Proving Parallel Vectors: Cross Product Question Explained

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spoc21
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Homework Statement



Prove that If [tex]\vec{a}[/tex]x [tex]\vec{b}[/tex] = 0, then [tex]\vec{a}[/tex]is parallel to [tex]\vec{b}[/tex].

Homework Equations


The Attempt at a Solution


I tried attempting the solution by using the following:

[tex]\vec{a}[/tex] = [a1, a2, a3]
[tex]\vec{b}[/tex] = [b1, b2, b3]When I took the cross product of a x b I got::

[a2b3 - b2a3 a3b1 - b3a1, a1b2 - b1a2]

and we can make this equal to 0, but I am confused here; I have no idea on how to prove that when a x b is 0, vectors a and b are parallel.

any help is appreciated!
 
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Imagine the zero as the zero vector, defined in R3 as 0=[0,0,0]

Edit: You could view the geometric interpretation. In which case you will have a parallellogram with area equal to zero.
 
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thanks, so this is what I end up with:

a2b3 = b2a3...(1)

a3b1 = b3a1...(2)

a1b2 = b1a2...(3)

Does this suggest that the two vectors, a and b are parallel?
 
spoc21 said:
Bump..any one?

thanks

Its it simply


Let [tex]\vec{a},\vec{b} \neq 0[/tex] and let

[tex]\vec{a} \times \vec{b} = |\vec{a}| \cdot |\vec{b}| \cdot sin(\theta}) \cdot \mathrm{n}[/tex] be the definition of the cross product where [tex]0 \leq \theta \leq \pi[/tex]

then for [tex]\theta = 0[/tex] [tex]\vec{a} \times \vec{b} = |\vec{a}| \cdot |\vec{b}| \cdot sin(0}) \cdot \mathrm{n} = 0[/tex]

q.e.d.
 
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dot product? I don't understand it, could anyone please help..
 
Hi spoc21! :smile:
spoc21 said:
a2b3 = b2a3...(1)

a3b1 = b3a1...(2)

a1b2 = b1a2...(3)

Does this suggest that the two vectors, a and b are parallel?

Yes, because they become

a2/a3 = b2/b3...(1)

a3/a1 = b3/b1...(2)

a1/a2 = b1/b2...(3) :wink:
 
tiny-tim said:
Hi spoc21! :smile:Yes, because they become

a2/a3 = b2/b3...(1)

a3/a1 = b3/b1...(2)

a1/a2 = b1/b2...(3) :wink:

Thanks tiny-tim So basically if I write the vectors in this form, I am showing that they are parallel, and this would be enough for the proof? Also, just for my knowledge could you please elaborate on this..
Thank you very much :smile:
 
Hi spoc21! :smile:
spoc21 said:
Thanks tiny-tim So basically if I write the vectors in this form, I am showing that they are parallel, and this would be enough for the proof?

Yes, that's enough to do it. :wink:

To elaborate …two vectors are parallel if one is a scalar times the other …

and you can easily check that that means that the ratios of their coordinates must be the same. :smile: