Chapter 10, Problem 5RE

To determine

To find: The component form of the vector obtained by rotating the vector $〈0,1〉$ through an angle of $\frac{2\pi }{3}$ radians.

Answer to Problem 5RE

The component form of the vector obtained by rotating the vector $〈0,1〉$ through an angle of $\frac{2\pi }{3}$ radians is $〈-\frac{\sqrt{3}}{2},-\frac{1}{2}〉$ .

Explanation of Solution

Given information:

The vector $〈0,1〉$ is rotated through an angle of $\frac{2\pi }{3}$ radians.

Formula used:

Component form of a vector:

A two-dimensional vector $v$ is an ordered pair of real numbers, denoted in component form as

  $〈a,b〉$ . The numbers $a$ and $b$ are the components of the vector $v$ .

Polar coordinates of a vector:

A polar form of a vector $v=〈a,b〉$ is denoted by  $\left(r,\theta \right)$  , where $r$ represents the distance of the vector $v$ from the origin and $\theta$ represents the angle measured from the ?? -axis or the directional angle of the vector.

The formula of conversion from rectangular to polar coordinates is:

  $a=r\cos \theta$

  $b=r\sin \theta$

  $\tan \theta =\frac{b}{a}$

Here, $r=\sqrt{a^2+b^2}$ is the magnitude of the vector $v=〈a,b〉$ .

The vector $〈a,b〉=〈r\cos \theta ,r\sin \theta 〉$ .

Calculation:

Let $〈a,b〉=〈0,1〉$ .

Apply the formula of Magnitude of vectors to obtain the magnitude of the vector $〈a,b〉=〈0,1〉$ , $r$ :

  $\begin{array}{c}r=\left|〈0,1〉\right|\\ =\sqrt{0^2+1^2}\\ =\sqrt{1^2}\\ =1\end{array}$

Let the directional angle of the vector $〈a,b〉$ be $\theta$ .

Substitute $〈a,b〉=〈0,1〉$ and $r=1$ in the formula of conversion from rectangular to polar coordinates to obtain the value of $\theta$:

  $\begin{array}{l}a=r\cos \theta \\ 0=1\cdot \cos \theta \\ \cos \theta =0\\ \theta =\frac{\pi }{2}\end{array}$

And,

  $\begin{array}{l}b=r\sin \theta \\ 1=1\cdot \sin \theta \\ \sin \theta =1\\ \theta =\frac{\pi }{2}\end{array}$

Therefore, the vector $〈a,b〉=〈0,1〉$ makes an angle $\frac{\pi }{2}$ radians with the $x$ -axis.

Now,

The vector $〈a,b〉$ is rotated through an angle of $\frac{2\pi }{3}$ radians.

After rotation, the directional angle, $\theta$ of the vector $〈a,b〉$ becomes:

  $\begin{array}{l}\theta =\frac{\pi }{2}+\frac{2\pi }{3}\\ \theta =\frac{3\pi +4\pi }{6}\\ \theta =\frac{7\pi }{6}\end{array}$

Apply the formula of Polar co-ordinates of vectors to obtain the vector $〈a,b〉$ after rotation:

Substitute $\theta =\frac{7\pi }{6}$ and $r=1$ :

  $\begin{array}{l}a=1\cdot \cos \frac{7\pi }{6}\\ a=\cos \frac{7\pi }{6}\\ a=\cos \left(\pi +\frac{\pi }{6}\right)\\ a=-\cos \frac{\pi }{6}\end{array}$

And,

  $\begin{array}{l}b=1\cdot \sin \frac{7\pi }{6}\\ b=\sin \frac{7\pi }{6}\\ b=\sin \left(\pi +\frac{\pi }{6}\right)\\ b=-\sin \frac{\pi }{6}\end{array}$

Simplify further to obtain:

  $a=-\frac{\sqrt{3}}{2}$

and

  $b=-\frac{1}{2}$

Thus, the component form of the vector obtained by rotating the vector $〈0,1〉$ through an angle of $\frac{2\pi }{3}$ radians is $〈-\frac{\sqrt{3}}{2},-\frac{1}{2}〉$ .