Example 4 from Algebra and Trigonometry, 10.8 Vectors

Showing That Two Vectors Are Equal

Show that vector v with initial point at $(5, −3)$ and terminal point at $(−1, 2)$ is equal to vector u with initial point at $(−1, −3)$ and terminal point at $(−7, 2) .$ Draw the position vector on the same grid as v and u. Next, find the magnitude and direction of each 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
As shown in Figure 7, draw the vector $v$ starting at initial $(5, −3)$ and terminal point $(−1, 2) .$ Draw the vector $u$ with initial point $(−1, −3)$ and terminal point $(−7, 2) .$ Find the standard position for each.
Next, find and sketch the position vector for v and u. We have
$\begin{array}{l} v = 〈 −1 - 5, 2 - (- 3) 〉 \\ = 〈 −6, 5 〉 \\ u = 〈 −7 - (−1), 2 - (−3) 〉 \\ = 〈 −6, 5 〉 \end{array}$
Since the position vectors are the same, v and u are the same.
An alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem.
$\begin{array}{l} | v | = \sqrt{{(−1 - 5)}^{2} + {(2 - (−3))}^{2}} \\ = \sqrt{{(−6)}^{2} + {(5)}^{2}} \\ = \sqrt{36 + 25} \\ = \sqrt{61} \\ | u | = \sqrt{{(−7 - (−1))}^{2} + {(2 - (−3))}^{2}} \\ = \sqrt{{(−6)}^{2} + {(5)}^{2}} \\ = \sqrt{36 + 25} \\ = \sqrt{61} \end{array}$
As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives
$\begin{array}{l} \tan θ = - \frac{5}{6} \Rightarrow θ = \tan^{- 1} (- \frac{5}{6}) \\ = - 39.8° \end{array}$
However, we can see that the position vector terminates in the second quadrant, so we add $180° .$ Thus, the direction is $- 39.8° + 180° = 140.2° .$

Figure 7

How did it go?

Solution

Solution