Part C of the inner product problem

  • Thread starter Dustinsfl
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  • #1
Dustinsfl
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I have attached the solutions of parts a, b, and what I have done for part c.

My part c isn't going to turn out correct and I don't know what is wrong.
 

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  • Inner product.pdf
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Answers and Replies

  • #2
Dick
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I'm not quite following what you are doing for part c, but the Pythagorean Law would be ||1||^2=||p||^2+||1-p||^2 or <1,1>=<p,p>+<1-p,1-p>. Since you know that p=3x/2 you could just compute all the inner products. Or if you had checked in part b that <p,1-p>=0 you could just write 1=(1-p)+p and take the inner product of each side with itself.
 
  • #3
Dustinsfl
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I didn't see I could use the p in part b. Instead I was using p proving from the law of cosine.
 
  • #4
Dick
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I didn't see I could use the p in part b. Instead I was using p proving from the law of cosine.

But you are using p=3x/2 in part c. But I don't see anything resembling a law of cosines or a proof there. Try it again. State clearly what you want to prove and give a reason for each step.
 
  • #5
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In part b you still need to show that p and 1 - p are orthogonal, which is pretty easy to do. From that you'll have p and 1 - p being the legs of a right triangle, and 1 being the hypotenuse.

For the c part do as Dick said.
 

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