Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
0 votes

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

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

Can you answer this question?

1 Answer

0 votes
  • 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
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App