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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Area of a rectangle having vertices $A, B, C$ and $D$ with position vectors $ -\hat i + \large\frac{1}{2}$$ \hat j + 4\hat k, \: \hat i +\large \frac{1}{2}$$ \hat j + 4\hat k, \: \hat i -\large \frac{1}{2}$$ \hat j + 4\hat k$ and $-\hat i -\large \frac{1}{2}$$ \hat j + 4\hat k$ respectively is

\[ \begin{array}{1 1} (A) \;\frac{1}{2} \qquad &(B) \;1 \qquad \\ (C) \;2 \qquad & (D)\;4 \end{array} \]

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1 Answer

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Toolbox:
  • Area of the parallelogram $ABCD$ is given by $|\overrightarrow AB\times \overrightarrow AD|$
  • $\overrightarrow a\times\overrightarrow b=\begin{vmatrix}\hat i &\hat j &\hat k\\a_1&a_2& a_3\\b_1 &b_2 & b_3\end{vmatrix}$
Step 1:
Let $\overrightarrow {OA}=-\hat i+\large\frac{1}{2}$$\hat j+4\hat k$
$\quad\;\;\overrightarrow {OB}=\hat i+\large\frac{1}{2}$$\hat j+4\hat k$
$\quad\;\;\overrightarrow {OC}=\hat i-\large\frac{1}{2}$$\hat j+4\hat k$
$\quad\;\;\overrightarrow {OD}=-\hat i-\large\frac{1}{2}$$\hat j+4\hat k$
We know rectangle is a parallelogram.
Area of a parallelogram is $|\overrightarrow {AB}\times \overrightarrow {AD}|$
Step 2:
Let us first determine $\overrightarrow {AB}=\overrightarrow {OB}-\overrightarrow {OA}$
$\qquad\qquad\qquad\qquad\;\;\;=(\hat i+\large\frac{1}{2}$$\hat j+4\hat k)-(-\hat i+\large\frac{1}{2}$$\hat j+4\hat k)$
$\qquad\qquad\qquad\qquad\;\;\;=2\hat i$
Step 3:
Next let us first determine $\overrightarrow {AD}=\overrightarrow {OD}-\overrightarrow {OA}$
$\qquad\qquad\qquad\qquad\qquad\;\;\;=(-\hat i-\large\frac{1}{2}$$\hat j+4\hat k)-(-\hat i+\large\frac{1}{2}$$\hat j+4\hat k)$
$\qquad\qquad\qquad\qquad\qquad\;\;\;=-\hat j$
Step 4:
$\overrightarrow {AB}\times\overrightarrow {AD}=\begin{vmatrix}\hat i &\hat j &\hat k\\2&0& 0\\0 &-1 & 0\end{vmatrix}$
$\qquad\;\;\;\;\;=\hat i(0)-\hat j(0)+\hat k(-2)$
$\qquad\;\;\;\;\;=-2\hat k$
Step 5:
$|\overrightarrow {AB}\times \overrightarrow {AD}|=\sqrt{(-2)^2}=2$
Hence the area is $2$ sq. units.
The correct option is C.
answered May 23, 2013 by sreemathi.v
 

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