Question

The two adjacent sides of a parallelogram are $2\hat{i}-4\hat{j}+5\hat{k}$ and $\hat{i}-2\hat{j}-3\hat{k}$
Statement I The unit vector parallel to its diagonal is $\frac{3}{5}\hat{i}-\frac{6}{5}\hat{j}+\frac{2}{5}\hat{k}$
Statement II Area of parallelogram is $11\sqrt{5}sq$ units.

• A. Only statement I is true
• B. Only statement II is true
• C. Both statements are true
• D. Both statements are false

Answer 1

Answer: (B)

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
$v=a+b\text{. }$
$(\because$ 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 $ABCD=|a\times b|=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -4& 5\\ 1& -2& -3\end{array}\right|$
$=|\hat{i}(12+10)-\hat{j}(-6-5)+\hat{k}(-4+4)|=|22\hat{i}+11\hat{j}+0\hat{k}|$
$=\sqrt{(22{)}^2+(11{)}^2+0^2}=\sqrt{(11{)}^2\left(2^2+1^2\right)}=11\sqrt{5}$ sq units