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

Mar 13, 2018

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

#### 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 $\vec{A} \otimes \vec{B}$ is usually equal and opposite in magnitude of $\vec{B} \otimes \vec{A}$.

As a quick refresher, a cross-product of $\vec{A} \otimes \vec{B}$ 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:

$\left({A}_{y} \times {B}_{z} - {A}_{z} \times {B}_{y}\right) i + \left({A}_{z} \times {B}_{x} - {A}_{x} \times B z\right) j + \left({A}_{x} \times {B}_{y} - {A}_{y} \times {B}_{x}\right) k$

For the two solutions, lets set:
$\vec{A} = \left[- i + j + k\right]$
$\vec{B} = \left[3 i + 2 j - 3 k\right]$

Let's look at both solutions:

1. $\vec{A} \otimes \vec{B}$

As stated above:

$\vec{A} \otimes \vec{B} = \left({A}_{y} \times {B}_{z} - {A}_{z} \times {B}_{y}\right) i + \left({A}_{z} \times {B}_{x} - {A}_{x} \times B z\right) j + \left({A}_{x} \times {B}_{y} - {A}_{y} \times {B}_{x}\right) k$

$\vec{A} \otimes \vec{B} = \left(1 \times \left(- 3\right) - 1 \times 2\right) i + \left(1 \times 3 - \left(- 1\right) \times \left(- 3\right)\right) j + \left(- 1 \times 2 - 1 \times 3\right) k$

$\vec{A} \otimes \vec{B} = \left(- 3 - 2\right) i + \left(3 - 3\right) j + \left(- 2 - 3\right) k$

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

1. $\vec{B} \otimes \vec{A}$

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

$| i j k |$
$| {B}_{x} {B}_{y} {B}_{z} |$
$| {A}_{x} {A}_{y} {A}_{z} |$

$\vec{B} \otimes \vec{A} = \left({A}_{z} \times {B}_{y} - {A}_{y} \times {B}_{z}\right) i + \left({A}_{x} \times {B}_{z} - {A}_{z} \times B x\right) j + \left({A}_{y} \times {B}_{x} - {A}_{x} \times {B}_{y}\right) k$

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

$\vec{B} \otimes \vec{A} = \left(1 \times 2 - 1 \times \left(- 3\right)\right) i + \left(\left(- 1\right) \times \left(- 3\right) - 1 \times 3\right) j + \left(1 \times 3 - \left(- 1\right) \times 2\right) k$

$\vec{B} \otimes \vec{A} = \left(2 - \left(- 3\right)\right) i + \left(3 - 3\right) j + \left(3 - \left(- 2\right)\right) k$
color(blue)(vecBoxvecA=5i+0j+5k