# Vector Proof

by SeReNiTy
Tags: proof, vector
 P: 171 Hi, i was required to show that -1 < $$\frac{a.b}{\|{a}\|\|{b}\|}}$$ > -1 I did this by using the cosine rule which is $$c^2 = a^2 + b^2 - 2a.b\cos{\vartheta}$$ How ever our teacher did it by a scharts proof which i don't quite understand, , Now my question is why can't i prove it using the cosine rule and could somebody explain the schwartz proof a bit better?
 Quote by SeReNiTy Hi, i was required to show that -1 < $$\frac{a.b}{\|{a}\|\|{b}\|}}$$ > -1 I did this by using the cosine rule which is $$c^2 = a^2 + b^2 - 2a.b\cos{\vartheta}$$ How ever our teacher did it by a scharts proof which i don't quite understand, , Now my question is why can't i prove it using the cosine rule and could somebody explain the schwartz proof a bit better?
Has your teacher defined the angle between two vectors to be the arccosine of your expression, or is he using the inequality 1 > $$\frac{a.b}{\|{a}\|\|{b}\|}}$$ > -1 to motivate such a definition for general vector spaces (not just Euclidean space) ? In the latter case, it is understandable that he does not equate it with what he is trying to motivate yet.
 P: 171 Vector Proof yes the inequality is suppose to be -1< something < 1 Also i'm suppose to prove that the expression $$\frac{a.b}{\|{a}\|\|{b}\|}}$$ lies within the domain [-1,1] I did this using the cosine rule because cos(theta) is within that domain, how do i do it with the schwartz proof?