Orthogonal Vectors Problem

[DJfox]

The vectors (a+4b) and (3a-b) are orthogonal. If the magnitude of b is three times that of a, and vectors a and b are non-zero, calculate the angle between a and b.

I'm a bit stumped by this question, it was on a recent quiz our calc & vectors class did, and the correct answer was marked as 0 degrees, but that shouldn't be possible given that 2 parallel vectors couldn't combine to create 2 orthogonal vectors. Is anyone willing to take a shot at this question and get out a valid answer? I've tried a few different strategies, but I'm mostly just shooting in the dark. I would also be very appreciative if anyone could also explain what makes this question weirder than most other similar questions. Thanks!

 

[Steven G]

Since u=(a+4b) and v= (3a-b) are orthogonal then you know that

cos(t) = [u*v]/[ ||u|| ||v|| ], where t is the angle between u and v.

It must be necessary cos(t) = 0, hence [u*v]/[ ||u|| ||v|| = 0. Now a fraction equals 0 whenever the numerator is 0 and the denominator is not. It should be clear that ||u|| ||v|| isn't 0.

Now u*v = 3a^2 - 4b^2 + 11ab = 0. Also ||b|| = 3||a||.

See what you can do from here.

 

[Dr.Peterson]

The vectors (a+4b) and (3a-b) are orthogonal. If the magnitude of b is three times that of a, and vectors a and b are non-zero, calculate the angle between a and b.

I'm a bit stumped by this question, it was on a recent quiz our calc & vectors class did, and the correct answer was marked as 0 degrees, but that shouldn't be possible given that 2 parallel vectors couldn't combine to create 2 orthogonal vectors. Is anyone willing to take a shot at this question and get out a valid answer? I've tried a few different strategies, but I'm mostly just shooting in the dark. I would also be very appreciative if anyone could also explain what makes this question weirder than most other similar questions. Thanks!

I set the dot product of (a+4b) and (3a-b) equal to 0, and used the fact that ||b|| = 3||a||, and found that, in fact, the angle is 0.

But in that case, it turns out that 3a-b is the zero vector, which can't be orthogonal to anything. So this is sort of an extraneous solution; you are wise to have seen that the answer is impossible for a different reason. It is possible that the author of the problem didn't observe this; you should discuss it with your teacher (and perhaps get extra credit).