Question

On a coordinate plane, vector u has the origin (7, 3) and terminal point (1, 11). Given vector u, what is the magnitude, |u|, and directional angle, θ, in standard position?

a) |u| = 5.3, θ = 126.9°
b) |u| = 5.3, θ = 143.1°
c) |u| = 10, θ = 126.9°
d) |u| = 10, θ = 143.1°

Answer 1

The magnitude of vector u is 10, and the directional angle θ in the standard position is 126.9°, which corresponds to answer choice c) |u| = 10, θ = 126.9°.

Step-by-step explanation:

To find the magnitude and directional angle of vector u, we must first compute the change in the x and y coordinates. The change in x (Δx) is calculated as the x-coordinate of the terminal point minus the x-coordinate of the origin, so Δx = 1 - 7 = -6. Similarly, the change in y (Δy) is 11 - 3 = 8. The magnitude of vector u is found using the Pythagorean theorem:

|u| = √((Δx)^2 + (Δy)^2) = √((-6)^2 + (8)^2) = √(36 + 64) = √100 = 10.

Next, we determine the directional angle θ in the standard position (measured from the positive x-axis). We use the tangent of θ since it relates the opposite side (Δy) over the adjacent side (Δx) in a right triangle:

θ = arctan(Δy/Δx) = arctan(8/(-6)).

However, since the resulting angle will be in the second quadrant (because Δx is negative while Δy is positive), we must add 180° to find the correct angle measured counterclockwise from the positive x-axis:
θ = arctan(8/(-6)) + 180° ≈ 126.9° + 180° = 306.9°.

But, if we need the directional angle less than 180°, we subtract 180° from 306.9° to get θ = 126.9°. Therefore, the correct answer is c) |u| = 10, θ = 126.9°.