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Here's my attempt:

Let $𝒙 = (x_1, x_2, x_3)$ and $𝒚 = (y_1, y_2, y_3)$

The cross product of $𝒙, 𝒚$ is $𝒙⨯𝒚=(x_2y_3-x_3y_2, x_3y_1 - x_1y_3, x_1y_2 - x_2y_1)$

And linear independence of $𝒙, 𝒚$ means that if $a𝒙 + b𝒚 = 0$, then $a = b = 0$, i.e. $a(x_1, x_2, x_3) + b(x_1, x_2, x_3) = 0 ⇒ a = b = 0$

So if cross product is nonzero, then $ x_2y_3-x_3y_2 + x_3y_1 - x_1y_3 + x_1y_2 - x_2y_1 ≠ 0, i.e. x_1(y_2-y_3)+x_2(y_3-y_1) + x_3(y_1 - y_2) ≠ 0$

And then I'm just getting confused.

I can't seem to connect the two in a formal proof.

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  • $\begingroup$ Quite possibly it's not available but this follows immediately from the identity $|{\bf x} \times {\bf y}| = |{\bf x}|\,|{\bf y}| \sin \theta$, where $\theta \in [0, \frac{\pi}{2}]$ is the angle between $\bf x$ and $\bf y$. $\endgroup$ – Travis Willse Feb 12 '15 at 0:59
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Your problem is equivalent to the following:

$x$ and $y$ are linearly dependent iff $x\times y=(0,0,0)$.

To prove this assume first that $x$ and $y$ are linearly dependent, i.e. there is some $a\in\mathbb R$ such that $x=ay$. Then it is easy to see that $x\times y=(0,0,0)$ (just plug in).

For the converse assume now that $x\times y=(0,0,0)$. Then $x_2y_3=x_3y_2$, $x_3y_1=x_1y_3$ and $x_1y_2=x_2y_1$. If $y\neq0$, then one entry is $\neq0$, say $y_3\neq0$. Then $x_2=\frac{x_3y_2}{y_3}$, $x_1=\frac{x_3y_1}{y_3}$. Thus, $(x_1,x_2,x_3)=\frac{x_3}{y_3}\cdot(y_1,y_2,y_3)$, so $x$ and $y$ are linearly dependent.


Here is an alternative proof (for the converse) using the identity $v\cdot(x\times y)=\det(v,x,y)$ for each $v\in\mathbb R^3$, i.e. assume $x\times y=(0,0,0)$. Take a vector $v\notin\text{span}\{x,y\}$. If $x$ and $y$ were linearly independent, then $\{v,x,y\}$ were a basis of $\mathbb R^3$ and therefore $\det(v,x,y)\neq0$, contradicting $\det(v,x,y)=v\cdot(x\times y)=0$.

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If,$$ {\bf x} \times {\bf y} \neq 0 $$ Then, for any $$ a {\bf x} + b {\bf y} = 0 $$ We have, $$ (a {\bf x} + b {\bf y}) \times {\bf x} = 0 $$ $$ a ({\bf x} \times {\bf x}) + b ({\bf y} \times {\bf x}) = 0 $$ $$ 0 + b( {\bf y} \times {\bf x}) = 0 $$ Since, $ {\bf x} \times {\bf y} \neq 0 $, then there must be $b=0$.

Similarly, Let, $$ (a {\bf x} + b {\bf y}) \times {\bf y} = 0 $$ We have, $a=0$

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All the answers here are algebraic. Consider a more geometric approach. The area of a parallelogram spanned by the vectors v and w is given by the length of v x w. This is either a standard theorem or part of your definition of cross product. But if you need to prove it, it will be the bulk of the proof of your desired theorem as the rest is easy (see below).

If the area is zero, you can scale and rotate so that v and w lie in the xy-plane with v being a unit vector along the x axis. If you argue by continuity, you will be able to see the only case when the signed area is zero is when w is colinear with v.

If v and w are colinear, then you can simply compute the area and see it's zero.

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