11j 2k x 5j 3k

I beleve this person wants the cross-product  ....(0, 11, 2) x (0, 5, 3)

This will form a vector that is perpendicular to both of these

To get the x coordinate, we have

(11* 3) - (2 * 5)  =  33 - 10 = 23

To get the y coordinate, we have

(2*0) - (0* 3)  = 0

To get the z coordinate, we have 

(0 * 5) - (11* 0)   = 0

So....our cross-product vector is  (23, 0, 0)

This makes sense......since both of the given vectors lie in the y-z plane......the vector perpendicular to both should just lie along the x axis....!!!

Here's a pic.....

Yuo can read more about this proceedure, here ......http://www.mathsisfun.com/algebra/vectors-cross-product.html

55j²6k²