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,-6><−2,−1,−6> and <7,3,2><7,3,2>
This gives a 3xx33×3 matrix:
abs((veci,vecj,veck),(-2,-1,-6),(7,3,2))
To find the cross product, first imagine covering up the i column (or actually do so if possible), and take the cross product of the j and k 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 the product of the second number and its diagonal. This is your new i component.
(-1*2)-(3*-6)=-2-(-18)=16
=>16veci
Now imagine covering up the j column. Similarly to above, take the cross product of the i and k columns. However, this time, whatever your answer is, you will multiply it by -1.
-[(-2*2)-(7*-6)]=-[-4-(-42)]=-38
=>-38vecj
Finally, imagine covering up the k column. Now, take the cross product of the i and j columns.
(-2*3)-(-1*7)=-6-(-7)=1
=>1veck
Therefore, the cross product is < 16, -38, 1 > or, equivalently, (16i-38j+k).
