Given vectors uu and vv, the cross product of these two vectors, uxxvu×v is given by:
![enter image source here]()
Where
uxxv=(u_2v_3-u_3v_2)veci-(u_1v_3-u_3v_1)vecj+(u_1v_2-u_2v_1)vecku×v=(u2v3−u3v2)→i−(u1v3−u3v1)→j+(u1v2−u2v1)→k
This process may look rather complicated but in reality isn't so bad once you get the hang of it.
We have vectors <2,-1,2><2,−1,2> and <3,-1,2><3,−1,2>
This gives a 3xx33×3 matrix in the form of:
![enter image source here]()
To find the cross product, first imagine covering up the ii column (or actually do so if possible), and take the cross product of the jj and kk columns, similar to as you would using cross multiplication with proportions. In the clockwise direction, starting with the number at the top left, multiply the first number by its diagonal, then subtract from that product the product of the second number and its diagonal. This is your new ii component.
(-1*2)-(2*-1)=-2-(-2)=0(−1⋅2)−(2⋅−1)=−2−(−2)=0
=>0veci⇒0→i
Now imagine covering up the jj column. Similarly to above, take the cross product of the ii and kk columns. However, this time, whatever your answer is, you will multiply it by -1−1.
-1[(2*2)-(3*2)]=2−1[(2⋅2)−(3⋅2)]=2
=>2vecj⇒2→j
Finally, imagine covering up the kk column. Now, take the cross product of the ii and jj columns.
(2*-1)-(-1*3)=-2-(-3)=1(2⋅−1)−(−1⋅3)=−2−(−3)=1
=>1veck⇒1→k
Thus, the cross product is (0i+2j+1k)(0i+2j+1k) or <0,2,1><0,2,1>.
