Linear algebra, vector product HELP

[green_tea]

Hi! I need some help to understand this... It's the definition of the vector product of vectors a and b.



a and b are vectors.
v is the angle between a and b.
a is orthogonal to the plane.
b' is the orthogonal projection of b on the plane. |b'| = |b| sin v
b'' is orthogonal to a and to b'

Now |b''| = |b'|= |b|sin v (1)

And my book says that a x b = |a|b''
Whyyy? :?: :?: :?:

I thouth that they may get this from (1) by multiplying both sides of the equation with |a|. But then it would be likes this:

|a||b''| = |a||b|sinv = |a x b|

and now I'm very confused with the absolute walues because I think that if |a||b''|=|a x b |, then |ab''|=|a x b | and then if i get rid of the absolute values on both sides I get ab''=a x b, but it should be |a|b''=|a x b|

So obviously I'm thinking wrong...
Can anyone explain this to me, pleeeeease.... I got my exams on this soon and I need to understand this....

 

[Deleted member 4993]

Hi! I need some help to understand this... It's the definition of the vector product of vectors a and b.



a and b are vectors.
v is the angle between a and b.
a is orthogonal to the plane.
b' is the orthogonal projection of b on the plane. |b'| = |b| sin v
b'' is orthogonal to a and to b'

Now |b''| = |b'|= |b|sin v (1)

And my book says that a x b = |a|b''
Whyyy? :?: :?: :?:

I thouth that they may get this from (1) by multiplying both sides of the equation with |a|. But then it would be likes this:

|a||b''| = |a||b|sinv = |a x b|

and now I'm very confused with the absolute walues because I think that if |a||b''|=|a x b |, then |ab''|=|a x b | and then if i get rid of the absolute values on both sides I get ab''=a x b, but it should be |a|b''=|a x b|

So obviously I'm thinking wrong...
Can anyone explain this to me, pleeeeease.... I got my exams on this soon and I need to understand this....

In vectors, |a| - means magnitude of - not "absolute value of".

\(\displaystyle |A| \, = \, \sqrt{A \cdot A}\)

and

\(\displaystyle |A\, X \, B| = |A||B|\cdot \sin \theta\)

 

[Unco]

Hi Green_tea,

Hi! I need some help to understand this... It's the definition of the vector product of vectors a and b.

a and b are vectors.
v is the angle between a and b.
a is orthogonal to the plane.
b' is the orthogonal projection of b on the plane. |b'| = |b| sin v
b'' is orthogonal to a and to b'

Now |b''| = |b'|= |b|sin v (1)

And my book says that a x b = |a|b''
Whyyy? :?: :?: :?:

Since axb is orthogonal to a and b, we must have that axb = kb'' where k is some scalar. Furthermore, from your diagram it appears that b'' is chosen so that k is non-negative. (*)

This implies that |a x b| = k|b''|. But |b''|= |b|sin(v) so we have |a x b| = k|b|sin(v). (**)

We also know that |a x b| = |a||b|sin(v) (***).

So, equating (**) and (***) we see that k = |a|.

Thus (*) becomes axb = |a|b''.

---
Hey Sub,

\(\displaystyle |A\, X \, B| = |A||B|\cdot \sin \theta\)

It's probably best to avoid using \cdots between scalars when there are vectors also floating around. You can also use \times for the cross product.

 

[green_tea]

Hi Green_tea,

Hi! I need some help to understand this... It's the definition of the vector product of vectors a and b.

a and b are vectors.
v is the angle between a and b.
a is orthogonal to the plane.
b' is the orthogonal projection of b on the plane. |b'| = |b| sin v
b'' is orthogonal to a and to b'

Now |b''| = |b'|= |b|sin v (1)

And my book says that a x b = |a|b''
Whyyy? :?: :?: :?:

Since axb is orthogonal to a and b, we must have that axb = kb'' where k is some scalar. Furthermore, from your diagram it appears that b'' is chosen so that k is non-negative. (*)

This implies that |a x b| = k|b''|. But |b''|= |b|sin(v) so we have |a x b| = k|b|sin(v). (**)

We also know that |a x b| = |a||b|sin(v) (***).

So, equating (**) and (***) we see that k = |a|.

Thus (*) becomes axb = |a|b''.

---
Hey Sub,
[quote="Subhotosh Khan":2yv9f5b6]\(\displaystyle |A\, X \, B| = |A||B|\cdot \sin \theta\)

It's probably best to avoid using \cdots between scalars when there are vectors also floating around. You can also use \times for the cross product.[/quote:2yv9f5b6]


Thanx for help guys, now I understand this perfectly!!!