Question

The two adjacent sides of a parallelogram are $2\hat{i}-4\hat{j}+5\hat{k}$ and $\hat{i}-2\hat{j}-3\hat{k}$. Find the unit vector parallel to its diagonal. Also, find its area.

Answer 1

Adjacent sides of a parallelogram are given as $a=2\hat{i}-4\hat{j}+5\hat{k}$ and $b=\hat{i}-2\hat{j}-3\hat{k}$
Then, the diagonal of a parallelogram is given by v = a + b.
($\because$ From the figure, it is clear that resultant of adjacent sides of a parallelogram is given by the diagonal)
$\therefore v=2\hat{i}-4\hat{j}+5\hat{k}+\hat{i}-2\hat{j}-3\hat{k}=(2+1)\hat{i}+(-4-2)\hat{j}+(5-3)\hat{k}=3\hat{i}-6\hat{j}+2\hat{k}$
Comparing with $X=x\hat{i}+y\hat{j}+z\hat{k}$, we get
x = 3, y = -6, z = 2
$\therefore |v|=\sqrt{x^2+y^2+z^2}=\sqrt{(3{)}^2+(-6{)}^2+(2{)}^3}=\sqrt{9+36+4}=\sqrt{49}=7$
Thus, the unit vector parallel to the diagonal is
$\frac{v}{|v|}=\frac{3\hat{i}-6\hat{j}+2\hat{k}}{7}=\frac{3}{7}\hat{i}-\frac{6}{7}\hat{j}+\frac{2}{7}\hat{k}$
Also, area of parallelogram ABCD, $|a\times b|=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -4& 5\\ 1& -2& -3\end{array}\right|=|\hat{i}(12+10)-\hat{j}(-6-5)+\hat{k}(-4+4)|=|22\hat{i}+11\hat{j}+0\hat{k}|=\sqrt{(22{)}^2+(11{)}^2+0^2}=\sqrt{\left(11{)}^2\right(2^2+1^2)}=11\sqrt{5}sq.unit$

Note If $d_1$ and $d_2$ are the diagonals of a parallelogram, then area of parallelogram $=\frac{1}{2}|d_1\times d_2|$