Chapter 9.7, Problem 44AYU

To determine

To find: The vector $\text{u} \times \left(\text{j} + \text{k}\right)$ which is orthogonal to both

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

Answer to Problem 44AYU

Solution:

The vector $- 4\text{i} - 2\text{j} + 2\text{k}$ is orthogonal to both $\text{u}$ and $\text{j} + \text{k}$,

Explanation of Solution

Given:

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

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 $\text{u} \times \left(\text{j} + \text{k}\right)$

$\begin{array}{lll}\text{u} \times \left(\text{j} + \text{k}\right)\hfill & = \hfill & \left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 2& - 3& 1\\ 0& 1& 1\end{array}\right|\hfill \\ \hfill & = \hfill & \left(- 3 - 1\right)\text{i} - \left(2 - 0\right)\text{j} + \left(2 - 0\right)\text{k}\hfill \\ \hfill & = \hfill & - 4\text{i} - 2\text{j} + 2\text{k}\hfill \end{array}$

The vector $- 4\text{i} - 2\text{j} + 2\text{k}$ is orthogonal to both $\text{u}$ and $\text{j} + \text{k}$,

Check:

Two vectors are orthogonal if theit dot product is zero,

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

$\begin{array}{lll}\text{u} \cdot \left(\text{u} \times \left(\text{j} + \text{k}\right)\right)\hfill & = \hfill & \left(2\text{i} - 3\text{j} + \text{k}\right) \cdot \left(- 4\text{i} - 2\text{j} + 2\text{k}\right)\hfill \\ \hfill & = \hfill & - 8 + 6 + 2 = 0\hfill \\ \left(\text{j} + \text{k}\right).\left(\text{u} \times \left(\text{j} + \text{k}\right)\right)\hfill & = \hfill & \left(\text{j} + \text{k}\right) \cdot \left(- 4\text{i} - 2\text{j} + 2\text{k}\right)\hfill \\ \hfill & = \hfill & 0 - 2 + 2 = 0\hfill \end{array}$