The Cross Product of Two Vectors

Definition:

u x v = (u₂v₃ – u₃v₂)i – (u₁v₃ – u₃v₁)j + (u₁v₂ – u₂v₁)k

Setting up a determinant with cofactor expansion with i j k as the first row, u as the second row, and v as the third row does a nice job of calculating the cross product.

Example:

(a)  Find u x v and v x u when
      u = 2i – 3j + k
      v = -i + 4j – 3k

     Answer:
    Set up a determinant:  |  i     j    k |
                                      | 2  -3    1 |
                                      |-1   4   -3 |

    Now calculate the determinant, using the first row as your cofactors:
    (9 - 4)i - (-6 - (-1))j + (8 - 3)k  =  5i - (-5)j + 5k  =  5i + 5j + 5k

    The cross product is the vector 5i + 5j + 5k

(b)  Find v x v above

     Answer:
     Set up a determinant:  |  i     j    k |
                                       |-1   4   -3 |
                                       |-1   4   -3 |

     Now calculate the determinant, using the first row as your cofactors:
     (-12 - (-12))i - (3 - 3)j + (-4 - (-4)k  =  0i - 0j + 0k

     The cross product is the zero vector.

(c)  Find i x j.

    Answer:
     Set up a determinant:  |  i     j    k |
                                       | 1    0   0  |
                                       | 0    1   0  |

     Now calculate the determinant, using the first row as your cofactors:
     (0 - 0)i - (0 - 0)j + (1 - 0)k  =  0i - 0j + k

     The cross product is the other standard unit vector.
 

Properties of the cross product:

(1)  u x v = - (v x u)
(2)  u x ( v + w ) = (u x v) + (u x w)
(3)  c(u x v) = (cu) x v = u x (cv)
(4)  u x 0 = 0 x u = 0
(5)  u x u = 0
(6)  u × (v x 2) = (u x v) × w

Geometric Properties:
(1)  u x v is orthogonal to both u and v
(2)  ||u x v||   =   sin q ||u|| ||v||
(3)  u x v = 0 if and only if u and v are scalar multiples of each other
(4)  ||u x v|| = area of parallelogram having u and v as adjacent sides.

Examples:

(a)  Find a unit vector that is orthogonal to both u = 7i – 14j + 5k  and  v = 14i + 28j – 15k

       Answer:
       Orthogonal vectors can be found by doing the cross product:  u x v
       |  i       j       k |
       | 7    -14      5 |   =   (210 - 140) i  -  (-105 - 70) j  +  (196 - 196) k  =  70 i + 175 j + 0k
       | 14   28   -15 |

       Now convert the answer to its unit vector.  The magnitude of the answer is \/(70² + 175² + 0²) = 35 \/29:
      /     70   ,   175   ,   0  \
      \ 35\/29   35\/29        /
 

(b)  Verify that A(3,5,0), B(-1,8,5), C(1,3,11), and D(5,0,6) are the vertices of a parallelogram and find its area.

       Answer:
       In a parallelogram opposite sides are parallel.  Show that AB // CD  and that BC // AD.
       Find component forms:  AB = <-4, 3, 5> and CD  = <4, -3, -5>.  Thus AB = (-1)CD
              BC = <2, -5, 6>  and  AD = <2, -5, 6>.  Thus BC // AD and ABCD is a parallelogram.

       ||u x v|| = area of parallelogram having u and v as adjacent sides.

       So  find the cross product of vectors AB × BC.
                           |   i    j   k |
       AB × BC =  | -4   3   5 |   =  (18 - (-25)) i  -  (-24 - 10) j  +  (20 - 6) k  =  43 i  +  34 j  +  14 k
                           |  2  -5   6 |

       Now find the magnitude of the cross product:
       \/(432 + 342 + 142) = \/3201  »  56.577
 

Triple Scale Product:

                      |   u₁    u₂   u₃ |
u × (v x w) =  |   v₁    v₂   v₃ |
                      | w₁    w₂   w₃|

This gives the volume of a parallelepiped with vectors u, v, and w as adjacent edges.

Example:
Find the volume of the parallelepiped with vertices
 A(3,0,0), B(4,1,2), C(3,-1,4), D(2,-2,2), E(-1,5,4), F(0,6,6), G(-1,4,8), H(-2,3,6).

Answer:
AB, AD, and AE are three vectors forming the parallepiped's length, width, and height.
So find the determinant with these three vectors as the entries.
|   1    1   2 |
| -1   -2   2 |  =  (-8 - 8 - 10) - (16 + 10 - 4)  =  -26 - 22 = - 48
| -4    5   4 |