Angle Between Vectors A & B: Solve the Mystery!

[mrowcow]

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.

 

[Curious3141]

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?

 

[SammyS]

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 ?

 

[mrowcow]

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!

 

[SammyS]

Glad to be of help.

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

 

[cepheid]

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.

 

[Curious3141]

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.

 

[mrowcow]

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