This post will be about the cross product. This topic can be found in a high school Calculus and Vectors course (that I know of in Ontario, Canada) or in a introductory linear algebra course.
In here, we will deal with vectors in a three dimensional space. Also, it is easier to compute products if one knows how to compute (2 by 2) determinants. With that being said, I assume that the reader is familiar with determinants.
Suppose we have two vectors \(\mathbf{u} = (u_1, u_2, u_{3})\) and \(\mathbf{v} = (v_1, v_2, v_{3})\) in three dimensions. The cross product $ ( ) $ of two vectors is a (different) vector which is perpendicular to the two vectors.
There are two ways to compute cross products. The first formula does not assume the reader knows determinants. This formula is given by:
\[ \mathbf{u} \times \mathbf{v} = (u_{2}v_{3} - u_{3}v_{2}, u_{3}v_{1} - u_{1}v_{3}, u_{1}v_{2} - u_{2}v_{1} )\]
The second formula is actually the same as the above but in determinant notation. This formula is:
Source: http://quicklatex.com/cache3/c4/ql_926219b87045b16e28f37e299ca0d7c4_l3.png
Recall that the 2 by 2 determinant is equal to \(ad - bc\).
After looking at those two formulas you might be wondering how do I memorize all that stuff!? With some close observations, you can spot a few patterns and come up with some memory aids.
With the first formula, all the terms are in the \(uv\) format. If you look closely at the numbered subscripts, the numbers are mirrored. In the first part we have \(u_{2}v_{3} - u_{3}v_{2}\). The first subscript is a 2 which goes with the u and the second subscript is a 3 with the v. After the minus sign the subscripts are mirrored in the order of 3 and 2. The second part has subscripts of 3,1,1,3 and the last part has 1,2,2,1.
For the first formula, one possible memory aid would be 2332, 3113 and 1221. An alternative would be 23, 31 and 12 since we have \(u_{2}v_{3}\) representing 23 and \(- u_{3}v_{2}\), 31 for \(u_{3}v_{1} - u_{1}v_{3}\) and 12 for \(u_{1}v_{2} - u_{2}v_{1}\).
The second formula assumes that the reader is familiar with the linear algebra concept of determinants. Notice how the top row in each determinant is for the \(\textbf{u}\) vector and the bottom row is for the \(\textbf{v}\) vector. One possible memory aid would be 23, -13, and 12 as the subscripted numbers are used as we go from the leftmost column to the rightmost column. The negative for -13 represents the negative determinant.
Another way to arrive to the second formula is through this multi-part method.
Form a 2 by 3 matrix in the format of
The first component of the cross product can be computed by deleting the first column of the above matrix and take the resulting 2 by 2 determinant.
The second component consists of deleting the second column of the 2 by 3 matrix and taking the determinant of
The final and third component of the cross product \(\textbf{u} \times \textbf{v}\) would be removing the last column of the matrix and computing the determinant
The result would be the second formula turned into the first formula.
Like with any new math operation like the cross product, there are rules associated with them. In grade school, there were rules for addition, subtraction, multiplication and division.
Here are the properties of the cross product given the vectors \(\textbf{u}\), \(\textbf{v}\) and \(\textbf{w}\) in 3-space with \(k\) as a numeric scalar. (\(\textbf{0}\) is the zero vector in 3-space here)
\[ \textbf{u} \times \textbf{v} = - (\textbf{v} \times \textbf{u})\]
\[ \textbf{u} \times (\textbf{v} + \textbf{w}) = (\textbf{u} \times \textbf{v}) + (\textbf{u} \times \textbf{w}) \text{ (Multiply From Left) }\]
\[(\textbf{u} + \textbf{v}) \times \textbf{w} = (\textbf{u} \times \textbf{w}) + (\textbf{v} \times \textbf{w}) \text{ (Multiply From Right) }\]
\[k(\textbf{u} \times \textbf{v}) = (k \textbf{u}) \times \textbf{v} = \textbf{u} \times (k \textbf{v}) \]
\[\textbf{u} \times \textbf{0} = \textbf{0} \times \textbf{u} \]
\[\textbf{u} \times \textbf{u} = \textbf{0} \]
Example One
Given \(\textbf{u} = (0, -3, 2)\) and \(\textbf{v} = (5, -2, 1)\), Find \(\textbf{u} \times \textbf{v}\) and \(\textbf{v} \times \textbf{u}.\)
We can use the formula \[\textbf{u} \times \textbf{v} = (u_{2}v_{3} - u_{3}v_{2}, u_{3}v_{1} - u_{1}v_{3}, u_{1}v_{2} - u_{2}v_{1} )\] or the determinant notation of the formula. Here, the determinant notation of the formula will be used.
We also compute \(\textbf{v} \times \textbf{u}\).
You can also verify the property \(\textbf{u} \times \textbf{v} = - (\textbf{v} \times \textbf{u})\). In this case \(\textbf{u} \times \textbf{v} = (1, 10, 15 ) = - (-1, -10, -15) = - (\textbf{v} \times \textbf{u})\).
Example Two
Suppose that there are vectors \(\textbf{u} = (-3, -1, 0)\) and \(\textbf{v} = (2, 0, -1)\). Find a vector which is perpendicular/orthogonal to both vectors \(\textbf{u}\) and \(\textbf{v}\).
The vector cross product of a two vectors is a (third) vector which is perpendicular to both the two vectors.
We find the cross product \(\textbf{u} \times \textbf{v}\) as follows:
The vector that is perpendicular to vectors \(\textbf{u}\) and \(\textbf{v}\) is \(\textbf{u} \times \textbf{v} = (1, -3, 2)\).
Here are few practice problems. Solutions are in the next section.
Given two vectors \(\textbf{u} = (1, 2, 3)\) and \(\textbf{v} = (2, 3, 4)\). What is the cross product \(\textbf{u} \times \textbf{v}\)?
Given \(\textbf{u} = (2, -4, 7)\) and \(\textbf{v} = (1, 0, 2)\). What is the cross product \(\textbf{v} \times \textbf{u}\)?
Suppose you have the vector \(\textbf{u} = (34, 23, -12)\). Verify that \(\textbf{u} \times \textbf{u} = \textbf{0}\) where is the zero vector in 3-space.
In this question we have three vectors. These vectors are \(\textbf{u} = (1, 1, 1)\), \(\textbf{v} = (0, 3, -2)\) and \(\textbf{w} = (8, 2, 1)\). What is the cross product \((\textbf{u} \times \textbf{v}) \times \textbf{w}\)?
\(\textbf{u} \times \textbf{v} = (-1, 2, -1)\)
\(\textbf{v} \times \textbf{u} - (8, -3, -4)\)
The numbers look scary but once the formula is applied you will find that you will get the zero vector.
\(\textbf{u} \times \textbf{v} = (-5, 2, 3)\) and \(\textbf{u} \times \textbf{v}) \times \textbf{w} = (-4, 29, -26)\)
