Chapter 9.2, Problem 9E

To determine

The value of $\text{u}\text{.v}$ .

Answer to Problem 9E

The desired value is $\text{u}\text{.v =}-1$ .

Explanation of Solution

Given :

  $\begin{array}{l}\text{u = }\langle 3,-2\rangle \\ \text{v = }\langle 1,2\rangle \end{array}$

Calculation:

Definition of dot product:

  $\begin{array}{l}\text{u =}\langle {\text{a}}_{\text{1}}{\text{,b}}_{\text{1}}\rangle \text{and}\text{v =}\langle {\text{a}}_{\text{2}}{\text{,b}}_{\text{2}}\rangle \\ \text{u}{\text{.v = a}}_1{\text{a}}_2+{\text{b}}_{\text{1}}{\text{b}}_{\text{2}}\end{array}$

So,

  $\begin{array}{l}\text{u}\text{.v = }\left(3\right)\left(1\right)+\left(-2\right)\left(2\right)\\ \text{u}\text{.v = }3-4\\ \text{u}\text{.v =}-1\end{array}$

To determine

The angle between u and v.

Answer to Problem 9E

The desired value is $\text{θ =}97°$ .

Explanation of Solution

Given :

  $\begin{array}{l}\text{u = }\langle 3,-2\rangle \\ \text{v = }\langle 1,2\rangle \end{array}$

Calculation:

Angle between two vector is given by:

  $\cos \text{θ = }\frac{\text{u}\text{.v}}{\left|\text{u}\right|\left|\text{v}\right|}$

So,

  $\begin{array}{l}\text{u}\text{.v = }\left(3\right)\left(1\right)+\left(-2\right)\left(2\right)\\ \text{u}\text{.v = }3-4\\ \text{u}\text{.v =}-1\end{array}$

And,

  $\begin{array}{l}\left|\text{u}\right|=\sqrt{{\left(3\right)}^2+{\left(-2\right)}^2}=\sqrt{13}\\ \left|\text{v}\right|=\sqrt{1^2+2^2}=\sqrt{5}\\ \cos \text{θ = }\frac{\text{-1}}{\sqrt{13}\sqrt{5}}\\ \cos \text{θ}=\frac{-1}{8.06}\\ \text{θ = }{\cos }^{-1}\left(-0.124\right)\\ \text{θ =}97°\end{array}$