# vector product

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vector product definition: which is correct and why?

a x b =|a||b|sin(theta)

or

a x b =|a||b|sin(theta) n^

why is the n^ there in some of the definitions

a x b =|a||b|sin(theta)

or

a x b =|a||b|sin(theta) n^

why is the n^ there in some of the definitions

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#2

Vector product is a vector (it's in the name ). So 2nd definition is correct as it defines a vector.

The first definition is clearly wrong. Where on earth have you seen that?

The first definition is clearly wrong. Where on earth have you seen that?

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#3

(Original post by

Vector product is a vector (it's in the name ). So 2nd definition is correct as it defines a vector.

The first definition is clearly wrong. Where on earth have you seen that?

**DFranklin**)Vector product is a vector (it's in the name ). So 2nd definition is correct as it defines a vector.

The first definition is clearly wrong. Where on earth have you seen that?

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**DFranklin**)

Vector product is a vector (it's in the name ). So 2nd definition is correct as it defines a vector.

The first definition is clearly wrong. Where on earth have you seen that?

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#5

(Original post by

can you pls explain why theres an n^ in the correct definition cause shouldnt it be a x b =|a||b|sin(theta) = n^

**localmemelord**)can you pls explain why theres an n^ in the correct definition cause shouldnt it be a x b =|a||b|sin(theta) = n^

**a**and

**b**.

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(Original post by

Because the vector product defines a vector perpendicular to both

**DFranklin**)Because the vector product defines a vector perpendicular to both

**a**and**b**.
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#7

(Original post by

no i understand that thanks. but i dont understand why the unit vector is within the equation because ive seen questions where they dont incorperate n^ when calculating a x b for example

**localmemelord**)no i understand that thanks. but i dont understand why the unit vector is within the equation because ive seen questions where they dont incorperate n^ when calculating a x b for example

**a**x

**b**|).

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(Original post by

If you have, the question is wrong. (Although it's far more likely you're missing something out, such as they actually calculated |

**DFranklin**)If you have, the question is wrong. (Although it's far more likely you're missing something out, such as they actually calculated |

**a**x**b**|).
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#9

(Original post by

im still confused as to what n^ or n in comparison to vectors a and b

**localmemelord**)im still confused as to what n^ or n in comparison to vectors a and b

**n**is a unit vector perpendicular to both

**a**and

**b.**

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#11

(Original post by

ok so if a x b = c and c is the vector perpendicular to a and b then does that mean that vector n in n^ = n/|n| equal to vector c

**localmemelord**)ok so if a x b = c and c is the vector perpendicular to a and b then does that mean that vector n in n^ = n/|n| equal to vector c

If a x b = c, then n=c and n^ = c/|c|

Last edited by RDKGames; 7 months ago

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(Original post by

He meant that n^ is the unit vector. Thats what the 'hat' on top of n means.

If a x b = c, then n=c and n^ = c/|c|

**RDKGames**)He meant that n^ is the unit vector. Thats what the 'hat' on top of n means.

If a x b = c, then n=c and n^ = c/|c|

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