What is the cross product of #<-3,0, 8 ># and #<-1, -2, 9 >#?

For a hand-made explanation on vector multiplication, click here (you can scroll down to the vector/cross product section); it explains basically all you need to know.

If you followed the link, you'll find (somewhere in the cross products section) an explanation about how cross products are found.

In short, you can use the following equation to find its components, using components of the two vectors, which we'll call #vecA# and #vecB#, and the vector product #vecC#:

#C_x = A_yB_z - A_zB_y#

#C_y = A_zB_x - A_xB_z#

#C_z = A_xB_y - A_yB_x#

Plugging in the values, we have

#C_x = (0)(9) - (8)(-2) = color(red)(16#

#C_y = (8)(-1) - (-3)(9) = color(blue)(19#

#C_z = (-3)(-2) - (0)(-1) = color(green)(6#

In unit vectors, the cross product is

#color(red)(16)hati + color(blue)(19)hatj + color(green)(6)hatk#

Or alternatively,

#color(darkorange)(<16,19,6>#

Multiply the major diagonals:

#| (color(red)(hati),color(blue)(hatj),color(yellow)(hatk),hati,hatj), (-3,color(red)(0),color(blue)(8),color(yellow)(-3),0), (-1,-2,color(red)(9),color(blue)(-1),color(yellow)(-2)) | =#

#color(red)((0)(9)hati)+ color(blue)((8)(-1)hatj)+ color(yellow)((-3)(-2)hatk)#

Multiply the minor diagonals:

#| (hati,hatj,color(yellow)(hatk),color(red)(hati),color(blue)(hatj)), (-3,color(yellow)(0),color(red)(8),color(blue)(-3),0), (color(yellow)(-1),color(red)(-2),color(blue)(9),-1,color(yellow)(-2)) | =#

#color(red)((0)(9)hati)+ color(blue)((8)(-1)hatj)+ color(yellow)((-3)(-2)hatk)- color(red)((8)(-2)hati) - color(blue)((-3)(9)hatj) - color(yellow)((0)(-1)hatk) = #

#16hati+19hatj+6hatk#

In the original notation #<16,19,6>#