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# Vector Operations in 3D - Problem 2

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Let's find the angle between two vectors; A which has components 3, 0, 4, and B which has components -1, 3, 4. I've drawn the two vectors here, and so what we're looking for is this angle between them. So let's call that angle theta. The cosine theta is A.B divided by the magnitude of A times the magnitude of B.

So we're going to need to calculate the magnitudes. Let's do the magnitude of A first. It's the square root of 3², 9 plus 0², plus 4² is 16. So 9 plus 0, plus 16 that's 25 and the square root of 25 is 5. What about the magnitude of B? It's the square root of 1 plus 9, plus 16. 1 plus 9 plus 16 that's root 26. So let's start filling this in. Cosine of theta equals A.B, what's A.B? 3 times -1, -3, plus 0 times 3, 0, plus 4 times 4, 16, all over 5 times root 26. So that means cosine theta equals 13 over 5 root 26.

Let's use our calculators to calculate theta. Theta is going to be inverse cosine of 13 over 5, root 26. So the inverse cosine 13 divided by 5 root 26, 59.3 degrees. So the angle between these two vectors is 59.3 degrees.

Now remember, you find the angle between vectors the same way in three dimensions as in two dimensions using this formula.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

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