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

The Correct Option is C

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.