Chapter 9.7, Problem 51AYU

To determine

To find: The area of the parallelogram with one corner at ${\text{p}}_1$ and adjacent sides → ${\text{p}}_1{\text{p}}_2$ and ${\text{p}}_1{\text{p}}_3$

For, ${\text{p}}_1 = \left(1,2, - 1\right),{\text{p}}_2 = \left(4,2, - 3\right),{\text{p}}_3 = \left(6, - 5,2\right),{\text{p}}_4 = \left(9, - 5,0\right)$

Answer to Problem 51AYU

Solution:

The area of the parallelogram is $\sqrt{998}$

Explanation of Solution

Given:

${\text{p}}_1 = \left(1,2, - 1\right),{\text{p}}_2 = \left(4,2, - 3\right),{\text{p}}_3 = \left(6, - 5,2\right),{\text{p}}_4 = \left(9, - 5,0\right)$

Formula used:

The area of the parallelogram is $\Vert u \times v\Vert$

Calculation:

Two adjacent sides of the parallelogram are

$\begin{array}{l}\text{u} = {\text{p}}_1{\text{p}}_2 = \left(4 - 1\right)\text{i} + \left(2 - 2\right)\text{j} + \left(- 3 + 1\right)\text{k} = 3\text{i} - 2\text{k};\\ \text{v} = {\text{p}}_1{\text{p}}_3 = \left(6 - 1\right)\text{i} + \left(- 5 - 2\right)\text{j} + \left(2 + 1\right)\text{k} = 5\text{i} - 7\text{j} + \text{3k}\\ \text{u} \times \text{v} = \left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 3& 0& - 2\\ 5& - 7& 3\end{array}\right|\\ = \left(0 - 14\right)\text{i} - \left(9 + 10\right)\text{j} + \left(- 21 + 0\right)\text{k}\\ = - 14\text{i} - 19\text{j} - 21\text{k }\end{array}$

The area of the parallelogram = $\Vert u \times v\Vert = \sqrt{196 + 361 + 441}9 = \sqrt{998}$