Chapter 6.4, Problem 42E

To determine

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

Answer to Problem 42E

The angle between vectors is $\theta =76.865977$ degree.

Explanation of Solution

Given information:

Vectors are $u=2i-3j\text{and}v=8i+3j$ .

Used concept:

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

  $u=2i-3j\text{and}v=8i+3j$ .

Now from the above concept, the angle is

  $\begin{array}{l}\cos \theta =\frac{u\cdot v}{\left|u\right|\left|v\right|}\\ \cos \theta =\frac{\left(2i-3j\right)\cdot \left(8i+3j\right)}{\sqrt{2^2+{\left(-3\right)}^2}\cdot \sqrt{{\left(8\right)}^2+3^2}}\\ \cos \theta =\frac{\left(2\cdot 8\right)+\left(-3\cdot 3\right)}{\sqrt{4+9}\cdot \sqrt{64+9}}\\ \cos \theta =\frac{16-9}{\sqrt{13}\cdot \sqrt{73}}\end{array}$

And, now

  $\begin{array}{c}\cos \theta =\frac{16-9}{\sqrt{13}\cdot \sqrt{73}}\\ \cos \theta =\frac{7}{\left(3.60555\right)\cdot \left(8.54400\right)}\\ \cos \theta =\frac{7}{30.8058}\\ \cos \theta =0.227229\\ \theta =\arccos \left(0.227229\right)\\ \theta =76.865977\end{array}$

So, the angle between vectors is $\theta =76.865977$ degree.