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Exercises Trigonometry

Example 9 from Algebra and Trigonometry, 10.8 Vectors

Finding the Unit Vector in the Direction of v

Find a unit vector in the same direction as $v$ $= 〈 −5, 12 〉.$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
First, we will find the magnitude.

$\begin{array}{l} | v | = \sqrt{{(- 5)}^{2} + {(12)}^{2}} \\ = \sqrt{25 + 144} \\ = \sqrt{169} \\ = 13 \end{array}$
Then we divide each component by $| v |,$ which gives a unit vector in the same direction as v:
$\frac{v}{| v |} = - \frac{5}{13} i + \frac{12}{13} j$
or, in component form

$\frac{v}{| v |} = 〈 - \frac{5}{13}, \frac{12}{13} 〉$
See Figure 15.

Figure 15

Verify that the magnitude of the unit vector equals 1. The magnitude of $- \frac{5}{13} i + \frac{12}{13} j$ is given as
$\begin{array}{l} \sqrt{{(- \frac{5}{13})}^{2} + {(\frac{12}{13})}^{2}} = \sqrt{\frac{25}{169} + \frac{144}{169}} \\ = \sqrt{\frac{169}{169}} = 1 \end{array}$
The vector u $= \frac{5}{13}$ i $+ \frac{12}{13}$ j is the unit vector in the same direction as v $= 〈 - 5, 12 〉 .$

How did it go?

PreviousFinding the Components of the Vector Next Writing a Vector in Terms of i and j

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 10.8, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.