Dot Product Question: How to Solve (2a-5b)dot(b+3a) with Unit Vectors?

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Homework Statement



I'm really at a loss here, if anyone could help me out I'd really appreciate it.


Given 'a' and 'b' unit vectors,

if |a+b| = root3, determine (2a-5b)dot(b+3a)
 
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|a + b|^2 = (a + b)\bullet(a + b)
and |a + b| = sqrt(3) ==> |a + b|^2 = 3

Now, use the fact that the dot product is associative, distributive, and commutative and the two equations above to see if you can evaluate (2a - 5b) )\bullet (b + 3a).
 
well I end up getting 13ab + 6a^2 - 5b^2

The answer is -11/2

I just can't seem to figure out how to get there :s
 
Work with (a + b) \cdot (a + b) = 3. You also know that a and b are unit vectors, which means that a \cdot a = 1 and b \cdot b = 1.
 
Would I do like

(a+b)dot(a+b)=3

1 + 2ab + 1 = 3

ab = 1/2

then sub 1/2 into the ab and then get 6.5 + 6a^2 - 5b^2 and solve from there?
 
Sort of, except that what you show as 6a^2 and -5b^2 is really 6a\cdot a and -5b\cdot b.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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