Chapter 6.4, Problem 92E

a.

To determine

The type of angle between two vectors when $\overrightarrow{u}\cdot \overrightarrow{v}=0$ .

Answer to Problem 92E

  ${90}^0$

Explanation of Solution

Given:

The vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ are shown in Figure-1 here with the condition $\overrightarrow{u}\cdot \overrightarrow{v}=0$ .

  

Formula/ concept used:

The scalar product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\cdot \left|\overrightarrow{b}\right|\cos \theta$ , $\theta$ angle between them

Calculations:

The scalar product of two perpendicular vectors $\overrightarrow{u}$ and $\overrightarrow{v}$

  $\overrightarrow{u}\cdot \overrightarrow{v}=\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\cos {90}^0$

  $\Rightarrow \overrightarrow{u}\cdot \overrightarrow{v}=\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\times 0=0$ . That is the given condition.

Hence, the angle between vectors in this case is right angle equals ${90}^0$ .

b.

To determine

The type of angle between two vectors when $\overrightarrow{u}\cdot \overrightarrow{v}>0$ .

Answer to Problem 92E

  $\left(0^0<\theta <{90}^0\right)$

Explanation of Solution

Given:

The vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ are shown in part (a) with the condition $\overrightarrow{u}\cdot \overrightarrow{v}>0$ .

Formula/ concept used:

The scalar product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\cdot \left|\overrightarrow{b}\right|\cos \theta$ , $\theta$ angle between them

Calculations:

The scalar product of two perpendicular vectors $\overrightarrow{u}$ and $\overrightarrow{v}$

  $\overrightarrow{u}\cdot \overrightarrow{v}=\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\cos \theta$

Now, $\overrightarrow{u}\cdot \overrightarrow{v}>0$

  $\Rightarrow \left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\cos \theta >0$ .

  $\Rightarrow \cos \theta >0$ , as $\left|\overrightarrow{u}\right|>0$ and $\left|\overrightarrow{v}\right|>0$

  $\Rightarrow 0^0<\theta <{90}^0$ .

Hence, the angle between vectors in this case is acute angle $\left(0^0<\theta <{90}^0\right)$ .

Conclusion:

Therefore, the angle between vectors is acute angle $\left(0^0<\theta <{90}^0\right)$ .

c.

To determine

The type of angle between two vectors $\overrightarrow{u}\cdot \overrightarrow{v}<0$ .

Answer to Problem 92E

Obtuse angle $\left({90}^0<\theta <{180}^0\right)$ .

Explanation of Solution

Given:

The vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ are shown in part (a) with the condition $\overrightarrow{u}\cdot \overrightarrow{v}<0$ .

  

Formula/ concept used:

The scalar product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\cdot \left|\overrightarrow{b}\right|\cos \theta$ , $\theta$ angle between them

Calculations:

The scalar product of two perpendicular vectors $\overrightarrow{u}$ and $\overrightarrow{v}$

  $\overrightarrow{u}\cdot \overrightarrow{v}=\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\cos \theta$

Now, $\overrightarrow{u}\cdot \overrightarrow{v}<0$

  $\Rightarrow \left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\cos \theta <0$ .

  $\Rightarrow \cos \theta <0$ , as $\left|\overrightarrow{u}\right|>0$ and $\left|\overrightarrow{v}\right|>0$

  $\Rightarrow {90}^0<\theta <{180}^0$ .

Hence, the angle between vectors in this case is obtuse angle ${90}^0<\theta <{180}^0$ .

Conclusion:

Therefore, the angle between vectors is obtuse angle $\left({90}^0<\theta <{180}^0\right)$ .