Chapter 9.7, Problem 40AYU

In Problems 23-44, use the given vectors $u,v,\text{ and }w$ to find each expression.

$u\text{ = 2}i-\text{ 3}j\text{ + }kv\text{ = }-\text{3}i+\text{ 3}j\text{ + 2}kw\text{ = }i+j\text{ + 3}k$

$\left(w\times w\right)\times v$

To determine

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

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 40AYU

Solution:

$\left(\text{w} \times \text{w}\right) \times \text{v} = 0$

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:

$\begin{array}{lll}\text{w} \times \text{w}\hfill & = \hfill & \left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 1& 1& 3\\ 1& 1& 3\end{array}\right|\hfill \\ \hfill & = \hfill & \left(3 - 3\right)\text{i} - \left(3 - 3\right)\text{j} + \left(1 - 1\right)\text{k}\hfill \\ \hfill & = \hfill & 0\text{i} - 0\text{j} + 0\text{k}\hfill \\ \hfill & = \hfill & 0\hfill \end{array}$

$\begin{array}{lll}\left(\text{w} \times \text{w}\right) \times \text{v}\hfill & = \hfill & \left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& 0\\ - 3& 3& 2\end{array}\right|\hfill \\ \hfill & = \hfill & \text{0}\hfill \end{array}$