Chapter 6.4, Problem 32E

To determine

To find the angle $\theta$ between the vectors.

Answer to Problem 32E

The angle between both vectors is $\theta =0.124362$ radians.

Explanation of Solution

Given information:

Vectors are $u=2i-3j\text{and}v=i-2j$ .

Used concept:

If $\theta$ is the angle between two nonzero vectors $a=a_{1}i+a_{2}j\text{and}b=b_{1}i+b_{2}j$, then

  $\begin{array}{l}\cos \theta =\frac{a\cdot b}{\left|a\right|\left|b\right|}\\ \cos \theta =\frac{\left(a_{1}i+a_{2}j\right)\cdot \left(b_{1}i+b_{2}j\right)}{\left|a\right|\left|b\right|}\\ \cos \theta =\frac{a_1\cdot b_1+a_2\cdot b_2}{\sqrt{a_1{}^2+a_2{}^2}\cdot \sqrt{b_1{}^2+b_2{}^2}}\end{array}$

Calculation:

  $u=2i-3j\text{and}v=i-2j$ .

Now from the above concept, the angle is

  $\begin{array}{l}\cos \theta =\frac{u\cdot v}{\left|u\right|\left|v\right|}\\ \cos \theta =\frac{\left(2i-3j\right)\cdot \left(i-2j\right)}{\sqrt{2^2+{\left(-3\right)}^2}\cdot \sqrt{1^2+{\left(-2\right)}^2}}\\ \cos \theta =\frac{\left(2\cdot 1\right)+\left(-3\cdot -2\right)}{\sqrt{4+9}\cdot \sqrt{1+4}}\\ \cos \theta =\frac{2+6}{\sqrt{13}\cdot \sqrt{5}}\end{array}$

Simplify,

  $\begin{array}{l}\cos \theta =\frac{2+6}{\sqrt{13}\cdot \sqrt{5}}\\ \cos \theta =\frac{8}{\sqrt{65}}\\ \cos \theta =\frac{8}{8.062257748}\\ \cos \theta =0.992277\end{array}$

And,

  $\begin{array}{c}\cos \theta =0.992277\\ \theta =\arccos \left(0.992277\right)\\ \theta =0.124362\end{array}$

So, the angle between both vectors is $\theta =0.124362$ radians.