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Exercises Trigonometry

Example 3 from Algebra and Trigonometry, 10.8 Vectors

Finding the Magnitude and Direction of a Vector

Find the magnitude and direction of the vector with initial point $P (- 8, 1)$ and terminal point $Q (- 2, - 5) .$ Draw the vector.

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, find the position vector.
$\begin{array}{l} u = 〈 −2, - (−8), −5 -1 〉 \\ = 〈 6, - 6 〉 \end{array}$
We use the Pythagorean Theorem to find the magnitude.
$\begin{array}{l} | u | = \sqrt{{(6)}^{2} + {(- 6)}^{2}} \\ = \sqrt{72} \\ = 6 \sqrt{2} \end{array}$
The direction is given as

$\begin{array}{l} \tan θ = \frac{−6}{6} = −1 \Rightarrow θ = \tan^{−1} (−1) \\ = - 45° \end{array}$
However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positive angle. Thus, $- 45° + 360° = 315° .$ See Figure 6.

Figure 6

How did it go?

PreviousDrawing a Vector with the Given Criteria and Its Equivalent Position Vector Next Showing That Two Vectors Are Equal

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.