Chapter 9, Problem 96RE

To determine

To calculate: whether the vectors are parallel, orthogonal, or neither of following vectors,

  $\text{v}=-2\text{i}-\text{j}\text{,}\text{w = 2i}\text{}+\text{j}\text{}\text{}$

Answer to Problem 96RE

The vectors $\text{v}=-2\text{i}-\text{j}\text{,}\text{w = 2i}\text{}+\text{j}\text{}\text{}$ are parallel.

Explanation of Solution

Given information:

The vectors are given as:

  $\text{v}=-2\text{i}-\text{j}\text{,}\text{w = 2i}\text{}+\text{j}\text{}\text{}$

Formula used:

let $\text{v}={\text{a}}_{\text{1}}\text{i}{\text{+ b}}_{\text{1}}{\text{j+ c}}_{\text{1}}\text{k}$ and ${\text{w = a}}_{\text{2}}\text{i}\text{}\text{+}\text{}{\text{b}}_{\text{2}}\text{j}\text{}\text{}\text{+}{\text{c}}_{\text{3}}\text{k}$ are two vectors,

then,

Two vectors $v\text{}\text{and}\text{}\text{}w$ are parallel if there exit a nonzero $\alpha$ such that $v=\alpha w$

If Two vectors $v\text{}\text{and}\text{}\text{}w$ parallel the angle between $v\text{}\text{and}\text{}\text{}w$ is $0\text{or}\text{}\pi$

angle: $\cos \theta =\text{}\frac{\text{v}\cdot \text{w}}{\Vert \text{v}\Vert \Vert \text{w}\Vert }$

where $\Vert \text{v}\Vert \text{=}\sqrt{{\text{a}}_{\text{1}}^{\text{2}}{\text{+b}}_{\text{1}}^{\text{2}}{\text{+c}}_{\text{1}}^{\text{2}}},\Vert \text{w}\Vert =\sqrt{{\text{a}}_2^{\text{2}}{\text{+b}}_2^{\text{2}}{\text{+c}}_2^{\text{2}}}$

dot product: $\text{v}\cdot \text{w}\text{=}{\text{a}}_1{\text{a}}_{\text{2}}\text{}{\text{+b}}_{\text{1}}{\text{b}}_{\text{2}}{\text{+c}}_{\text{1}}{\text{c}}_{\text{2}}$

Two vectors $v\text{}\text{and}\text{}\text{}w$ are orthogonal if and only if $v\text{}\cdot w=0$

Calculation:

Consider the vectors,

  $\text{v}=-2\text{i}-\text{j}\text{,}\text{w = 2i}\text{}+\text{j}\text{}\text{}$

Two vectors $v\text{}\text{and}\text{}\text{}w$ are parallel if there exit a nonzero $\alpha$ such that $v=\alpha w$ So,

  $\begin{array}{c}-2\text{i}-\text{j}=\alpha \left(\text{2i}\text{}+\text{j}\text{}\text{}\right)\\ -2\text{i}-\text{j}=-1\left(\text{2i}\text{}+\text{j}\text{}\text{}\right)\\ -2\text{i}-\text{j}=-2\text{i}-\text{j}\end{array}$

Since, $v=-1w$

Further,

Recall that for the dot product of vectors,

  $\text{v}\cdot \text{w}\text{=}{\text{a}}_1{\text{a}}_{\text{2}}\text{}{\text{+b}}_{\text{1}}{\text{b}}_{\text{2}}{\text{+c}}_{\text{1}}{\text{c}}_{\text{2}}$

The value of ${\text{a}}_{\text{1}}{\text{=-2,b}}_{\text{1}}{\text{=-1,c}}_{\text{1}}{\text{=0,a}}_{\text{2}}{\text{=2,b}}_{\text{2}}\text{=1}\text{and}\text{}{\text{c}}_{\text{2}}\text{=0}$

Therefore, the dot product is given by,

  $\begin{array}{c}\text{v}\cdot \text{w}\text{=}{\text{a}}_1{\text{a}}_{\text{2}}\text{}{\text{+b}}_{\text{1}}{\text{b}}_{\text{2}}{\text{+c}}_{\text{1}}{\text{c}}_{\text{2}}\\ =-2\cdot 2+-1\cdot 1+0\cdot 0\\ =-4-1\\ =-5\end{array}$

Recall that for the angle between the vectors,

  $\cos \theta =\text{}\frac{\text{v}\cdot \text{w}}{\Vert \text{v}\Vert \Vert \text{w}\Vert }$

where $\Vert \text{v}\Vert \text{=}\sqrt{{\text{a}}_{\text{1}}^{\text{2}}{\text{+b}}_{\text{1}}^{\text{2}}{\text{+c}}_{\text{1}}^{\text{2}}},\Vert \text{w}\Vert =\sqrt{{\text{a}}_2^{\text{2}}{\text{+b}}_2^{\text{2}}{\text{+c}}_2^{\text{2}}}$

Therefore, the angle between the vectors are given by,

  $\begin{array}{c}\cos \theta =\text{}\frac{\text{v}\cdot \text{w}}{\Vert \text{v}\Vert \Vert \text{w}\Vert }\\ \cos \theta =\text{}\frac{\left(-2\right)\cdot \left(2\right)+\left(-1\right)\cdot \left(1\right)+\left(0\right)\cdot \left(0\right)}{\sqrt{{\left(-2\right)}^2\text{+}{\left(-1\right)}^2\text{+}{\left(0\right)}^2}\sqrt{{\left(2\right)}^2\text{+}{\left(1\right)}^2\text{+}{\left(0\right)}^2}}\\ \cos \theta =\frac{-5}{\sqrt{5}\cdot \sqrt{5}}\\ \cos \theta =-1\end{array}$

So, the value of $\theta \text{}\text{is}\pi$

Therefor vectors $\text{v}=-2\text{i}-\text{j}\text{,}\text{w = 2i}\text{}+\text{j}\text{}\text{}$ are parallel.

Two vectors $v\text{}\text{and}\text{}\text{}w$ are orthogonal if and only if $v\text{}\cdot w=0$ since, dot product of vectors is not 0 so the vectors are not orthogonal.