- #1

prashant singh

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First one is more important please help.

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- #1

prashant singh

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First one is more important please help.

- #2

blue_leaf77

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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.why do we take cross product of A X B as a line normal to the plane which contains A and B.

Consider the cosine rules for the addition and the difference between two vectors.I also need a proof of A.B = |A||B|cos(theta)

$$

|\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

FactChecker

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- #4

pasmith

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\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

prashant singh

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- #6

SammyS

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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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