Chapter 9, Problem 92RE

To determine

To calculate: The angle and dot product of following vectors,

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

Answer to Problem 92RE

The dot product of vectors is $\text{v}\cdot \text{w}\text{=}2$ and angle between vectors is $\cos \theta =\frac{2}{3\sqrt{2}}$ .

Explanation of Solution

Given information:

The vectors are given as:

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

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,

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

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

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

Calculation:

Consider the vectors,

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

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 ${\text{a}}_{\text{1}}{\text{=1,b}}_{\text{1}}{\text{=-1,c}}_{\text{1}}{\text{=1,a}}_{\text{2}}{\text{=2,b}}_{\text{2}}\text{=1}\text{and}\text{}{\text{c}}_{\text{2}}\text{=1}$

Therefore, the dot product is given by,

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

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,

  $\begin{array}{c}\cos \theta =\text{}\frac{\text{v}\cdot \text{w}}{\Vert \text{v}\Vert \Vert \text{w}\Vert }\\ \cos \theta =\text{}\frac{\left(1\right)\cdot \left(2\right)+\left(-1\right)\cdot \left(1\right)+\left(1\right)\cdot \left(1\right)}{\sqrt{{\left(1\right)}^2\text{+}{\left(\text{-1}\right)}^2\text{+}{\left(\text{1}\right)}^2}\sqrt{{\left(2\right)}^2\text{+}{\left(\text{1}\right)}^2\text{+}{\left(\text{1}\right)}^2}}\\ \cos \theta =\frac{2}{\sqrt{6}\cdot \sqrt{3}}\\ \cos \theta =\frac{2}{3\sqrt{2}}\end{array}$

Thus, the dot product of vectors is $\text{v}\cdot \text{w}\text{=}2$ and angle between vectors is $\cos \theta =\frac{2}{3\sqrt{2}}$ .