# I Scalar product and vector product

1. May 1, 2016

### prashant singh

why do we take cross product of A X B as a line normal to the plane which contains A and B. I also need a proof of A.B = |A||B|cos(theta), I have seen many proves but they have used inter product ,A.A = |A|^2, which is a result of dot product with angle = 0, we can't use this too prove the dot product formula.

2. May 1, 2016

### blue_leaf77

You asked why, well it's because a vector product is defined that way. It's defined such that the result of the product is perpendicular to any linear combination of A and B.
Consider the cosine rules for the addition and the difference between two vectors.
$$|\vec{A}+\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos\theta \\ |\vec{A}-\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}| \cos\theta \\$$

3. May 1, 2016

### FactChecker

Regarding the first question: We need some fundamental operation that reflects rotations in R3 from direction A to direction B. A nice operation would be linear in both A and B and would be anti-commutative ( AxB = -BxA ). The definition of the cross product fits the bill.

4. May 1, 2016

### pasmith

In euclidean 2D space, if $\mathbf{a} = (a \cos \phi, a \sin \phi)$ and $\mathbf{b} = (b \cos \alpha, b \sin \alpha)$ then by basic trigonometry $$\mathbf{a} \cdot \mathbf{b} = (a \cos \phi, a \sin \phi) \cdot (b \cos \alpha, b \sin \alpha) = ab \cos \phi \cos\alpha + ab \sin \phi \sin \alpha = ab \cos(\phi - \alpha),$$ or $$\mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\|\|\mathbf{b}\| \cos\theta$$ where $\theta = \phi - \alpha$ is the angle between $\mathbf{a}$ and $\mathbf{b}$.

In arbitrary inner product spaces, $\|a\| = (a \cdot a)^{1/2}$ is the definition of $\|a\|$, and after one has proven from basic properties of the inner product that $|a \cdot b| \leq \|a\|\|b\|$ one can then define $\theta$ by $a \cdot b = \|a\|\|b\| \cos \theta$.

5. May 1, 2016

### prashant singh

So you are saying that ||a|| = (a.a)^1/2 is the defination given by the founders and there is no proof for this . But why there is no proof .

6. May 1, 2016

### SammyS

Staff Emeritus
In general, there is no proof for a definition. That would not be logical.

The definition tells the some meaning of something.