Chapter 9.2, Problem 6E

a.

To determine

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

Answer to Problem 6E

The desired value is 0 .

Explanation of Solution

Given :

  $\begin{array}{l}\text{u = i + }\sqrt{\text{3}}\text{j}\\ \text{v = }-\sqrt{\text{3}}\text{i + j}\end{array}$

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,

  $\begin{array}{l}\text{u}\text{.v = }\left(1\right)\left(-\sqrt{3}\right)+\left(\sqrt{3}\right)\left(1\right)\\ \text{u}\text{.v = }-\sqrt{3}+\sqrt{3}=0\end{array}$

b.

To determine

The angle between u and v.

Answer to Problem 6E

The desired value is $\text{θ =}90°$ .

Explanation of Solution

Given :

  $\begin{array}{l}\text{u = i + }\sqrt{\text{3}}\text{j}\\ \text{v = }-\sqrt{\text{3}}\text{i + j}\end{array}$

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,

  $\begin{array}{l}\text{u}\text{.v = }\left(1\right)\left(-\sqrt{3}\right)+\left(\sqrt{3}\right)\left(1\right)\\ \text{u}\text{.v = }-\sqrt{3}+\sqrt{3}=0\end{array}$

And,

  $\begin{array}{l}\left|\text{u}\right|=2\\ \left|\text{v}\right|=2\\ \cos \text{θ = }\frac{\text{0}}{4}=0\\ \text{θ = }{\cos }^{-1}\left(0\right)\\ \text{θ =}90°\end{array}$