Scalar product and vector product

  • #1
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.
First one is more important please help.
 

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  • #2
blue_leaf77
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why do we take cross product of A X B as a line normal to the plane which contains A and B.
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.
I also need a proof of A.B = |A||B|cos(theta)
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
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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
pasmith
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In euclidean 2D space, if [itex]\mathbf{a} = (a \cos \phi, a \sin \phi)[/itex] and [itex]\mathbf{b} = (b \cos \alpha, b \sin \alpha)[/itex] then by basic trigonometry [tex]\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),[/tex] or [tex]
\mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\|\|\mathbf{b}\| \cos\theta[/tex] where [itex]\theta = \phi - \alpha[/itex] is the angle between [itex]\mathbf{a}[/itex] and [itex]\mathbf{b}[/itex].

In arbitrary inner product spaces, [itex]\|a\| = (a \cdot a)^{1/2}[/itex] is the definition of [itex]\|a\|[/itex], and after one has proven from basic properties of the inner product that [itex]|a \cdot b| \leq \|a\|\|b\|[/itex] one can then define [itex]\theta[/itex] by [itex]a \cdot b = \|a\|\|b\| \cos \theta[/itex].
 
  • #5
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
SammyS
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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 .
In general, there is no proof for a definition. That would not be logical.

The definition tells the some meaning of something.
 
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