Chapter 9, Problem 3T

(a)

To determine

The graph of vector $u=\langle -4\sqrt{3},4\rangle$ in coordinate plane with initial point $\left(0,0\right)$.

Explanation of Solution

The initial point is $\left(0,0\right)$ for the vector $u=\langle -4\sqrt{3},4\rangle$, therefore the terminal point of the vector $u=\langle -4\sqrt{3},4\rangle$ is $\left(-4\sqrt{3},4\right)$.

The graph of vector $u=\langle -4\sqrt{3},4\rangle$ with initial point $\left(0,0\right)$ and terminal point $\left(-4\sqrt{3},4\right)$ is given below,

Figure (1)

Thus, the Figure (1) shows the graph of vector $u=\langle -4\sqrt{3},4\rangle$ in coordinate plane with initial point $\left(0,0\right)$.

(b)

To determine

To evaluate: The length and direction of the vector u.

Answer to Problem 3T

The length of vector u is 8 and the direction of u is $150°$.

Explanation of Solution

Given:

The vector u is given below,

$\langle -4\sqrt{3},4\rangle$

Formula used:

(a). The magnitude and length of a vector $v=\langle a_1,a_2\rangle$ is given below,

$\left|v\right|=\sqrt{a_1^2+a_2^2}$ (1)

(b). The formula to calculate the direction of a vector $v=\langle a_1,a_2\rangle$ is given below,

$\tan \theta =\frac{a_2}{a_1}$ (2)

Where,

• $\theta$ is the direction of the vector v.

Calculation:

Section (a):

Substitute $-4\sqrt{3}$ for $a_1$ and 4 for $a_2$ in equation (1), to find the length of the vector u,

$\begin{array}{c}\left|u\right|=\sqrt{{\left(-4\sqrt{3}\right)}^2+{\left(4\right)}^2}\\ =\sqrt{{\left(4\right)}^2{\left(\sqrt{3}\right)}^2+16}\\ =\sqrt{16\left(3\right)+16}\end{array}$

Simplify the above expression,

$\begin{array}{c}\left|u\right|=\sqrt{48+16}\\ =\sqrt{64}\\ =8\end{array}$

Thus, the length of the vector u is 8.

Section (b):

Substitute $-4\sqrt{3}$ for $a_1$ and 4 for $a_2$ in equation (2), to find the direction of the vector u,

$\begin{array}{c}\tan \theta =\frac{4}{-4\sqrt{3}}\\ =-\frac{1}{\sqrt{3}}\end{array}$

From the above expression the reference angle $\theta$ is $\frac{\pi }{6}$ and the vector u lies in quadrant (II), therefore the direction of the vector u is given below,

$\begin{array}{c}\theta =\pi -\frac{\pi }{6}\\ =\frac{5\pi }{6}\end{array}$

The direction of vector u in degree is given below,

$\begin{array}{c}\theta =\frac{5\pi }{6}\times \frac{180}{\pi }\\ =150°\end{array}$

Thus, the direction of the vector u is $150°$.