Question

Area of a rectangle having vertices $A,B,C,and \space D$ with position vectors$-\hat{i}+\frac{1}{2}\hat{j}+4\hat{k},\hat{i}+\frac{1}{2}\hat{j}+4\hat{k},\hat{i}-\frac{1}{2}\hat{j}+4\hat{k}\space and -\hat{i}-\frac{1}{2}\hat{j}+4\hat{k}$ respectively is

• A. $(\frac{1}{2})$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (C)

$2$

Answer 2

The position vectors of vertices $A,B,C,and\space D$ of rectangle $ABCD$ are given as:
$\overrightarrow{OA}=-\hat{i}+\frac{1}{2}\hat{j}+4\hat{k},\overrightarrow{OB}=\hat{i}+\frac{1}{2}\hat{j}+4\hat{k},\overrightarrow{OC}=\hat{i}-\frac{1}{2}\hat{j}+4\hat{k},\overrightarrow{OD}=-\hat{i}-\frac{1}{2}\hat{j}+4\hat{k}$
The adjacent sides $\overrightarrow{AB}\space and\space \overrightarrow{ BC}$ of the given rectangle are given as:
$\overrightarrow{AB}=(1+1)\hat{i}+(\frac{1}{2}-\frac{1}{2})\hat{j}+(4-4)\hat{k}=2\hat{i}$
$\overrightarrow{BC}=(1-1)\hat{i}+(-\frac{1}{2}-\frac{1}{2})\hat{j}+(4-4)\hat{k}=-\hat{j}$
$∴\overrightarrow{AB}\times\overrightarrow{AC}\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\\ 2 & 0 & 0\\\0&-1&0 \end{vmatrix}=\hat{k}(-2)=-2\hat{k}$
|$\overrightarrow{AB}\times \overrightarrow{AC}$|$=\sqrt{(-2)^{2}}=2$
Now,it is known that the area of a parallelogram whose adjacent sides are $\vec{a}$ and $\vec{b}$ is |$\vec{a}\times\vec{b}$|.
Hence,the area of the given rectangle is |$\overrightarrow{AB}\times\overrightarrow{BC}$|$=2$ square units.
The correct answer is C.