Chapter 9, Problem 93RE

To determine

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

  $\text{v}=4\text{i}-\text{ j}\text{+ 2k,}\text{w = i}\text{}-2\text{j}\text{}\text{}-3\text{k}$

Answer to Problem 93RE

The dot product of vectors is $\text{v}\cdot \text{w}\text{=}0$ and angle between vectors is $\cos \theta =0$ .

Explanation of Solution

Given information:

The vectors are given as:

  $\text{v}=4\text{i}-\text{ j}\text{+ 2k,}\text{w = i}\text{}-2\text{j}\text{}\text{}-3\text{k}$

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,

  $\text{v}=4\text{i}-\text{ j}\text{+ 2k,}\text{w = i}\text{}-2\text{j}\text{}\text{}-3\text{k}$

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 ${\text{a}}_{\text{1}}{\text{=4,b}}_{\text{1}}{\text{=-1,c}}_{\text{1}}{\text{=2,a}}_{\text{2}}{\text{=1,b}}_{\text{2}}\text{=-2}\text{and}\text{}{\text{c}}_{\text{2}}\text{=-3}$

Therefore, the dot product is given by,

  $\begin{array}{c}\text{v}\cdot \text{w}\text{=}{\text{a}}_1{\text{a}}_{\text{2}}\text{}{\text{+b}}_{\text{1}}{\text{b}}_{\text{2}}{\text{+c}}_{\text{1}}{\text{c}}_{\text{2}}\\ =4\cdot 1+-1\cdot -2+2\cdot -3\\ =4+2-6\\ =0\end{array}$

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,

  $\begin{array}{c}\cos \theta =\text{}\frac{\text{v}\cdot \text{w}}{\Vert \text{v}\Vert \Vert \text{w}\Vert }\\ \cos \theta =\text{}\frac{\left(4\right)\cdot \left(1\right)+\left(-1\right)\cdot \left(-2\right)+\left(2\right)\cdot \left(-3\right)}{\sqrt{{\left(4\right)}^2\text{+}{\left(\text{-1}\right)}^2\text{+}{\left(2\right)}^2}\sqrt{{\left(1\right)}^2\text{+}{\left(-2\right)}^2\text{+}{\left(-3\right)}^2}}\\ \cos \theta =\frac{0}{\sqrt{21}\cdot \sqrt{14}}\\ \cos \theta =0\end{array}$

Thus, the dot product of vectors is $\text{v}\cdot \text{w}\text{=}0$ and angle between vectors is $\cos \theta =0$ .