Cross product of two vectors is same despite transposing one of them

Question

Just wondering why the cross product is the same in both case:

const first = [1,2,1];
const second = [2,4,2];

console.log(math.cross(first,math.transpose(second)));
// [ 0, 0, 0 ]


console.log(math.cross(first,second));
// [ 0, 0, 0 ]

should they be the same? How does that work if one is transposed?

(Ultimately I am just trying to discover if the vectors are linearly independent or not.)

Answer 1

$\mathbf{x}\times 2 \mathbf{x}=0$

why?

using indicial tensor notation and

Einstein's summation convention (repeated index is a sum)

we have:

$(\mathbf{x}\times \mathbf{2x})_i=2\epsilon_{ijk}x_jx_k$

Where $\epsilon_{ijk}$ is the alternating tensor.

So exchanging indices $j,k$ and using the antisymmetry of $\epsilon_{ijk}$ we obtain the zero result.

ln a more intuitive way, think of a cross product as rotating one vector counterclockwise to another vector.

If the angle between the vectors is 0, then there is no rotation.

Hence, the product is 0.