Chapter 9.7, Problem 38AYU

To determine

To find: The value of $\left(\text{v} \times \text{u}\right) \cdot \text{w}$

for, $\text{u} = 2\text{i} - 3\text{j} + \text{k},\text{v} = - 3\text{i} + 3\text{j} + 2\text{k},\text{w} = \text{i} + \text{j} + 3\text{k}$

Answer to Problem 38AYU

Solution:

$\left(\text{v} \times \text{u}\right) \cdot \text{w} = 25$

Explanation of Solution

Given:

$\text{u} = 2\text{i} - 3\text{j} + \text{k},\text{v} = - 3\text{i} + 3\text{j} + 2\text{k},\text{w} = \text{i} + \text{j} + 3\text{k}$

Formula used:

If $\text{v} = {\text{a}}_1\text{i} + {\text{b}}_1\text{j} + {\text{c}}_1\text{k}$, and $\text{w} = {\text{a}}_2\text{i} + {\text{b}}_2\text{j} + {\text{c}}_2\text{k}$ are two vectors in space, the cross product $\text{v} \times \text{w}$ is

$\text{v} \times \text{w} = \left({\text{b}}_1{\text{c}}_2 - {\text{b}}_2{\text{c}}_1\right)\text{i} - \left({\text{a}}_1{\text{c}}_2 - {\text{a}}_2{\text{c}}_1\right)\text{j} - \left({\text{a}}_1{\text{b}}_2 - {\text{a}}_2{\text{b}}_1\right)\text{k}$

Calculation:

$\begin{array}{lll}\text{v} \times \text{u}\hfill & = \hfill & \left|\begin{array}{ccc}i& j& k\\ - 3& 3& 2\\ 2& - 3& 1\end{array}\right|\hfill \\ \hfill & = \hfill & \left(3 + 6\right)\text{i} - \left(- 3 - 4\right)\text{j} + \left(9 - 6\right)\text{k}\hfill \\ \text{v} \times \text{u}\hfill & = \hfill & 9\text{i} + 7\text{j} + 3\text{k}\hfill \end{array}$

now, we compute, $\left(\text{v} \times \text{u}\right) \cdot \text{w}$

$\begin{array}{lll}\left(\text{v} \times \text{u}\right) \cdot \text{w}\hfill & = \hfill & \left(9\text{i} + 7\text{j} + 3\text{k}\right) \cdot \left(\text{i} + \text{j} + 3\text{k}\right)\hfill \\ \hfill & = \hfill & 9 + 7 + 9 = 25\hfill \end{array}$