# Homework Help: Basic Vectors: If vector A and C are given what is the magnitude of vector B?

1. Sep 1, 2011

### DavidAp

Vector A has a magnitude of 14.0m and is angled 38.0 degrees counterclockwise from the +x direction. Vector C has a magnitude of 13.2m and is angled at 22.4 degrees counterclockwise to the -x direction. If vector A + vector B = vector C what is (a) the magnitude (b) and angle - relative to +x direction - of vector B? State your angle as a positive number.

(a) 26.9m
(b) 210 degrees

Relevant equations:
I will use vA as a shorthand to represent vector A and ||vA|| to represent the magnitude of vector A.

Ax = ||vA||cos(theta)
Ay = ||vA||sin(theta)
||vA|| = sqrt(Ax^2 + Ay^2)

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My reasoning was that if I could find the vectors of vA and vC I could subtract vA from vC to obtain the vector vB.

vA + vB = vC
vB = vC - vA = <Cx-Ax, Cy-Ay> = <Bx, By>

From there I would use the equation for the magnitude of the vector to obtain the answer for part a.

||vB|| = sqrt(Bx^2 + By^2)

However, when I plug in the numbers I keep getting the incorrect answer, 23.5m.

Ax = 11.0m Ay = 8.62m vA = <11.0m, 8.62m>
Cx = -12.2m Cy = 5.03m vC = <-12.2m, 5.03m>
vB = <-12.2m-11.0m, 5.03m-8.62m> = <-23.2m, -3.59m>
||vB|| = sqrt((-23.2^2 + (-3.59)^2)) = sqrt(551.13) = 23.5m

I haven't attempted part (b) yet due to the fact that there is an error to the first part of the question. What did I do wrong? I felt fairly confident in what I was doing and have been stumped on this question for nearly an hour. Thank you in advance for reading my problem.

Last edited: Sep 1, 2011
2. Sep 1, 2011

Cy=-5.03.

ehild

3. Sep 1, 2011

### DavidAp

It gave me the right answer! Amazing! But, how did you get the negative? In my calculator sin(157.6) = 0.38. Multiply that by 13.2 and I got 5.03. Where did the negative come from?

4. Sep 2, 2011

### ehild

The angle is not 180-22.4 but 180+22.4. It is counter clockwise to the -x direction, in the third quadrant.

ehild