Question

If $a=i+j+k,b=i+3j+3k$ and $c=7i+9j+11k,$

then the area of parallelogram having diagonals $a+b$ and $b+c$ is$?$

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

Answer 1

The correct option is A

$4\sqrt{6}$units

Explanation for the correct option:

Step 1. Find the value of a vector $A=a+b$

Given,

$a=i+j+k,$

$b=i+3j+5k$ and

$c=7i+9j+11k$

$A=(\mathrm{i}+j+\mathrm{k})+(\mathrm{i}+3\mathrm{j}+5\mathrm{k})$

$=i+j+k+i+3j+5k$

$\Rightarrow$ $A=2i+4j+6k$

Step 2: Find the value of vector $\begin{array}{rcl}B& =& b+c\end{array}$

$\begin{array}{rcl}B& =& b+c\end{array}$

$\begin{array}{rcl}& =& (\mathrm{i}+3\mathrm{j}+5\mathrm{k})+(7\mathrm{i}+9\mathrm{j}+11\mathrm{k})\end{array}$

$=i+3j+5k+7i+9j+11k$

$\therefore$$B=8i+12j+16k$

Step 3:Find area of parallelogram

Therefore, Area of parallelogram$\mathbf{=}\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.\mathbf{|}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{|}$

Area$=\frac{1}{2}\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 2& 4& 6\\ 8& 12& 16\end{array}\right|$

$\Rightarrow$ Area $=\frac{1}{2}\left|i\left[\left(4\right)\left(16\right)-\left(6\right)\left(12\right)\right]-j\left[\left(2\right)\left(16\right)-\left(8\right)\left(6\right)\right]+k\left[\left(2\right)\left(12\right)-\left(8\right)\left(4\right)\right]\right|$

$\Rightarrow$ Area$=\frac{1}{2}|\mathrm{i}(64-72)-\mathrm{j}(32-48)+\mathrm{k}(24-32)|$

$\Rightarrow$ Area$=\frac{1}{2}|-8\mathrm{i}+16\mathrm{j}-8\mathrm{k}|$

$\Rightarrow$ Area$=\left|-4i+8j-4k\right|$

$\Rightarrow$ Area$=\sqrt{{(-4)}^2+{\left(8\right)}^2+{(-4)}^2}$

$\Rightarrow$ Area$=\sqrt{96}$

$\Rightarrow$ Area$=4\sqrt{6}units$

Hence, option(A) is correct.