Chapter 9.7, Problem 50AYU

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

Answer to Problem 50AYU

Solution:

The area of the parallelogram is $\sqrt{64}$

Explanation of Solution

Given:

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

Formula used:

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

Calculation:

Two adjacent sides of the parallelogram are

$\begin{array}{l}\text{u} = {\text{p}}_1{\text{p}}_2 = \left(2 - 2\right)\text{i} + \left(3 - 1\right)\text{j} + \left(1 - 1\right)\text{k} = 2\text{j};\\ \text{v} = {\text{p}}_1{\text{p}}_3 = \left(- 2 - 2\right)\text{i} + \left(4 - 1\right)\text{j} + \left(1 - 1\right)\text{k} = - 4\text{i} + 3\text{j}\\ \text{u} \times \text{v} = \left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 2& 0\\ - 4& 3& 0\end{array}\right|\\ = \left(0 - 0\right)\text{i} - \left(0 - 0\right)\text{j} + \left(0 + 8\right)\text{k}\\ = 8\text{k }\end{array}$

The area of the parallelogram = $\Vert u \times v\Vert = \sqrt{64}$