,4,1),B(2,3,-1),C(4,5,0) and D(2,6,2) Let \overrightarrow{OA}=0 \hat i+4 \hat j+\hat k\;;\; \overrightarrow{OB}=2 \hat i+3 \hat j-\hat k\;;\; \overrightarrow{OC}=4 \hat i+5 \hat j+0 \hat k \;;\; and \: \overrightarrow{OD}=2\hat i+6 \hat j+2 \hat k Let us determine \overrightarrow {AC} \overrightarrow {AC}=\overrightarrow {OC}-\overrightarrow {OA} \qquad =(4 \hat i+5 \hat j+0 \hat k )-(0 \hat i+4 \hat j+ \hat k) \qquad =4 \hat i+ \hat j- \hat k Let us determine \overrightarrow {BD} \overrightarrow {BD}=\overrightarrow {OD}-\overrightarrow {OB} \qquad =(2 \hat i+6 \hat j+2 \hat k )-(2 \hat i+3 \hat j- \hat k) \qquad =0 \hat i+3 \hat j+3 \hat k Step 2 Therefore \overrightarrow{AC} x \overrightarrow{BD}=\begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & 1 & -1 \\ 0 & 3 &3 \end {vmatrix} On expanding we get, =\hat i(3+3)-\hat j (12+0)+\hat k(12-0) =6 \hat i-12 \hat j+12 \hat k =6( \hat i-2 \hat j+2 \hat k) |\overrightarrow{AC} x \overrightarrow{BD}|=6 \times \sqrt {1^2+(-2)^2+(2)^2} \qquad\qquad\quad=6 \times 3 \qquad\qquad\quad=18 Area of a quadilateral ABCD is = \large\frac{1}{2}$$| \overrightarrow{AC}$ x $\overrightarrow{BD}|$
$=\large\frac{1}{2}$$\times 18$
$=9 sq.units$
Hence $A$ is the correct option.