Chapter 9, Problem 89RE

To determine

To calculate:The Dot product $v\cdot w$ and angle between $v$ and $w$ .To use the vectors $v=i-3j$ and

  $w=-i+j$ .

Answer to Problem 89RE

The Dot product $v\cdot w$ and angle between $v$ and $w$ is $-4$ . And the angle between them is $\theta ={\cos }^{-1}\left(\frac{-4}{\sqrt{2}\sqrt{10}}\right)$ .

Explanation of Solution

Given information:

The vectors:

  $v=i-3j$ and $w=-i+j$

Formula used:

The dot product of the vectors are solved by the formula: $v\cdot w=a_1{a}_2+b_1{b}_2+c_1{c}_2$ .

The formula for finding the angle between two vectors are:

  $\begin{array}{c}\cos \theta =\frac{v\cdot w}{\Vert v\Vert \Vert w\Vert }\\ \theta ={\cos }^{-1}\frac{v\cdot w}{\Vert v\Vert \Vert w\Vert }\end{array}$

The formula to find the magnitude of a vector is as:

  $\begin{array}{l}v=ai+bj+ck\\ \Vert v\Vert =\sqrt{a^2+b^2+c^2}\end{array}$

Calculation:

Consider the vectors:

  $v=i-3j$ and $w=-i+j$

First Take a dot product of $v$ and $w$ .

Now apply the Dot product formula of the matrix:

  $\begin{array}{c}v\cdot w=a_1{a}_2+b_1{b}_2+c_1{c}_2\\ v\cdot w=\left(i-3j\right)\cdot \left(-i+j\right)\\ =1\cdot -1+\left(-3\right)\cdot 1\\ =-1-3\\ =-4\end{array}$

Now apply the formula to find the angle between the both vectors $v$ and $w$ :

  $\begin{array}{c}\cos \theta =\frac{v\cdot w}{\Vert v\Vert \Vert w\Vert }\\ \theta ={\cos }^{-1}\frac{v\cdot w}{\Vert v\Vert \Vert w\Vert }\end{array}$

Here first find the value of magnitude of $\Vert v\Vert$ .

  $\begin{array}{c}v=i-3j\\ \Vert v\Vert =\sqrt{{\left(1\right)}^2+{\left(-3\right)}^2}\\ \Vert v\Vert =\sqrt{1+9}\\ =\sqrt{10}\end{array}$

Now find the value of magnitude of $\Vert w\Vert$ .

  $\begin{array}{c}w=-i+j\\ \Vert w\Vert =\sqrt{{\left(-1\right)}^2+{\left(1\right)}^2}\\ \Vert w\Vert =\sqrt{1+1}\\ \Vert w\Vert =\sqrt{2}\end{array}$

Now apply the values of magnitude of the both vectors $v$ and $w$ .

Apply the values in the formula:

  $\begin{array}{c}\cos \theta =\frac{v\cdot w}{\Vert v\Vert \Vert w\Vert }\\ =\frac{-4}{\sqrt{10}\cdot \sqrt{2}}\\ =\frac{-4}{\sqrt{2}\sqrt{10}}\end{array}$

Thus, The Dot product $v\cdot w$ and angle between $v$ and $w$ is $-4$ . And the angle between them is $\theta ={\cos }^{-1}\left(\frac{-4}{\sqrt{2}\sqrt{10}}\right)$ .