Question

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

• A. $4\sqrt{6}$ units
• B. $\frac{1}{2\sqrt{21}}$ sq units
• C. $\frac{\sqrt{6}}{2}$ 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 area of parallelogram:

Given, $a=i+j+k,b=i+3j+5k$ and $c=7i+9j+11k$

Let

$\begin{array}{rcl}A& =& a+b\\ & =& (i+j+k)+(i+3j+5k)\\ & =& 2i+4j+6k\end{array}$

$\begin{array}{rcl}B& =& b+c\\ & =& (i+3j+5k)+(7i+9j+11k)\\ & =& 8i+12j+16k\end{array}$

Step 2. Area of parallelogram $=\frac{1}{2}$(product of diagonals)

$$\begin{gathered} =\frac{1}{2}|AxB| \\ =\frac{1}{2}\left|\right(2i+4j+6k)x(8i+12j+16k) \\ =\frac{1}{2}\mid i(64-72)-j(32-48)+k(24-32)\mid \\ =\frac{1}{2}\mid -8i+16j-8k\mid \\ =\sqrt{\left[{(-4)}^2+8^2+{(-4)}^2\right]} \\ =\sqrt{96} \\ =\mathbf{4}\sqrt{\mathbf{6}} \end{gathered}$$

Hence, Option ‘A’ is Correct.