New Reply

Given the dot product and cross product of two vectors, find the angle between them?

 
Share Thread
Jan15-12, 10:14 PM   #1
 

Given the dot product and cross product of two vectors, find the angle between them?


1. The problem statement, all variables and given/known data
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?


2. Relevant equations
A \dot B = ||A|| ||B|| cos θ and
A \cross B = ||A|| ||B|| sin θ


3. 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.
PhysOrg.com science news on PhysOrg.com

>> City-life changes blackbird personalities, study shows
>> Origins of 'The Hoff' crab revealed (w/ Video)
>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
Jan15-12, 10:44 PM   #2
 
Recognitions:
Homework Helper Homework Help
Quote by mrowcow View Post
1. The problem statement, all variables and given/known data
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?


2. Relevant equations
A \dot B = ||A|| ||B|| cos θ and
A \cross B = ||A|| ||B|| sin θ


3. 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?
Jan15-12, 10:47 PM   #3
 
Mentor
Quote by mrowcow View Post
1. The problem statement, all variables and given/known data
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?


2. Relevant equations
A \dot B = ||A|| ||B|| cos θ and
A \cross B = ||A|| ||B|| sin θ


3. 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 ?
Jan15-12, 11:13 PM   #4
 

Given the dot product and cross product of two vectors, find the angle between them?


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!
Jan15-12, 11:18 PM   #5
 
Mentor
Glad to be of help.

BTW: The angle between two vectors is ≤ 180° .
Jan15-12, 11:21 PM   #6
 
Mentor
Quote by mrowcow View Post
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.
Jan15-12, 11:56 PM   #7
 
Recognitions:
Homework Helper Homework Help
Quote by cepheid View Post
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.
Jan16-12, 01:48 AM   #8
 
Thank you so much! I understand now. The hints about the sign of cos and sin were super helpful. You are amazing people.
New Reply

Tags
angle, basic physics, cross product, dot product, vectors

Similar discussions for: Given the dot product and cross product of two vectors, find the angle between them?
Thread Forum Replies
Use the cross product to find the sin of an angle then prove it Calculus & Beyond Homework 4
Angle between 2 vectors using 1) Dot product and 2) cross product gives diff. answer? Calculus & Beyond Homework 8
Using cross product to find angle between two vectors Calculus & Beyond Homework 5
Find theta from the cross product and dot product of two vectors Calculus & Beyond Homework 11
Find angle to eliminate cross product term Calculus & Beyond Homework 1