Question

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}+3\hat{j}+5\hat{k}$ and $\overrightarrow{c}=7\hat{i}+9\hat{j}+11\hat{k}$ then the area of parallelogram having diagonals $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{b}+\overrightarrow{c}$ is

• A. $4\sqrt{6}$ sq units
• B. $\frac{1}{2}\sqrt{21}$ sq units
• C. $\frac{\sqrt{6}}{2}$ sq units
• D. $\sqrt{6}$ sq units

Answer 1

Answer: (A)

$4\sqrt{6}$ sq units

$\overline{a}+\overline{b}=2\stackrel{\wedge }{i}+4\stackrel{\wedge }{j}+6\stackrel{\wedge }{k}$

$\overline{b}+\overline{c}=8\stackrel{\wedge }{i}+12\stackrel{\wedge }{j}+16\stackrel{\wedge }{k}$

$⟹\overline{b}+\overline{c}=4(2\stackrel{\wedge }{i}+3\stackrel{\wedge }{j}+4\stackrel{\wedge }{k})$

Area of parallelogram=$\frac{1}{2}pqsin\theta$

where $p,q$ are lenghts of diagonals and $\theta$ is the angle between them.

Hence in this case,

$Area=\frac{1}{2}\left|\right(\overline{a}+\overline{b})\times (\overline{b}+\overline{c}\left)\right|$

$=\frac{1}{2}\left|2\right(\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}+3\stackrel{\wedge }{k})\times 4(2\stackrel{\wedge }{i}+3\stackrel{\wedge }{j}+4\stackrel{\wedge }{k}\left)\right|$

$=\frac{8}{2}\left[\right(\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}+3\stackrel{\wedge }{k})\times (2\stackrel{\wedge }{i}+3\stackrel{\wedge }{j}+4\stackrel{\wedge }{k}\left)\right]$

$=4\left|\begin{array}{ccc}\stackrel{\wedge }{i}& \stackrel{\wedge }{j}& \stackrel{\wedge }{k}\\ 1& 2& 3\\ 2& 3& 4\end{array}\right|$

$=4\left|\right[\stackrel{\wedge }{i}(8-9)-\stackrel{\wedge }{j}(4-6)+\stackrel{\wedge }{k}(3-4)\left]\right|$

$=4|\stackrel{\wedge }{-i}+2\stackrel{\wedge }{j}-\stackrel{\wedge }{k}|$

$=4\sqrt{1+4+1}$

$=4\sqrt{6}sq.units$