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 $ABCD =|\overrightarrow{ AB }||\overrightarrow{ AD }|=2 \times 1=2$