# Would The Cross Product of x and x be z0?

You will have to define what 'x' and 'z0' are for your question to make any sense. In summary, the cross product of A x A will always be zero in 3D space because to define a plane, you need two linearly independent vectors and two parallel vectors do not determine a unique plane. The result can be found directly from the definition of the cross product and can be understood geometrically in terms of an area.

Just like the title says, would that technically be true?

I know the cross product is normal to the plane of the two vectors being crossed, which would make it z. However, since the angle between two vectors is 0, sin (0) = 0...

You will have to define what 'x' and 'z0' are for your question to make any sense.

The cross product of A x A will always be zero if we are working in R^3. The basic reason is because to define a plane in 3D space, you need two vectors that are linearly independent. If you have two vectors that are linearly dependent, then all you get is a line and a unique plane can not be formed from a single line.

I think he means if you take a vector v in cartesian coords: ##\vec{v}=x\vec{\imath}## and is asking the result of taking the cross product with itself is a zero length vector pointing in the z direction - i.e. ##\vec{v}\times\vec{v}=0\vec{k}## ...which, as soon as you write it out, you see, the answer is "yes and no".

Chiro is correct - and the result is found directly from the definition of the cross product.
OP seems to be trying to understand it geometrically in terms of an area.

The trick is to ask if it makes sense to talk about the direction of a zero-length vector.
A vector zero in the z direction would be (x,y,z)=(0,0,0) and zero in the x direction would be just the same, after all.

@Meadman23: your recent questions have been getting more and more to do with things that are normally covered by entry-level college classes. Some of your earlier questions indicate you are taking electrical engineering at college level - is that the case?