Finding angles between vectors

[adrimare]

Homework Statement

I have two related questions related because they both ask to find angles between vectors.

The first one says to determine the angles between the sides of a parallelogram determined by the vectors OA = (3,2,-6) and OB = (-6,6,-2).

The second one asks to determine the angle between OP and AE, where OP = (3,4,5) and AE = (-3,4,5).

Homework Equations

?

The Attempt at a Solution

For the first one, I tried to use the cosine law, but got 1.5 degrees instead of 84.4 degrees.
These are my steps.

magnitude of OA = \sqrt{}(3² + 2² + -6²)= \sqrt{}49 = 7

magnitude of OB = \sqrt{}](-6² + 6² + -2²)= \sqrt{}76

vector AB = (-9,4,4)

magnitude of AB = \sqrt{}(-9² + 4² + 4²) = \sqrt{}113

\Theta = cos^-1 ((7² + \sqrt{}76 - \sqrt{}113) / (2*7*\sqrt{}76)) = approximately 1.5 degrees

I'm confused as to what to do here.

The next question has me confused as well, as I seem to be missing something key.

Please help!

 

[Mark44]

Homework Statement

I have two related questions related because they both ask to find angles between vectors.

The first one says to determine the angles between the sides of a parallelogram determined by the vectors OA = (3,2,-6) and OB = (-6,6,-2).

The second one asks to determine the angle between OP and AE, where OP = (3,4,5) and AE = (-3,4,5).

Homework Equations

?

The Attempt at a Solution

For the first one, I tried to use the cosine law, but got 1.5 degrees instead of 84.4 degrees.
These are my steps.

magnitude of OA = \sqrt{}(3² + 2² + -6²)= \sqrt{}49 = 7

Don't put [ sup] or [ sub] tags inside [ tex] tags. The reason should be obvious.

magnitude of OB = \sqrt{}](-6² + 6² + -2²)= \sqrt{}76

vector AB = (-9,4,4)

magnitude of AB = \sqrt{}(-9² + 4² + 4²) = \sqrt{}113

\Theta = cos^-1 ((7² + \sqrt{}76 - \sqrt{}113) / (2*7*\sqrt{}76)) = approximately 1.5 degrees

I'm confused as to what to do here.

The next question has me confused as well, as I seem to be missing something key.

Please help!

Your mixed tex and sup stuff is too hard to read, so I won't try to pick out the exact problem. Using the Law of Cosines (there's an easier way if you know about the dot product) and simplifying things a bit, you should get cos(theta) = 6/(7sqrt(76)). From this I get theta ~ 84.4 degrees. Make sure your calculator is in degree mode.

 

[adrimare]

thanks. It was in radian mode. whoops. But what about my second related question? Can I do the same thing? And I do know about dot product, but I'm supposed to do the question without it.

 

[Mark44]

Sure, the 2nd problem is almost exactly the same. The two sides adjacent to your angle are OP and AE.