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 $\vec a$=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 $\vec a$+$\vec b$.
$\vec a$+$\vec b$=(2+1)$\hat{i}$+(-4-2)$\hat j$+(5-3)$\hat k$=3$\hat{i}$-6$\hat j$+2$\hat k$
Thus, the unit vector parallel to the diagonal is
$\frac{\vec a+\vec b}{|\vec a+\vec b|}$=$\frac{3\hat i-6\hat j+2\hat k}{\sqrt{32}}$+(-6)2+22=3$\hat{i}$-6$\hat j$+2\hat k$$\sqrt{9+36+4}=$\frac{3\hat i-6\hat j+2\hat k}{7}$=$\frac{3}{7}\hat i-\frac{6}{7}\hat j+\frac{2}{7}\hat k$
∴Area of parallelogram ABCD =$\hat{i}$(12+10)-$\hat j$(-6-5)+$\hat k$(-4+4)
=22$\hat i$+11$\hat j$
=11(2$\hat i$+$\hat j$)
∴|$\vec a\times \vec b$|=11$\sqrt{22}$+12=11$\sqrt{5}$
Hence, the area of a parallelogram is 11$\sqrt{5}$ square units.