# Thread: Angle in a 3-D Parallelogram

1. ## Angle in a 3-D Parallelogram

Hello! I am in Calc III, and I have a question. The answer sheet gives me a different answer than I have, but I feel that I am correct. Could you all point me one way or another?

Parallelogram with points: A(2, -1, 4); B(1,0,-1); C(1,2,3); D(2,1,8)

The question is.... what is the interior angle at B?

The answer sheet says use vectors AB and BC and comes up with 140.8 degrees.

I used vectors BA and BC and got 39.23 degrees.

Which one is correct? Thanks!

2. Originally Posted by SSVicious
Hello! I am in Calc III, and I have a question. The answer sheet gives me a different answer than I have, but I feel that I am correct. Could you all point me one way or another?

Parallelogram with points: A(2, -1, 4); B(1,0,-1); C(1,2,3); D(2,1,8)

The question is.... what is the interior angle at B?

The answer sheet says use vectors AB and BC and comes up with 140.8 degrees.

I used vectors BA and BC and got 39.23 degrees.

Which one is correct? Thanks!
Here is the problem the angles AB BA are no equal (why?) but
$180-39.23=...$

3. Originally Posted by SSVicious
Hello! I am in Calc III, and I have a question. The answer sheet gives me a different answer than I have, but I feel that I am correct. Could you all point me one way or another?
Originally Posted by SSVicious

Parallelogram with points: A(2, -1, 4); B(1,0,-1); C(1,2,3); D(2,1,8)

The question is.... what is the interior angle at B?

The answer sheet says use vectors AB and BC and comes up with 140.8 degrees.

I used vectors BA and BC and got 39.23 degrees.

Which one is correct? Thanks!
I did the calculation and got 140.768 or 140.8. Heres what I did, I found the vector from $B$ to $A$ and called it vector $U$. I found $U$ by subtracting each corresponding point in $B$ from each point in $A$ to get vector $U = (-1, 1, -5)$.

I then did the same thing with from points $B$ to $C$ and called this vector $V$. The subtraction resulted in: $V = (0, 2, 4)$.

Then I computed the dot product, divided it by the product of the magnitudes of $U$ and $V$ and took the inverse cosine of this value, to find the angle between the vectors, like this:

$cos(\theta) = \frac{VU}{|{V}||{U}|}$

I found the angle to be 140.8

I could be wrong, but check my work to make sure, dont assume I am right. Hope this helps