Question

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

a) $\frac{1}{2}$
b) 1
c) 2
d) 4

Answer 1

The position vectors of vertices A, B, C and D of rectangle ABCD are given.
First, we compute vectors AB and AD
Now, AB = PV of B - PV of A $=(\hat{i}+\frac{1}{2}\hat{j}+4\hat{k})-(-\hat{i}+\frac{1}{2}\hat{j}+4\hat{k})$
$=[1-(-1\left)\right]\hat{i}+(\frac{1}{2}-\frac{1}{2})\hat{j}+(4-4)\hat{k}=2\hat{i}$
and AD = PV of D - PV of A $=(-\hat{i}-\frac{1}{2}\hat{j}+4\hat{k})-(-\hat{i}+\frac{1}{2}\hat{j}+4\hat{k})$
$=[-1-(-1\left)\right]\hat{i}+(-\frac{1}{2}-\frac{1}{2})\hat{j}+(4-4)\hat{k}=-\hat{j}AB\times AD=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 0& 0\\ 0& -1& 0\end{array}\right|=-2\hat{k}$
Area of rectangle ABCD $=|AB\times AD|=\sqrt{(-2{)}^2}=2sq.unit$
[Now, it is known that the area of a parallelogram whose adjacent sides are a and b is $|a\times b|$]
Hence, the area of the rectangle is $|AB|\times |AD|=2sq.unit$.
Note If a and b are the adjacent sides of a parallelogram, then area $=|a\times b|$ if a and b are the diagonals of a parallelogram, then area $=\frac{1}{2}|a\times b|$