Exercise 32 Page 606

Let's find the magnitude and the direction of the given vector.

Magnitude

The magnitude of a vector is the length from the its initial point to its terminal point. Having been given the component form of a vector $⟨x_1,y_1⟩,$ we can find the magnitude by substituting $(x_1,y_1)=(4,\text{-}8)$ and $(x_2,y_2)=(0,0)$ into the Distance Formula. The magnitude of $\stackrel{\rightharpoonup }{d}$ is $\sqrt{80}.$

Direction

The direction of a vector can be expressed as the angle it forms with the horizontal axis if the initial point is placed at the origin. Let's graph $\stackrel{\rightharpoonup }{d},$ its horizontal component, and its vertical component.

The direction of our vector will be the angle $\alpha .$ Note that $\alpha$ and $\theta$ add up to be a $360^{\circ }$ angle. Therefore, we can find $\alpha$ by subtracting $\theta$ from $360^{\circ }.$ We can find $\theta$ using trigonometry. The tangent of $\theta$ is equal to the ratio of the length of the opposite side divided by the length of the adjacent side. To find $\theta$ we can use the inverse tangent ratio and solve with a calculator. Finally, we can substitute $\theta \approx {63.4}^{\circ }$ in our expression for $\alpha .$ The direction of $\stackrel{\rightharpoonup }{d}$ is about $296.6^{\circ }.$