How Do You Calculate the Magnitude of Vector C in Vector Subtraction?

In summary: So, what is the magnitude and direction of that vector?The magnitude is the magnitude of A - the magnitude of B, and the direction is the direction of A - the direction of B. To find the direction, find the angle in the third quadrant that is 90° larger than the angle you have and then subtract it from the angle you have. Then it will be correct in the third quadrant. (The direction for B is 180° larger than the angle you have, so subtract that from the angle you have, and you will have the angle for the vector you must add to B to get A. What is the direction of that vector?In summary, the magnitude of vector C, the difference
  • #1
T54
4
0

Homework Statement


https://smart.physics.illinois.edu/images/phys101/Homework/02/vector.gif
Two vectors A and B are shown in the x-y plane. Vector A has a magnitude of 5 m and makes an angle of 55° with the positive x-axis vector B has a magnitude of 2 m and makes an angle of 33° with the negative x-axis. (See the figure above.) Vector C (not shown in the diagram) is the difference of A and B (C = A - B).
What is the magnitude of C?

Homework Equations


[v] = sqrt(vx2-vy2)

The Attempt at a Solution


A=5cos55deg+5sin55deg
B=-2cos33deg+2sin33deg
So I plugged it into the equation above and got 6.391
But apparently my answer is not correct. I am wondering where I went wrong
 
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  • #2
Welcome to PF!

Hmm, I get an answer a little lower than yours for the magnitude of C. Maybe you made a calculator error? Show us what you've done.
 
  • #3
T54 said:

The Attempt at a Solution


A=5cos55deg+5sin55deg
B=-2cos33deg+2sin33deg

In what quadrant is -B?
 
  • #4
TaxOnFear said:
Welcome to PF!

Hmm, I get an answer a little lower than yours for the magnitude of C. Maybe you made a calculator error? Show us what you've done.

sqrt(2.868+4.096)2-(-1.677+1.089)2)
sqrt(48.497-.346)
= 6.989
This time I got a little higher than my previous answer.
 
  • #5
Rapier said:
In what quadrant is -B?
Qaud II
 
  • #6
I think I was making it to complicated.
The angle between vector A and vector B is 90°, so if I just the pythagorean theorem
I would get magnitude of vector C = sqrt(5² + 2²) = sqrt29
which is 5.385
 
  • #7
T54 said:
Qaud II

There is part of your problem. B is in Quad II. When you take the inverse of a vector, like -B, you extend the line right out of his butt in the opposite direction and of the same magnitude. -B is in Quad IV.

EDIT: I want to add something to make sure this is clear. A - B = A + (-B), just as if these were regular variables.

If you draw that line of B backwards into Quad IV, you have -B and the angle is -33° (or 327°, if you prefer)...and keeping that 33° reference angle.

When you are dealing with vectors, there is two methods to deal with signs. Either you reduce everything to reference angles and then tack on the signs for the appropriate quadrant or you ignore the signs entirely and let the angles take care of it.

In Quad I: cos 33° is positive and sin 33° is positive
In Quad IV: cos 33° is negative and sin 33° is negative, so you either need to tack on the negatives yourself or use the -33° or 327° angle measures.

Once you understand A-B the rest is pretty easy peasy.

The magnitude of the combined vector will be the square root of the squares of the sums of the X and Y components. Which is a fancy way of saying that:

(A-B)=√( (Ʃx)^2 + (Ʃy)^2)

Try recalculating with the new information and give a shout if something is still unclear.
 
Last edited:
  • #8
T54 said:
I think I was making it to complicated.
The angle between vector A and vector B is 90°, so if I just the pythagorean theorem
I would get magnitude of vector C = sqrt(5² + 2²) = sqrt29
which is 5.385

The angle between A and B is NOT 90°. Add up the angles you were given (33° + 55° = 88°). We know that the total angle measures from 0 to ∏ is 180°. 180°-88°=92° between the two angles.

I know it's very tempting, but you need to avoid slamming numbers into equations. You can bang your head against the wall for quite a while that way. Draw your sketch and label your vectors and then create your equations based upon the geometry, not by the utility of any particular equation.
 
  • #9
T54 said:

Homework Statement


https://smart.physics.illinois.edu/images/phys101/Homework/02/vector.gif
Two vectors A and B are shown in the x-y plane. Vector A has a magnitude of 5 m and makes an angle of 55° with the positive x-axis vector B has a magnitude of 2 m and makes an angle of 33° with the negative x-axis. (See the figure above.) Vector C (not shown in the diagram) is the difference of A and B (C = A - B).
What is the magnitude of C?

Homework Equations


[v] = sqrt(vx2-vy2)


The Attempt at a Solution


A=5cos55deg+5sin55deg
B=-2cos33deg+2sin33deg
So I plugged it into the equation above and got 6.391
But apparently my answer is not correct. I am wondering where I went wrong

Note: if C = A - B, it follows that C + B = A

What vector must you add to B, in order to get A, for that is indeed C.
 

FAQ: How Do You Calculate the Magnitude of Vector C in Vector Subtraction?

1. What is vector subtraction?

Vector subtraction is a mathematical operation that involves finding the difference between two vectors. It is used to determine the direction and magnitude of the resulting vector when two vectors are added or subtracted.

2. How is the magnitude of a vector determined through subtraction?

The magnitude of a vector is determined through vector subtraction by finding the length of the resulting vector from the subtraction. This can be done using the Pythagorean theorem, where the magnitude is equal to the square root of the sum of the squares of the individual vector components.

3. Can vector subtraction result in a negative magnitude?

Yes, vector subtraction can result in a negative magnitude. This occurs when the two vectors are in opposite directions and have equal magnitudes. The resulting vector will have a magnitude of zero, and its direction will be in the opposite direction of the two subtracted vectors.

4. How is vector subtraction used in real-world applications?

Vector subtraction is used in various fields, such as physics, engineering, and navigation, to determine the resultant force or displacement of multiple vectors. It is also commonly used in computer graphics to represent motion and direction.

5. Can vector subtraction be performed with more than two vectors?

Yes, vector subtraction can be performed with more than two vectors. The process involves subtracting each individual vector's components and adding them together to determine the resultant vector's magnitude and direction.

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