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.