Chapter 6.2, Problem 21E

Solution

To Find:

The angle between the given vectors using an algebraic method. Approximate the exact answers using a calculator when needed.

  ${90.97}^{°}$ approximately.

Given:

The vectors $u$ and $v$:

  

Concepts Used:

For the points $A=\left(x_1,y_1\right)$ and $B=\left(x_2,y_2\right)$ , the vector with initial point $A$ and terminal point $B$ is represented in component form as $〈x_2-x_1,y_2-y_1〉$ .

If $u=〈a,b〉$ then $\left|u\right|=\sqrt{a^2+b^2}$ .

The dot product of the vectors $u=〈u_1,u_2〉$ and $v=〈v_1,v_2〉$ is defined as:

  $u\cdot v=u_1{v}_1+u_2{v}_2$

The smaller angle $\theta$ between two vectors $u$ and $v$ is given by ${\cos }^{-1}\left(\frac{u\cdot v}{\left|u\right|\left|v\right|}\right)$

Calculations:

Determine the component form of the vector $u$ . Observe that it has an initial point $\left(0,0\right)$ and terminal point $\left(8,5\right)$ . Write it in component form:

  $\begin{array}{c}u=〈8-0,5-0〉\\ =〈8,5〉\end{array}$

Determine the component form of the vector $v$ . Observe that it has an initial point $\left(0,0\right)$ and terminal point $\left(-3,4\right)$ . Write it in component form:

  $\begin{array}{c}v=〈-3-0,4-0〉\\ =〈-3,4〉\end{array}$

Determine the angle between the vectors $u$ and $v$ .

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{u\cdot v}{\left|u\right|\left|v\right|}\right)\\ ={\cos }^{-1}\left(\frac{〈8,5〉\cdot 〈-3,4〉}{\left|〈8,5〉\right|\left|〈-3,4〉\right|}\right)\\ ={\cos }^{-1}\left(\frac{\left(8\right)\left(-3\right)+\left(5\right)\left(4\right)}{\sqrt{{\left(8\right)}^2+{\left(5\right)}^2}\sqrt{{\left(-3\right)}^2+{\left(4\right)}^2}}\right)\\ ={\cos }^{-1}\left(\frac{-24+20}{\sqrt{64+25}\sqrt{9+16}}\right)\\ ={\cos }^{-1}\left(\frac{-4}{5\sqrt{89}}\right)\\ \approx {90.97}^{°}\end{array}$

Conclusion:

The angle between the given vectors is $\approx {90.97}^{°}$ .