Question

Diagonals of a parallelogram are given by $3\hat{i}-4\hat{j}-\hat{k}$ and $2\hat{i}+3\hat{j}-6\hat{k}$. Prove that parallelogram is rhombus and also find area.

Answer 1

$\overrightarrow{a}.\overrightarrow{b}=(3i-4j-k).(2i+3j-6k)=6-12+6=0$

$\because$ dot product is 0 $\Rightarrow$ the vectors are perpendicular i.e. the parallelogram is a rhombus

$\overrightarrow{a}=3i-4j-k$

$\overrightarrow{b}=2i+3j-6k$

Area $=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|$

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}i& j& k\\ 3& -4& -1\\ 2& 3& -6\end{array}\right|$

$=i\left(27\right)-j(-16)+k\left(17\right)$

$=27i+16j+17k$

$|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{{27}^2+{16}^2+{17}^2}$

$=\sqrt{1274}$

$\therefore Area=\frac{\sqrt{1274}}{2}$

$=\frac{35.6}{2}$

$=17.85$