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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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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} $$| \overrightarrow{AC}$ x $ \overrightarrow{BD}|$
Step 1
Let the given points be $A(0,4,1),B(2,3,-1),C(4,5,0)$ and $D(2,6,2)$
Let $ \overrightarrow{OA}=0 \hat i+4 \hat j+\hat k\;;\; \overrightarrow{OB}=2 \hat i+3 \hat j-\hat k\;;\; \overrightarrow{OC}=4 \hat i+5 \hat j+0 \hat k \;;\; and \: \overrightarrow{OD}=2\hat i+6 \hat j+2 \hat k$
Let us determine $\overrightarrow {AC}$
$\overrightarrow {AC}=\overrightarrow {OC}-\overrightarrow {OA}$
$\qquad =(4 \hat i+5 \hat j+0 \hat k )-(0 \hat i+4 \hat j+ \hat k)$
$\qquad =4 \hat i+ \hat j- \hat k$
Let us determine $\overrightarrow {BD}$
$\overrightarrow {BD}=\overrightarrow {OD}-\overrightarrow {OB}$
$\qquad =(2 \hat i+6 \hat j+2 \hat k )-(2 \hat i+3 \hat j- \hat k)$
$\qquad =0 \hat i+3 \hat j+3 \hat k$
Step 2
Therefore $\overrightarrow{AC}$ x $ \overrightarrow{BD}=\begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & 1 & -1 \\ 0 & 3 &3 \end {vmatrix}$
On expanding we get,
$=\hat i(3+3)-\hat j (12+0)+\hat k(12-0)$
$=6 \hat i-12 \hat j+12 \hat k$
$=6( \hat i-2 \hat j+2 \hat k)$
$|\overrightarrow{AC}$ x $ \overrightarrow{BD}|=6 \times \sqrt {1^2+(-2)^2+(2)^2}$
$\qquad\qquad\quad=6 \times 3$
Area of a quadilateral $ ABCD $ is $ = \large\frac{1}{2} $$| \overrightarrow{AC}$ x $ \overrightarrow{BD}|$
$=\large\frac{1}{2} $$\times 18$
$=9 sq.units$
Hence $A$ is the correct option.
answered Jun 12, 2013 by meena.p
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