
Vector Proof Help
First off, sorry if this is in the wrong section. I was unsure of where to put it, and if there's a more suitable location please point me to it and I'll ask a mod to move it. Anyways, this appeared on my exam today, and I couldn't figure it out on the exam nor could the people that I spoke to.
a and b are unit vectors.
If a dotted with b = 1, show that a = b
Also, it did not mention any dimension so the "brute force" method (solving for the general case <x,y,z> and <a,b,c> using the dot product definition) isn't ideal. Also, we were to find a method other than using the = abcos o formula.
I found a way to prove the opposite, as a vector dotted with itself is simply the length of the vector squared, in this case simply one, so if a = b then a dotted b = a dotted a =  (a dotted a) = 1. However, I couldn't find a way to link this logic in the other direction.

The vector becomes negative when it changes the direction. If a and b are in the same direction, the angle between them is zero. cos(0) = 1. so a.b = ab. when a and b are in the opposirte direction, the angle between them is 180 degrees. And cos(180) = 1. So a.b= abcos(180) = ab.

Thanks, but the problem specifically said not to use trig functions.
Any ideas?

You cannot define the dot or cross product of the vectors without using the trigonometric function.

That's not true. There are several methods of defining the dot product of two vectors. For example, in two dimensions, the dot product of <a, b> and <c, d> is ac+ bd. In higher dimensions it is common to use the dot product to define angles.
Lord Voldemort, look at the dot product of $\displaystyle \vec{a}+ \vec{b}$ with itself. $\displaystyle (\vec{a}+ \vec{b})\cdot(\vec{a}+ \vec{b})= \vec{a}\cdot\vec{a}+ 2\vec{a}\cdot\vec{b}+ \vec{b}\cdot\vec{b}$. Use the fact that $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ are unit vectors and that $\displaystyle \vec{a}\cdot\vec{b}= 1$. What does that give you and what can you conclude? Remember that this is the dot product of a vector with itself.

Thanks so much, I see it now.