Chapter 10, Problem 11CT

To determine

To check: The vectors $\text{u}=\text{i}-2\text{j}-\text{k}$ and $\text{v}=\text{j}+6\text{k}$ are orthogonal, parallel or neither.

Answer to Problem 11CT

The vectors $\text{u}=\text{i}-2\text{j}-\text{k}$ and $\text{v}=\text{j}+6\text{k}$ neither parallel nor orthogonal.

Explanation of Solution

Given information:

The vectors are $\text{u}=\text{i}-2\text{j}-\text{k}$ and $\text{v}=\text{j}+6\text{k}$ .

Calculation:

To check that two vectors $u$ and $v$ are parallel, check that $\text{v}=c\text{u}$ , for any value of $c$ .

Both the vectors $\text{u}=\text{i}-2\text{j}-\text{k}$ and $\text{v}=\text{j}+6\text{k}$ are linearly independent. Therefore, $\text{v}=c\text{u}$ is not possible for any value of $c$ . So, the vectors $\text{u}=\text{i}-2\text{j}-\text{k}$ and $\text{v}=\text{j}+6\text{k}$ are not parallel.

Check that two vectors $u$ and $v$ are orthogonal, check that $\text{u}\cdot \text{v}=0$

Calculate the dot product of both the vectors $\text{u}=\text{i}-2\text{j}-\text{k}$ and $\text{v}=\text{j}+6\text{k}$ .

  $\begin{array}{c}u\cdot v=\left(\text{i}-2\text{j}-\text{k}\right)\cdot \left(\text{j}+6\text{k}\right)\\ =1\times 0+\left(-2\right)\times 1+\left(-1\right)\times 6\\ =0-2-6\\ =-8\end{array}$

As the dot product of both the vectors $\text{u}=\text{i}-2\text{j}-\text{k}$ and $\text{v}=\text{j}+6\text{k}$ is non-zero. So, the vectors $\text{u}=\text{i}-2\text{j}-\text{k}$ and $\text{v}=\text{j}+6\text{k}$ are not orthogonal.

Therefore, the vectors $\text{u}=\text{i}-2\text{j}-\text{k}$ and $\text{v}=\text{j}+6\text{k}$ neither parallel nor orthogonal.