What is the unit vector that is orthogonal to the plane containing # (-i + j + k) # and # (3i + 2j - 3k) #?

When you take the cross product of two vectors, you are calculating the vector that is orthogonal to the first two. However, the solution of #vecAoxvecB# is usually equal and opposite in magnitude of #vecBoxvecA#.

As a quick refresher, a cross-product of #vecAoxvecB# builds a 3x3 matrix that looks like:

#|i j k|#
#|A_x A_y A_z|#
#|B_x B_y B_z|#

and you get each term by taking the product of the diagonal terms going from left to right, starting from a given unit vector letter (i, j, or k) and subtracting the product of diagonal terms going from right to left, starting from the same unit vector letter:

#(A_yxxB_z-A_zxxB_y)i+(A_zxxB_x-A_x xxBz)j+(A_x xxB_y-A_yxxB_x)k#

For the two solutions, lets set:
#vecA=[-i+j+k]#
#vecB=[3i+2j-3k]#

Let's look at both solutions:

1. #vecAoxvecB#

As stated above:

#vecAoxvecB=(A_yxxB_z-A_zxxB_y)i+(A_zxxB_x-A_x xxBz)j+(A_x xxB_y-A_yxxB_x)k#

#vecAoxvecB=(1xx(-3)-1xx2)i+(1xx3-(-1)xx(-3))j+(-1 xx2-1xx3)k#

#vecAoxvecB=(-3-2)i+(3-3)j+(-2-3)k#

#color(red)(vecAoxvecB=-5i+0j-5k#

1. #vecBoxvecA#

As a flip to the first formulation, take the diagonals again, but the matrix is formed differently:

#|i j k|#
#|B_x B_y B_z|#
#|A_x A_y A_z|#

#vecBoxvecA=(A_zxxB_y-A_yxxB_z)i+(A_x xxB_z-A_z xxBx)j+(A_y xxB_x-A_x xxB_y)k#

Notice that the subtractions are flipped around. This is what causes the 'Equal and opposite' form.

#vecBoxvecA=(1xx2-1xx(-3))i+((-1) xx(-3)-1 xx3)j+(1 xx3-(-1) xx2)k#

#vecBoxvecA=(2-(-3))i+(3-3)j+(3-(-2))k#
#color(blue)(vecBoxvecA=5i+0j+5k#