Chapter 11.2, Problem 46E

To determine

ToFind:The angle $\theta$ between the vectors $u=8j-20k$ and $v=10i-5k$

Answer to Problem 46E

Theangle between the given vectors is approximately $\theta \approx 1.14$

Explanation of Solution

Given:

The vectors $u=8j-20k$ and $v=10i-5k$

Concept Used:

If $\theta$ is the angle between two nonzero vectors $u$ and $v$ , then

  $\cos \left(\theta \right)=\frac{u·v}{\Vert u\Vert \Vert v\Vert }$

Calculation:

Given the vectors $u=8j-20k$ and $v=10i-5k$

That is

  $\begin{array}{c}u=0i+8j-20k\\ =\langle 0,8,-20\rangle \end{array}$

And

  $\begin{array}{c}v=10i+0j-5k\\ =\langle 10,0,-5\rangle \end{array}$

For and angle $\theta$ between two nonzero vectors $u$ and $v$ , we have $\cos \left(\theta \right)=\frac{u·v}{\Vert u\Vert \Vert v\Vert }$

Substituting the values of $u$ and $v$

  $\begin{array}{c}\cos \left(\theta \right)=\frac{\langle 0,8,-20\rangle ·\langle 10,0,-5\rangle }{\Vert \langle 0,8,-20\rangle \Vert \Vert \langle 10,0,-5\rangle \Vert }\\ =\frac{\left(0\right)\left(10\right)+\left(8\right)\left(0\right)+\left(-20\right)\left(-5\right)}{\left(\sqrt{{\left(0\right)}^2+{\left(8\right)}^2+{\left(-20\right)}^2}\right)\left(\sqrt{{\left(10\right)}^2+{\left(0\right)}^2+{\left(-5\right)}^2}\right)}\\ =\frac{0+0+100}{\left(\sqrt{0+64+400}\right)\left(\sqrt{100+0+25}\right)}\end{array}$

  $\begin{array}{c}=\frac{0+0+100}{\left(\sqrt{0+64+400}\right)\left(\sqrt{100+0+25}\right)}\\ =\frac{100}{\left(\sqrt{464}\right)\left(\sqrt{125}\right)}\\ =\frac{100}{\left(4\sqrt{29}\right)\left(5\sqrt{5}\right)}\\ =\frac{5}{\left(\sqrt{29}\right)\left(\sqrt{5}\right)}\end{array}$

  $=\frac{5}{\sqrt{145}}$

That is

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{5}{\sqrt{145}}\right)\\ \approx 1.14\end{array}$

Therefore, $\theta \approx 1.14$

Conclusion:

Theangle between the given vectors is approximately $\theta \approx 1.14$