McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
7. Vectors
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Exercise 33 Page 688

To find the magnitude, you will need the Distance Formula. To find the direction, begin by drawing the vector.

Magnitude: sqrt(45)
Direction: ≈ 243.4 ^(∘)

Practice makes perfect

Let's find the magnitude and the direction of the given vector. k = ⟨ -3,-6 ⟩

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)=(- 3,- 6) and (x_2,y_2)=( 0, 0) into the Distance Formula.
|k|=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
|k|=sqrt(( 0-(- 3))^2+( 0-(-6))^2)
Evaluate
|k|=sqrt((0+3)^2+(0+6)^2)
|k|=sqrt(3^2+6^2)
|k|=sqrt(9+36)
|k|=sqrt(45)

The magnitude of k is sqrt(45).

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 k, its horizontal component, and its vertical component.

The direction of our vector will be the angle α. Note that α is a sum of θ and 180 ^(∘). Therefore, we can find α by adding θ and 180 ^(∘). α= θ+180 ^(∘) We can find θ using trigonometry. The tangent of θ is equal to the ratio of the length of the opposite side divided by the length of the adjacent side. Tangent=Opposite/Adjacent ⇒ tan θ = 63 To find θ we can use the inverse tangent ratio and solve with a calculator.
θ = tan ^(-1)6/3
θ = 63.43494...
θ ≈ 63.4
Finally, we can substitute θ ≈ 63.4^(∘) in our expression for α. α ≈ 63.4+180 ^(∘) ⇔ α ≈ 243.4 ^(∘) The direction of k is about 243.4 ^(∘).