Chapter 14.2, Problem 16WE

a.

To determine

To find: The vector w in component form.

Answer to Problem 16WE

The vector w in component form is $-4i-3j$ .

Explanation of Solution

Given:

The following vectors:

  $\begin{array}{c}u=i-j\\ v=5i+2j\\ w=u-v\end{array}$

Calculation:

Substitute the values of vectors u and v in the expression $u-v$ to determine the value of vector w.

  $\begin{array}{c}w=u-v\\ =i-j-\left(5i+2j\right)\\ =\left(1-5\right)i+\left(-1-2\right)j\\ =-4i-3j\end{array}$

So, vector w in component form is $-4i-3j$ .

b.

To determine

To draw: A diagram showing the vectors u, v and w.

Answer to Problem 16WE

Figure 1 represents the diagram showing the vectors u, v and w.

Explanation of Solution

Given:

The following vectors:

  $\begin{array}{c}u=i-j\\ v=5i+2j\\ w=u-v\end{array}$

Calculation:

The diagram showing the vectors u, v and w is given below:

  

Figure 1

c.

To determine

To find: The value of $\Vert w\Vert$ .

Answer to Problem 16WE

The value of $\Vert w\Vert$ is 5.

Explanation of Solution

Given:

The vector $w=-4i-3j$

Calculation:

For a vector $z=ai+bj$ , the value of $\Vert z\Vert$ is given by the formula $\Vert z\Vert =\sqrt{a^2+b^2}$ .

Substitute -4 for a and -3 for b in the formula $\Vert w\Vert =\sqrt{a^2+b^2}$ to determine the value of $\Vert w\Vert$ .

  $\begin{array}{c}\Vert w\Vert =\sqrt{{\left(-4\right)}^2+{\left(-3\right)}^2}\\ =\sqrt{16+9}\\ =\sqrt{25}\\ =5\end{array}$

d.

To determine

To find: The angle $\gamma$ , $0°\le \gamma \le 360°$ , that w makes with the positive x -axis.

Answer to Problem 16WE

The angle $\gamma$ is $216.87°$ .

Explanation of Solution

Given:

The vector $w=-4i-3j$

Calculation:

For a vector $z=ai+bj$ , the angle $\gamma$ that vector z makes with the positive x -axis is calculated using the formula $\gamma ={\tan }^{-1}\frac{b}{a}$ .

Substitute -4 for a and -3 for b in the formula $\gamma ={\tan }^{-1}\frac{b}{a}$ to determine the value of $\gamma$ .

  $\begin{array}{c}\gamma ={\tan }^{-1}\frac{-3}{-4}\\ ={\tan }^{-1}\frac{3}{4}\\ \approx 36.87°\end{array}$

Since vector w is in the negative y direction, value of $\gamma$ becomes $180°+36.87°=216.87°$ .