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Suppose that a, b and c are vectors. (length of a is defined as |a|)

|a| = |b| = |c| = sqrt(2). Given that 2c(a + b) = ab + 6(eq1)

Prove a + b - 2c = 0(eq2)

Suppose eq1 is true. a + b = 2c. Substituting in eq1 I get

=> 2c*2c = ab + 6

=> 4 c^2 = ab + 6

=> 4 |c|^2 = ab + 6

=> 4 * 2 - 6 = ab

=> ab = 2.

I am stack in here. This is an exercise of a friend and my vector skills are kinda rusty. Any insights ?

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  • $\begingroup$ I'm not convinced by this. You're assuming $a+b = 2c$ which is exactly what you're trying to prove. $\endgroup$
    – user417289
    Nov 10, 2017 at 20:18
  • $\begingroup$ @AlgTop then I have no idea to be honest.. Can you guide me through this ? $\endgroup$
    – KostasRim
    Nov 10, 2017 at 20:20
  • $\begingroup$ I'm not sure exactly how to do this, but I'll look at it. What field is your vector space over? and what exactly do you mean by $|c|$? $\endgroup$
    – user417289
    Nov 10, 2017 at 20:21
  • $\begingroup$ @AlgTop |c| I mean length of a vector. I got only this, as I said the exercise is not mine and the last time I used vector proofs was 5-6 years ago $\endgroup$
    – KostasRim
    Nov 10, 2017 at 20:22
  • $\begingroup$ @AlgTop maybe a tip will do the trick, let me know if you get anything $\endgroup$
    – KostasRim
    Nov 10, 2017 at 20:23

1 Answer 1

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If $a + b = 2c$ and the lengths of $a$, $b$, and $c$ are the same, that means they're all the same vector, by the triangle inequality. So you just have to show that the vectors are all the same vector.

A promising approach would be to expand the all the dot products $vw$ in equation 1 into $|v||w|\space\text{cos}(\theta_v - \theta_w)$, then show that if the vectors $a$, $b$, and $c$ are not all the same, there is a contradiction.

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  • $\begingroup$ thanks will check it out :) $\endgroup$
    – KostasRim
    Nov 10, 2017 at 21:25

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