Question

Area of rectangle having vertices $A,B,C$ and $D$ with position vector
$\left(-\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}\right),\left(\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}\right),\left(\hat{i}-\frac{1}{2}\hat{j}+4\hat{k}\right)$ and $\left(-\hat{i}-\frac{1}{2}\hat{j}+4\hat{k}\right)$ is

• A. $\frac{1}{2}$ sq. units
• B. 1 sq. units
• C. 2 sq units
• D. 4 sq. units.

Answer 1

Answer: (C)

The position vectors of $A$ and $B$ are $-\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}$
and $\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}$
$\therefore \overrightarrow{AB}=[1-(-1)]\hat{i}+0\cdot \hat{j}+0\cdot \hat{k}$ $\therefore |\overrightarrow{AB}|=2$
The position vectors of $A$ and $D$ are
$-\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}$ and $-\hat{i}-\frac{1}{2}\hat{j}+4\hat{k}$
$\therefore \overrightarrow{AD}=(-\hat{i}+\hat{i})+\left(-\frac{1}{2}-\frac{1}{2}\right)\hat{j}+(4\hat{k}-4\hat{k})=-\hat{j}$
$\therefore |\overrightarrow{AD}|=1$
Area of rectangle $ABCD=|\overrightarrow{AB}||\overrightarrow{AD}|=2\times 1=2$