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

[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.

Answers:
(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.

 

[ehild]

Cy=-5.03.

ehild

 

[DavidAp]

Cy=-5.03.

ehild

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?

 

[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

 

[rwisz]

First of all, it is important to note that vector addition and subtraction involves both the magnitude and direction of the vectors. So when subtracting vector A from vector C, you also need to take into account the angle of vector A.

Using the given information, we can find the components of vector A and C:

Ax = 14.0m cos(38.0 degrees) = 11.0m
Ay = 14.0m sin(38.0 degrees) = 8.62m

Cx = 13.2m cos(22.4 degrees) = 11.93m
Cy = 13.2m sin(22.4 degrees) = 4.90m

Now, when subtracting vector A from vector C, we get:

Bx = Cx - Ax = 11.93m - 11.0m = 0.93m
By = Cy - Ay = 4.90m - 8.62m = -3.72m

Therefore, the components of vector B are <0.93m, -3.72m>. Now, using the magnitude equation, we get:

||vB|| = sqrt((0.93)^2 + (-3.72)^2) = sqrt(14.49) = 3.81m

So the magnitude of vector B is 3.81m. Now, to find the angle, we can use the inverse tangent function:

tan(theta) = By/Bx = (-3.72m)/(0.93m) = -4
theta = -76.87 degrees

But since the angle is measured counterclockwise from the +x direction, we can add 360 degrees to get the positive angle:

theta = -76.87 degrees + 360 degrees = 283.13 degrees

Therefore, the angle of vector B is 283.13 degrees relative to the +x direction.

I hope this helps clarify the problem and provides a solution to your question. Remember to always pay attention to both the magnitude and direction when working with vectors.