Chapter 9.7, Problem 42AYU

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

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

Find a vector orthogonal to both $u\text{ and }w$.

To determine

To find: The vector $\text{u} \times \text{w}$ which is orthogonal to both

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

Answer to Problem 42AYU

Solution:

The vector $- 10\text{i} - 5\text{j} + 5\text{k}$ is orthogonal to both $\text{u}$ and $\text{v}$

Explanation of Solution

Given:

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

Formula used:

Property:

Let $\text{u} = {\text{a}}_1\text{i} + {\text{b}}_1\text{j} + {\text{c}}_1\text{k}$, and $\text{v} = {\text{a}}_2\text{i} + {\text{b}}_2\text{j} + {\text{c}}_2\text{k}$ be two vectors in space, $\text{u} \times \text{v}$ is orthogonal to both $\text{u}$ and $\text{v}$

Calculation:

Based on property, such a vector is $\text{u} \times \text{w}$

$\begin{array}{lll}\text{u} \times \text{w}\hfill & = \hfill & \left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 2& - 3& 1\\ 1& 1& 3\end{array}\right|\hfill \\ \hfill & = \hfill & \left(- 9 - 1\right)\text{i} - \left(6 - 1\right)\text{j} + \left(2 + 3\right)\text{k}\hfill \\ \hfill & = \hfill & - 10\text{i} - 5\text{j} + 5\text{k}\hfill \end{array}$

The vector $- 10\text{i} - 5\text{j} + 5\text{k}$ is orthogonal to both $\text{u}$ and $\text{v}$

Check:

Two vectors are orthogonal if theit dot product is zero,

That is, we have to prove, $\text{u} \cdot \left(\text{u} \times \text{w}\right) = 0$ and $\text{w} \cdot \left(\text{u} \times \text{w}\right) = 0$

$\begin{array}{lll}\text{u} \cdot \left(\text{u} \times \text{w}\right)\hfill & = \hfill & \left(2\text{i} - 3\text{j} + \text{k}\right) \cdot \left(- 10\text{i} - 5\text{j} + 5\text{k}\right)\hfill \\ \hfill & = \hfill & - 20 + 15 + 5 = 0\hfill \\ \text{w} \cdot \left(\text{u} \times \text{w}\right)\hfill & = \hfill & \left(\text{i} + \text{j} + 3\text{k}\right) \cdot \left(- 10\text{i} - 5\text{j} + 5\text{k}\right)\hfill \\ \hfill & = \hfill & - 10 - 5 + 15 = 0\hfill \end{array}$