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December 20, 2014

December 20, 2014

Posted by **Aman** on Saturday, November 19, 2011 at 8:52am.

(1)3i+2j-6k,4i-3j+k

(2)2i-3j+k,3i-j-2k

- algebra -
**Reiny**, Saturday, November 19, 2011 at 9:58am1.

your 2 vectors are [3,2,-6] and [4,-3,1]

using the dot product

12 - 6 - 6 = √(9+4+36) √(16+9+1) cosŲ, where Ų is the angle between them

0 = .... , no point to go further.

I can see cosŲ = 0

Ų = 90° or π/2 radians

2.

6 + 3 - 2 = √(4+9+1) √(9+1+4) cosŲ

7 = √14√14cosŲ

7 = 14cosŲ

cosŲ = 1/2

Ų = 60° or π/3

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