Question

Find the area of the parallelogram with sides a=i-2j+k and b=2i+j+k

Answer 1

Answer on Question #68582, Math / Differential Equations

Find the area of the parallelogram with sides $\pmb{a} = \pmb{i} - 2\pmb{j} + \pmb{k}$ and $b = 2\pmb{i} + \pmb{j} + \pmb{k}$ Solution

The area of the parallelogram will be the length of the cross product of adjacent sides. We have

$$\begin{array}{l} \boldsymbol {a} \times \boldsymbol {b} = \left| \begin{array}{c c c} \boldsymbol {i} & \boldsymbol {j} & \boldsymbol {k} \\ 1 & - 2 & 1 \\ 2 & 1 & 1 \end{array} \right| = \\ = (- 1) ^ {1 + 1} \boldsymbol {i} \left| \begin{array}{c c} - 2 & 1 \\ 1 & 1 \end{array} \right| + (- 1) ^ {1 + 2} \boldsymbol {j} \left| \begin{array}{c c} 1 & 1 \\ 2 & 1 \end{array} \right| + (- 1) ^ {1 + 3} \boldsymbol {k} \left| \begin{array}{c c} 1 & - 2 \\ 2 & 1 \end{array} \right| = \\ = \boldsymbol {i} (- 2 - 1) - \boldsymbol {j} (1 - 2) + \boldsymbol {k} (1 - (- 4)) = - 3 \boldsymbol {i} + \boldsymbol {j} + 5 \boldsymbol {k} \\ \end{array}$$

Thus, the area of the parallelogram is

$$\left| - 3 \boldsymbol {i} + \boldsymbol {j} + 5 \boldsymbol {k} \right| = \sqrt {(- 3) ^ {2} + (1) ^ {2} + (5) ^ {2}} = \sqrt {3 5} (u n i t s ^ {2})$$

Answer: $\sqrt{35}$ units².

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