Question

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

Answer 1

There are two unit vectors here, depending on your order of operations. They are `(-5i +0j -5k)` and `(5i +0j 5k)`

Explanation:

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`