Chapter 6.4, Problem 33E

To determine

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

Answer to Problem 33E

The angle between vectors is $\theta =0.0$ radians.

Explanation of Solution

Given information:

Vectors are $u=2i-j\text{and}v=6i-3j$ .

Used concept:

If $\theta$ is the angle between two nonzero vectors $u=2i-j\text{and}v=6i-3j$, 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:

  $u=2i-j\text{and}v=6i-3j$ .

Now from the above concept, the angle is

  $\begin{array}{l}\cos \theta =\frac{u\cdot v}{\left|u\right|\left|v\right|}\\ \cos \theta =\frac{\left(2i-j\right)\cdot \left(6i-3j\right)}{\sqrt{2^2+{\left(-1\right)}^2}\cdot \sqrt{6^2+{\left(-3\right)}^2}}\\ \cos \theta =\frac{\left(2\cdot 6\right)+\left(-1\cdot -3\right)}{\sqrt{4+1}\cdot \sqrt{36+9}}\\ \cos \theta =\frac{12+3}{\sqrt{5}\cdot \sqrt{45}}\end{array}$

Simplify,

  $\begin{array}{l}\cos \theta =\frac{12+3}{\sqrt{5}\cdot \sqrt{45}}\\ \cos \theta =\frac{15}{\sqrt{225}}\\ \cos \theta =\frac{15}{\sqrt{{15}^2}}\\ \cos \theta =\frac{15}{15}\end{array}$

Cancel out common terms,

  $\begin{array}{c}\cos \theta =1\\ \theta =\arccos \left(1\right)\\ \theta =0.0\end{array}$

So, the angle between vectors is $\theta =0.0$ radians.