Surjective is onto. I cross product any u with the given vector and add u.

I wind up with a matrix times u. Equate that product to a vector k which

is any vector in real space. If I u_{1},u_{2},u_{3} can be soved for than the exprssion is surjective. This means the matrix determinant must not be zero.

the rows of the matrix I computed are:(1,-1,-1);(-1,1,0),;(-1,0,1)

Its determinant is not zero so expression is surjective.

injective means one to one releationship between any single u and the expression outcome.

set expression = to two vectors, l_{1} and l_{2}

show if l_{1 }≠ l_{2} then v cross u is unequal in some circumstance to v cross u ;;;;;;;;;;;

a contradiction so the expression is injective