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 $A B C D=| a \times b |=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & -4 & 5 \\ 1 & -2 & -3\end{vmatrix}$
$=|\hat{ i }(12+10)-\hat{ j }(-6-5)+\hat{ k }(-4+4)|=|22 \hat{ i }+11 \hat{ j }+0 \hat{ k }|$
$=\sqrt{(22)^2+(11)^2+0^2}=\sqrt{(11)^2\left(2^2+1^2\right)}=11 \sqrt{5}$ sq units