Chapter 11, Problem 36RE

To determine

Find the angle $\theta$ between the given vectors.

Answer to Problem 36RE

The angle $\theta$ between $\text{u}=\langle 3,1,–1\rangle$ and $\text{v}=\langle 4,5,2\rangle$ is $\theta =47.6°$

Explanation of Solution

Given:

The following vectors

  $\text{u}=\langle 3,1,–1\rangle$ and

  $\text{v}=\langle 4,5,2\rangle$

Consider a following vector a can be expressed as:

  $a=\langle x,y,z\rangle$

Then the magnitude of the vector a is:

  $\left|a\right|=\sqrt{x^2+y^2+z^2}$

Consider the vectors $u=\langle x_1,y_1,z_1\rangle$ and $v=\langle x_2,y_2,z_2\rangle$ then the angle between them is

  $cos\theta =\frac{\text{u}•\text{v}}{\Vert \text{u}\Vert \Vert \text{v}\Vert }$

Consider the following vectors

  $\text{u}=\langle 3,1,–1\rangle$ and

  $\text{v}=\langle 4,5,2\rangle$

Since $cos\theta =\frac{\text{u}•\text{v}}{\Vert \text{u}\Vert \Vert \text{v}\Vert }$ therefore, the angle $\theta$ between u and v is $\theta ={\cos }^{-1}\left(\frac{\text{u}•\text{v}}{\Vert \text{u}\Vert \Vert \text{v}\Vert }\right)$

To find the magnitude of $\text{u}=\langle 3,1,–1\rangle$ and $\text{v}=\langle 4,5,2\rangle$

  $\begin{array}{l}\left|\left|\text{u}\right|\right|=\sqrt{3^2+{\left(1\right)}^2+{\left(-1\right)}^2}\hfill \\ =\sqrt{9+1+1}\hfill \\ =\sqrt{11}\hfill \end{array}$

And

  $\begin{array}{l}\begin{array}{l}\left|\left|\text{v}\right|\right|=\sqrt{4^2+5^2+2^2}\hfill \\ =\sqrt{16+25+4}\hfill \end{array} \\ =\sqrt{45}\\ =3\sqrt{5}\end{array}$

Now to find the dot product of $\text{u}=\langle 3,1,–1\rangle$ and $\text{v}=\langle 4,5,2\rangle$

  $\begin{array}{l}\text{u•v}=\langle 3,1,–1\rangle •\langle 4,5,2\rangle \hfill \\ =3\left(4\right)+\left(1\right)\left(5\right)+\left(–1\right)\left(2\right)\hfill \\ =12+5–2\hfill \\ =15\hfill \end{array}$

Find,

  $\begin{array}{l}\frac{\text{u}•\text{v}}{\Vert \text{u}\Vert \Vert \text{v}\Vert }=\frac{15}{3\sqrt{5}\sqrt{11}}\\ =\frac{15}{3\sqrt{55}}\\ =\frac{15\sqrt{55}}{165}\\ =\frac{\sqrt{55}}{11}\end{array}$

And finally to find the angle $\theta$ between $\text{u}=\langle 3,1,–1\rangle$ and $\text{v}=\langle 4,5,2\rangle$

  $\begin{array}{l}\theta ={\cos }^{-1}\left(\frac{\text{u}•\text{v}}{\Vert \text{u}\Vert \Vert \text{v}\Vert }\right)\\ ={\cos }^{-1}\left(\frac{\sqrt{55}}{11}\right)\\ \theta =47.6°\end{array}$

Hence, the angle $\theta$ between $\text{u}=\langle 3,1,–1\rangle$ and $\text{v}=\langle 4,5,2\rangle$ is $\theta =47.6°$