Proving Parallel Vectors: Cross Product Question Explained

[spoc21]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I tried attempting the solution by using the following:

\vec{a} = [a₁, a₂, a₃]
\vec{b} = [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!

 

[Wingeer]

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.

 

[spoc21]

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]

Bump..any one?

thanks

 

[Susanne217]

Bump..any one?

thanks

Its it simply


Let \vec{a},\vec{b} \neq 0 and let

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

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

q.e.d.

 

[spoc21]

dot product? I don't understand it, could anyone please help..

 

[tiny-tim]

Hi spoc21!

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)

 

[spoc21]

Hi spoc21! Yes, because they become

a₂/a₃ = b₂/b₃...(1)

a₃/a₁ = b₃/b₁...(2)

a₁/a₂ = b₁/b₂...(3)

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

 

[tiny-tim]

Hi spoc21!

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.

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.

 

[spoc21]

Thanks!