Chapter 11, Problem 10CT

To determine

Determine whether u and v are orthogonal, parallel or neither.

Answer to Problem 10CT

The vectors $\text{u}=\langle 4,–1,6\rangle$ and $\text{v}=\langle –2,\frac{1}{2},–3\rangle$ are not orthogonal

Explanation of Solution

Given:

The vectors

  $\text{u}=\langle 4,–1,6\rangle$ and $\text{v}=\langle –2,\frac{1}{2},–3\rangle$

Consider the two vectors $\text{u}=\langle x_1,y_1,z_1\rangle$ and $\text{v}=\langle x_2,y_2,z_2\rangle$ then the dot product of u and v is

  $\begin{array}{l}\text{u•v}=\langle x_1,y_1,z_1\rangle •\langle x_2,y_2,z_2\rangle \hfill \\ =x_1\left(x_2\right)+y_1{\left(y_2\right)}_{}+z_1\left(z_2\right)\hfill \end{array}$

The two vectors $\text{u}=\langle x_1,y_1,z_1\rangle$ and $\text{v}=\langle x_2,y_2,z_2\rangle$ are parallel to each other as they have the relation $\text{u=cv}$ where c is constant.

The vectors $\text{u}=\langle x_1,y_1,z_1\rangle$ and $\text{v}=\langle x_2,y_2,z_2\rangle$ are orthogonal to each other if the dot product of $\text{u}=\langle x_1,y_1,z_1\rangle$ and $\text{v}=\langle x_2,y_2,z_2\rangle$ is 0 that is if $\text{u•v}=0$

Consider the following vectors

  $\text{u}=\langle 4,–1,6\rangle$ and $\text{v}=\langle –2,\frac{1}{2},–3\rangle$

The above two vectors u and v are parallel to each other as they have the relation u=cv where c is some constant.

Since,

  $\begin{array}{l}\text{u}=\langle 4,–1,6\rangle \hfill \\ =–2\langle –2,\frac{1}{2},–3\rangle \hfill \\ =–2\text{v}\hfill \end{array}$

Therefore,

  $\text{u= –2v}$

And hence the vectors $\text{u}=\langle 4,–1,6\rangle$ and $\text{v}=\langle –2,\frac{1}{2},–3\rangle$ are parallel

To check whether the vectors are orthogonal or not, first compute the dot product of u and v and if it is 0 then u and v will be orthogonal to each other but in this problem the dot product is not equal to 0 as

  $\begin{array}{l}\text{u•v}=\langle 4,–1,6\rangle •\langle –2,\frac{1}{2},–3\rangle \hfill \\ =\left(4\right)\left(–2\right)+\left(–1\right)\frac{1}{2}+\left(6\right)\left(–3\right)\hfill \end{array}$

Simplify,

  $\begin{array}{l}=–8–\frac{1}{2}–18\hfill \\ =\frac{-16-1-36}{2}\hfill \\ =-\frac{53}{2}\hfill \\ \ne 0\hfill \end{array}$

Hence the vectors $\text{u}=\langle 4,–1,6\rangle$ and $\text{v}=\langle –2,\frac{1}{2},–3\rangle$ are not orthogonal.