- #1

- 56

- 2

First one is more important please help.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter prashant singh
- Start date

- #1

- 56

- 2

First one is more important please help.

- #2

blue_leaf77

Science Advisor

Homework Helper

- 2,637

- 785

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

Science Advisor

Gold Member

- 6,749

- 2,768

- #4

pasmith

Homework Helper

- 2,155

- 773

\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

- 56

- 2

- #6

SammyS

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,420

- 1,056

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

The definition tells the some meaning of something.

Share: