Chapter 9.7, Problem 49AYU

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

Answer to Problem 49AYU

Solution:

The area of the parallelogram is $\sqrt{34}$

Explanation of Solution

Given:

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

Formula used:

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

Calculation:

Two adjacent sides of the parallelogram are

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

The area of the parallelogram = $\Vert u \times v\Vert = \sqrt{16 + 9 + 9}9 = \sqrt{34}$