# Finding angles between vectors

1. Apr 11, 2010

### adrimare

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

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).

2. Relevant equations

?

3. 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!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 11, 2010

### Staff: Mentor

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

3. Apr 11, 2010

### 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.

4. Apr 11, 2010

### Staff: Mentor

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

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