Chapter 6.4, Problem 35E

To determine

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

Answer to Problem 35E

The angle between vectors is $\theta =0.835481$ radians.

Explanation of Solution

Given information:

Vectors are $u=-6i-3j\text{and}v=-8i+4j$ .

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:

  $u=-6i-3j\text{and}v=-8i+4j$ .

Now from the above concept, the angle is

  $\begin{array}{l}\cos \theta =\frac{u\cdot v}{\left|u\right|\left|v\right|}\\ \cos \theta =\frac{\left(-6i-3j\right)\cdot \left(-8i+4j\right)}{\sqrt{{\left(-6\right)}^2+{\left(-3\right)}^2}\cdot \sqrt{{\left(-8\right)}^2+4^2}}\\ \cos \theta =\frac{\left(-6\cdot -8\right)+\left(-3\cdot 4\right)}{\sqrt{36+9}\cdot \sqrt{48+16}}\\ \cos \theta =\frac{48-12}{\sqrt{45}\cdot \sqrt{64}}\\ \cos \theta =\frac{36}{\sqrt{9\cdot 5}\cdot 8}\end{array}$

Cancel out common terms,

  $\begin{array}{l}\cos \theta =\frac{36}{\sqrt{9\cdot 5}\cdot 8}\\ \cos \theta =\frac{9}{\sqrt{9\cdot 5}\cdot 2}\\ \cos \theta =\frac{9}{2\cdot 3\sqrt{5}}\\ \cos \theta =\frac{3}{2\sqrt{5}}\end{array}$

And, now

  $\begin{array}{l}\theta =\arccos \left(\frac{3}{2\sqrt{5}}\right)\\ \theta =\arccos \left(\frac{3}{2\left(2.236067\right)}\right)\\ \theta =\arccos \left(\frac{3}{4.472135}\right)\\ \theta =\arccos \left(0.670820\right)\\ \theta =0.835481\end{array}$

So, the angle between vectors is $\theta =0.835481$ radians.