The diagonals of a parallelogram are represented by $R_{1} = 3 i + 2 j - 7 k$ and $R_{2} = 5 i + 6 j - 3 k$ . Find the area of parallelogram.

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Solution

Step 1. Determine the cross-product of the diagonal vectors.
The vectors of the diagonals of the parallelogram are $R_{1} = 3 i + 2 j - 7 k$ and $R_{2} = 5 i + 6 j - 3 k$ .
Determine the cross-product of the vectors:
$\begin{array}{rcl} R_{1} \times R_{2} & = & |\begin{matrix} i & j & k \\ 3 & 2 & - 7 \\ 5 & 6 & - 3 \end{matrix}| \\ \Rightarrow & R_{1} \times R_{2} = i (- 6 + 42) - j (- 9 + 35) + k (18 - 10) \\ \Rightarrow & R_{1} \times R_{2} = 36 i - 26 j + 8 k \end{array}$
So, the cross-product is $\begin{array}{rcl} R_{1} \times R_{2} & = & 36 i - 26 j + 8 k \end{array}$ .
Step 2: Determine the area of the parallelogram.
The area of the parallelogram is given by :
$\frac{1}{2} |\begin{array}{l} R_{1} \times R_{2} \end{array}|$ .
Use the above mentioned formula to find the area:
$\begin{array}{rcl} Area of parallelogram & = & \frac{1}{2} |R_{1} \times R_{2}| \\ \Rightarrow & Area of parallelogram = \frac{1}{2} |36 i - 26 j + 8 k| \\ \Rightarrow & Area of parallelogram \approx \frac{1}{2} \sqrt{36^{2} + {(- 26)}^{2} + 8^{2}} \\ \Rightarrow & Area of parallelogram \approx \frac{1}{2} \times 45.12 \end{array} \begin{array}{rcl} \Rightarrow & Area of parallelogram \approx 22.56 {unit}^{2} \end{array}$
So, the area of the parallelogram is $22.56 {unit}^{2}$

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