You can use the dot product to solve this problem. See http://en.wikipedia.org/wiki/Dot_product
The dot product is an operation on two vectors. There are two different definitions of dot product. Let \vec(A)=[A_1,A_2,...,A_n] be a vector and \vec(B)=[B_1,B_2,...,B_n] be another vector, then we have 2 formulas for dot product:
1) Algebraic definition:
\vec(A) \cdot \vec(B) = \sum_1^n A_i B_i = A_1 B_1 + A_2 B_2 + ... + A_n B_n
2) Geometric definition:
\vec(A) \cdot \vec(B) = ||\vec(A)||\ ||\vec(B)||\cos(\theta)
where \theta is the angle between \vec(A) and \vec(B), and ||\vec(A)|| denotes the magnitude of \vec(A) and has the formula:
||\vec(A)|| = \sqrt(A_1^2 + A_2^2 + ... + A_n^2)
We can solve many questions (such as the angle between two vectors) by combining the two definitions:
\sum_1^n A_i B_i = ||\vec(A)||\ ||\vec(B)||\cos(\theta)
or
A_1 B_1 + A_2 B_2 + ... + A_n B_n = (\sqrt(A_1^2 + A_2^2 + ... + A_n^2))(\sqrt(B_1^2 + B_2^2 + ... + B_n^2))\cos(\theta)
If we have two vectors, then the only unknown is \theta in the above equation, and thus we can solve for \theta, which is the angle between the two vectors.
Example:
Q: Given \vec(A) = [2, 5, 1], \vec(B) = [9, -3, 6], find the angle between them.
A:
From the question, we see that each vector has three dimensions. From above, our formula becomes:
A_1 B_1 + A_2 B_2 + A_3 B_3 = (\sqrt(A_1^2 + A_2^2 + A_3^2))(\sqrt(B_1^2 + B_2^2 + B_3^2))\cos(\theta)
Left side:
A_1 B_1 + A_2 B_2 + A_3 B_3 = (2)(9) + (5)(-3) + (1)(6) = 9
Right side:
||\vec(A)|| = \sqrt(A_1^2 + A_2^2 + A_3^2) = \sqrt(2^2 + 5^2 + 1^2) = \sqrt(30)
||\vec(B)|| = \sqrt(B_1^2 + B_2^2 + B_3^2) = \sqrt(9^2 + (-3)^2 + 6^2) = \sqrt(126)
\theta is unknown
Plug everything into the formula, we get:
9 = (\sqrt(30))(\sqrt(126))\cos(\theta)
Solve for \theta:
\cos(\theta) = \frac(9)((\sqrt(30))(\sqrt(126))
\theta = \cos^-1(\frac(9)((\sqrt(30))(\sqrt(126))))
Using a calculator, we get:
\theta = 81.58 degrees
See the following video of ...
Angle Between Vectors Example
