# Area of a rectangle having vertices A, B, C, and D with position vectors

Question:

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

(A) $\frac{1}{2}$

(B) 1

(C) 2

(D) 4

Solution:

The position vectors of vertices A, B, C, and D of rectangle ABCD are given as:

$\overrightarrow{\mathrm{OA}}=-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \overrightarrow{\mathrm{OB}}=\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \overrightarrow{\mathrm{OC}}=\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}, \overrightarrow{\mathrm{OD}}=-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$

The adjacent sides $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{BC}}$ of the given rectangle are given as:

$\overrightarrow{\mathrm{AB}}=(1+1) \hat{i}+\left(\frac{1}{2}-\frac{1}{2}\right) \hat{j}+(4-4) \hat{k}=2 \hat{i}$

$\overrightarrow{\mathrm{BC}}=(1-1) \hat{i}+\left(-\frac{1}{2}-\frac{1}{2}\right) \hat{j}+(4-4) \hat{k}=-\hat{j}$

$\therefore \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{BC}}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 0 & -1 & 0\end{array}\right|=\hat{k}(-2)=-2 \hat{k}$

$\Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{BC}}|=2$

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 $|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{BC}}|=2$ square units.

The correct answer is $\mathrm{C}$.