Angle Between Vectors A & B: Solve the Mystery!

In summary, the angle between two vectors with a scalar product of -6.00 and a vector product magnitude of 4.00 can be found by using the equations A \dot B = ||A|| ||B|| cos θ and A \cross B = ||A|| ||B|| sin θ. Since the cosine is negative, the angle must be between 90 and 270 degrees. However, since the angle between two vectors is always less than or equal to 180 degrees, the angle must be either 146.7 or 213.7 degrees. To determine the correct angle, we must use the sign of the sine, which in this case is positive. Therefore, the angle between the two vectors is
  • #1
mrowcow
3
0

Homework Statement


Vectors A and B have scalar product -6.00 and their vector product has magnitude 4.00. What is the angle between these two vectors?

Homework Equations


A \dot B = ||A|| ||B|| cos θ and
A \cross B = ||A|| ||B|| sin θ

The Attempt at a Solution


Then I reasoned that tan(θ) = -4/6 so θ = cot(-4/6) = -33.69. I entered -33.7 and +33.7 degrees into Mastering Physics. Both are wrong. I know the answer should be in degrees. I'm confused about what I'm doing wrong. Thanks for helping out.
 
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  • #2


mrowcow said:

Homework Statement


Vectors A and B have scalar product -6.00 and their vector product has magnitude 4.00. What is the angle between these two vectors?


Homework Equations


A \dot B = ||A|| ||B|| cos θ and
A \cross B = ||A|| ||B|| sin θ


The Attempt at a Solution


Then I reasoned that tan(θ) = -4/6 so θ = cot(-4/6) = -33.69. I entered -33.7 and +33.7 degrees into Mastering Physics. Both are wrong. I know the answer should be in degrees. I'm confused about what I'm doing wrong. Thanks for helping out.

Hint: the cosine is negative. What do you know about the signs of various trig ratios in the different quadrants?
 
  • #3


mrowcow said:

Homework Statement


Vectors A and B have scalar product -6.00 and their vector product has magnitude 4.00. What is the angle between these two vectors?


Homework Equations


A \dot B = ||A|| ||B|| cos θ and
A \cross B = ||A|| ||B|| sin θ


The Attempt at a Solution


Then I reasoned that tan(θ) = -4/6 so θ = cot(-4/6) = -33.69. I entered -33.7 and +33.7 degrees into Mastering Physics. Both are wrong. I know the answer should be in degrees. I'm confused about what I'm doing wrong. Thanks for helping out.
Hello mrowcow. Welcome to PF !

If the scalar product is negative, then cos(θ) is also negative. Correct ?

If cos(θ) is negative, what do you know about θ ?
 
  • #4


If the cosine is negative, then the angle is between 90 and 270 degrees. So the angle is 213.7 or 146.7 degrees? Was I completely off track originally then? I still feel like my original approach makes sense... except for the fact that your point makes my original answer definitively wrong. Thanks!
 
  • #5


Glad to be of help.

BTW: The angle between two vectors is ≤ 180° .
 
  • #6


mrowcow said:
If the cosine is negative, then the angle is between 90 and 270 degrees. So the angle is 213.7 or 146.7 degrees? Was I completely off track originally then? I still feel like my original approach makes sense... except for the fact that your point makes my original answer definitively wrong. Thanks!

You can narrow it down further. You've got it down to two quadrants in which the angle has to be in, but the sign of the sine can tell you which of those two it is.

Your original approach was fine. You just have to be careful with the inverse tan (which is called arctangent, NOT cotangent by the way) because it's not uniquely-valued. There are many different angles which have the same tangent. In particular, if your tangent is negative, it could be because the cos is negative and the sine is positive, but it could also be because the cos is positive and the sine is negative. To resolve this ambiguity, your calculator's arctan function, by convention, picks angles between 0 and 90 (in magnitude) to return. This may not be the right answer in all situations.
 
  • #7


cepheid said:
You can narrow it down further. You've got it down to two quadrants in which the angle has to be in, but the sign of the sine can tell you which of those two it is.

Actually, here it can't (since only the magnitude of the cross product is given, which is always positive).

But Sammy's further comment (angle is less than or equal to 180 deg.) is the clincher. :smile:
 
  • #8


Thank you so much! I understand now. The hints about the sign of cos and sin were super helpful. You are amazing people.
 

1. What is the definition of angle between two vectors?

The angle between two vectors is the measure of the rotation required to align one vector with the other, with both vectors sharing the same initial point.

2. How do you calculate the angle between two vectors?

The angle between two vectors can be calculated using the dot product formula: θ = cos^-1 (A · B / |A||B|), where A and B are the two vectors and |A| and |B| are their magnitudes.

3. What is the range of values for the angle between two vectors?

The angle between two vectors can range from 0° to 180°. When the two vectors are parallel, the angle is 0°, and when they are antiparallel, the angle is 180°.

4. Can the angle between two vectors be negative?

No, the angle between two vectors cannot be negative. It is always measured in a counterclockwise direction from the first vector to the second vector, and therefore, it will always be a positive value.

5. What is the significance of the angle between two vectors?

The angle between two vectors is important in various fields of science, such as physics, engineering, and mathematics. It helps in understanding the relationship between two vectors and can be used to solve problems related to forces, motion, and geometry.

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