A square matrix is a matrix that has the same number of rows and columns.
For every square matrix, we can calculate a determinant, which is a real number that has several important uses in linear mathematics.
For a {2 \mathord{\times} 2} matrix:
The determinant of the matrix { \hspace+2pt \left[\hspace+2pt \begin{array}{rr} \hfil a &\hfil\hspace-5pt b \cr \hfil c &\hfil\hspace-5pt d \cr \end{array} \hspace+2pt\right] \space } is { \space ad - bc. }
The determinant is an operator, and it’s written either like this:
{ \det \left[\hspace+2pt \begin{array}{rr} \hfil a &\hfil\hspace-5pt b \cr \hfil c &\hfil\hspace-5pt d \cr \end{array} \hspace+2pt\right] = ad - bc }
or like this:
{ \left| \hspace+2pt \begin{array}{rr} \hfil a &\hfil\hspace-5pt b \cr \hfil c &\hfil\hspace-5pt d \cr \end{array} \hspace+2pt \right| = ad - bc }
Notice that the pattern is to “cross multiply” the two diagonals, and then subtract the second from the first, moving this way:
We can use the determinant to find the area of a parallelogram, given that we know where its vertex points are located.
For example, we can define a parallelogram using two vectors {\mathbf{a}} and {\mathbf{b},} like this:
{\mathbf{b} = (2, 4)}
{\mathbf{a} \mathord{+} \mathbf{b} = (7, 5)}
The area of the parallelogram is:
{ \left| \det \left[\hspace+2pt \begin{array}{rr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \end{array} \hspace+2pt\right] \right| \ = \ \left| \det \left[\hspace+2pt \begin{array}{rr} \hfil 5 &\hfil\hspace-5pt 1 \cr \hfil 2 &\hfil\hspace-5pt 4 \cr \end{array} \hspace+2pt\right] \right| \ = \ |5 \mathord{\times} 4 - 1 \mathord{\times} 2| \ = \ |18| \ = \ 18 }
The reason we’re taking the absolute value of the determinant is because it sometimes produces a negative result, and we might not want the area to be negative.
A parallelogram has these four vertex points: {(0, 0),} {(3, 4),} {(1, 7),} and {(-2, 3).}
Use the {2 \mathord{\times} 2} determinant to find the area of this parallelogram:
There’s a different way to calculate the area of a parallelogram, using the
“base times height” formula. In this diagram, the base of the
parallelogram is {|\mathbf{a}|,} and its height is
Given {|\mathbf{b}|} and the interior angle {\theta,} we can calculate {h} like this:
[[ \sin \theta = {{h} \over {|\mathbf{b}|}} ]]
{h = |\mathbf{b}| \sin \theta}
We can now apply the “base times height” formula to find the area of the parallelogram:
{\text{area} \ = \ |\mathbf{a}| h \ = \ |\mathbf{a}| \, |\mathbf{b}| \sin \theta}
If we equate this result with the area that we got above from the determinant, we obtain this identity:
{ \left| \det \left[\hspace+2pt \begin{array}{rr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \end{array} \hspace+2pt\right] \right| \ = \ |\mathbf{a}| \, |\mathbf{b}| \, \sin \theta } {(}for {0° \le \theta \le 180°)}
Let’s remove the absolute value from the last equation in the previous section. This will give us an identity that applies to all determinants:
{ \det \left[\hspace+2pt \begin{array}{rr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \end{array} \hspace+2pt\right] \ = \ |\mathbf{a}| \, |\mathbf{b}| \, \sin \theta } {(}for all {\theta)}
The sign of {\theta} then indicates the sign of the determinant.
The resulting determinant measures the signed area of the parallelogram.
Signed area is the area of a shape that’s either positive or negative, depending on the orientation of the shape, or depending on whether the shape’s area is being added to or subtracted from another area.
The orientation of a parallelogram is defined by the sign of its determinant. The orientation changes sign if you swap its two row vectors in the determinant:
{ \det \left[\hspace+2pt \begin{array}{rr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \end{array} \hspace+2pt\right] \ = \ - \det \left[\hspace+2pt \begin{array}{rr} \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 \cr \end{array} \hspace+2pt\right] }
Given two 2D vectors, {\mathbf{a}} and {\mathbf{b},} we have the following two identities:
{\mathbf{a} _1 \mathbf{b} _1 + \mathbf{a} _2 \mathbf{b} _2 = |\mathbf{a}| \, |\mathbf{b}| \, \cos \theta \quad } (the dot product of {\mathbf{a}} and {\mathbf{b}}) {\mathbf{a} _1 \mathbf{b} _2 - \mathbf{a} _2 \mathbf{b} _1 = |\mathbf{a}| \, |\mathbf{b}| \, \sin \theta \quad } (the determinant { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \end{array} \hspace+1pt \right| })
Here’s an informal way to understand these two expressions:
{\mathbf{a} _1 \mathbf{b} _1 + \mathbf{a} _2 \mathbf{b} _2}
This is a measurement of the directional “similarity” of {\mathbf{a}} and{\mathbf{b}}:
- Two vectors in the same direction have a positive directional “similarity”.
- Two vectors in opposite directions have a negative directional “similarity”.
- Two vectors that are perpendicular have a directional “similarity” of zero.
{\mathbf{a} _1 \mathbf{b} _2 - \mathbf{a} _2 \mathbf{b} _1}
This is a measurement of the “perpendicularity” of {\mathbf{a}} and{\mathbf{b}}:
- Two vectors that are parallel have zero “perpendicularity”.
- Two vectors that are perpendicular have a “perpendicularity” that’s either positive or negative, depending on their orientation.
Here’s a summary of the cases where they evaluate to
Vectors {\mathbf{a}} and {\mathbf{b}} are perpendicular if and only if:{ \mathbf{a} \cdot \mathbf{b} \ = \ \mathbf{a} _1 \mathbf{b} _1 + \mathbf{a} _2 \mathbf{b} _2 \ = \ 0 }Vectors {\mathbf{a}} and {\mathbf{b}} are parallel if and only if:{ \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \end{array} \hspace+1pt \right| \ = \ {\mathbf{a} _1 \mathbf{b} _2 - \mathbf{a} _2 \mathbf{b} _1} \ = \ 0 }
For a {3 \mathord{\times} 3} matrix:
The determinant of the matrix { \left[\hspace+2pt \begin{array}{rrr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 &\hfil\hspace-5pt \mathbf{a} _3 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt \mathbf{c} _2 &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+2pt\right] }
is:{ \left| \hspace+1pt \begin{array}{rrr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 &\hfil\hspace-5pt \mathbf{a} _3 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt \mathbf{c} _2 &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+1pt \right| \ = \ \mathbf{a} _1 { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{b} _2 &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil \mathbf{c} _2 &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+1pt \right| } \ - \ \mathbf{a} _2 { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+1pt \right| } \ + \ \mathbf{a} _3 { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt \mathbf{c} _2 \cr \end{array} \hspace+1pt \right| } }
Here’s the pattern that shows how the three terms are constructed:
{ \left| \hspace+1pt \begin{array}{rrr} \hfil \boxed{\mathbf{a} _1} &\hfil\hspace-5pt {\class{gray}{\mathbf{a} _2}} &\hfil\hspace-5pt {\class{gray}{\mathbf{a} _3}} \cr \hfil {\class{gray}{\mathbf{b} _1}} &\hfil\hspace-5pt \mathbf{b} _2 &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil {\class{gray}{\mathbf{c} _1}} &\hfil\hspace-5pt \mathbf{c} _2 &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+1pt \right| } 🡆 { + \ \mathbf{a} _1 { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{b} _2 &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil \mathbf{c} _2 &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+1pt \right| } } { \left| \hspace+1pt \begin{array}{rrr} \hfil {\class{gray}{\mathbf{a} _1}} &\hfil\hspace-5pt \boxed{\mathbf{a} _2} &\hfil\hspace-5pt {\class{gray}{\mathbf{a} _3}} \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt {\class{gray}{\mathbf{b} _2}} &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt {\class{gray}{\mathbf{c} _2}} &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+1pt \right| } 🡆 { - \ \mathbf{a} _2 { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+1pt \right| } } { \left| \hspace+1pt \begin{array}{rrr} \hfil {\class{gray}{\mathbf{a} _1}} &\hfil\hspace-5pt {\class{gray}{\mathbf{a} _2}} &\hfil\hspace-5pt \boxed{\mathbf{a} _3} \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 &\hfil\hspace-5pt {\class{gray}{\mathbf{b} _3}} \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt \mathbf{c} _2 &\hfil\hspace-5pt {\class{gray}{\mathbf{c} _3}} \cr \end{array} \hspace+1pt \right| } 🡆 { + \ \mathbf{a} _3 { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt \mathbf{c} _2 \cr \end{array} \hspace+1pt \right| } }
We can use the {3 \mathord{\times} 3} determinant to find the volume of a parallelepiped, given the locations of its vertex points.
Let’s define a parallelepiped by starting with four vertex points: {(0,0,0),} {\mathbf{a},} and {\mathbf{b},} and {\mathbf{c},} and add them together in all combinations to obtain the four remaining vertex points:
The volume of this parallelepiped is:
{ V \ = \ \left| \hspace+1pt \begin{array}{rrr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 &\hfil\hspace-5pt \mathbf{a} _3 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt \mathbf{c} _2 &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+1pt \right| \ = \ \mathbf{a} _1 { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{b} _2 &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil \mathbf{c} _2 &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+1pt \right| } \ - \ \mathbf{a} _2 { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+1pt \right| } \ + \ \mathbf{a} _3 { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt \mathbf{c} _2 \cr \end{array} \hspace+1pt \right| } }
Given the definition of the determinant:
{ \left| \hspace+1pt \begin{array}{rrr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 &\hfil\hspace-5pt \mathbf{a} _3 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt \mathbf{c} _2 &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+1pt \right| \ = \ \mathbf{a} _1 { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{b} _2 &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil \mathbf{c} _2 &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+1pt \right| } \ - \ \mathbf{a} _2 { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _3 \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt \mathbf{c} _3 \cr \end{array} \hspace+1pt \right| } \ + \ \mathbf{a} _3 { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \hfil \mathbf{c} _1 &\hfil\hspace-5pt \mathbf{c} _2 \cr \end{array} \hspace+1pt \right| } }
Calculate the volume of the parallelepiped that has these eight vertex points:
{(0, 0, 0)}
{\mathbf{a} = (2, -2, 3)}
{\mathbf{b} = (3, 1, 2)}
{\mathbf{c} = (4, 2, 3)}
{\mathbf{a}+\mathbf{b}}
{\mathbf{a}+\mathbf{c}}
{\mathbf{b}+\mathbf{c}}
{\mathbf{a}+\mathbf{b}+\mathbf{c}}
The cross product is an operator that takes two vectors and combines them to produce a third vector. The cross product of {\mathbf{a}} and {\mathbf{b}} is written as {\mathbf{a} \times \mathbf{b}.}
Here’s the definition of {\mathbf{a} \times \mathbf{b}} for two
{ \mathbf{a} \times \mathbf{b} \ = \; \left( \, \mathbf{a} _2 \mathbf{b} _3 - \mathbf{a} _3 \mathbf{b} _2, \ \ \mathbf{a} _3 \mathbf{b} _1 - \mathbf{a} _1 \mathbf{b} _3, \ \ \mathbf{a} _1 \mathbf{b} _2 - \mathbf{a} _2 \mathbf{b} _1 \vphantom{\large X} \, \right) }
or, expressing it using {2 \mathord{\times} 2} determinants:
{ \mathbf{a} \times \mathbf{b} \ = \; \left( \ { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{a} _2 &\hfil\hspace-5pt \mathbf{a} _3 \cr \hfil \mathbf{b} _2 &\hfil\hspace-5pt \mathbf{b} _3 \cr \end{array} \hspace+1pt \right| } , \ { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{a} _3 &\hfil\hspace-5pt \mathbf{a} _1 \cr \hfil \mathbf{b} _3 &\hfil\hspace-5pt \mathbf{b} _1 \cr \end{array} \hspace+1pt \right| } , \ \ { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \end{array} \hspace+1pt \right| } \, \right) }
There’s another way of defining the cross product that’s based on the {3 \mathord{\times} 3} determinant:
{ \mathbf{a} \times \mathbf{b} \ = \left| \hspace+1pt \begin{array}{ccc} {\LARGE{\hat{\normalsize \mathbf{i}}}} &\hspace-5pt {\LARGE{\hat{\normalsize \mathbf{j}}}} &\hspace-5pt {\LARGE{\hat{\normalsize \mathbf{k}}}} \cr \mathbf{a} _1 &\hspace-5pt \mathbf{a} _2 &\hspace-5pt \mathbf{a} _3 \cr \mathbf{b} _1 &\hspace-5pt \mathbf{b} _2 &\hspace-5pt \mathbf{b} _3 \cr \end{array} \hspace+1pt \right| \ = \ { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{a} _2 &\hfil\hspace-5pt \mathbf{a} _3 \cr \hfil \mathbf{b} _2 &\hfil\hspace-5pt \mathbf{b} _3 \cr \end{array} \hspace+1pt \right| } \ {\LARGE{\hat{\normalsize \mathbf{i}}}} \ - \ { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _3 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _3 \cr \end{array} \hspace+1pt \right| } \ {\LARGE{\hat{\normalsize \mathbf{j}}}} \ + \ { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \end{array} \hspace+1pt \right| } \ {\LARGE{\hat{\normalsize \mathbf{k}}}} }
where:{{\LARGE{\hat{\normalsize \mathbf{i}}}} \; = (1, 0, 0)} (the unit vector in the direction of the positive {x} axis)and the {\text{“}\mathord{+}\text{”}} and {\text{“}\mathord{-}\text{”}} operators are vector addition and vector subtraction.
{{\LARGE{\hat{\normalsize \mathbf{j}}}} \; = (0, 1, 0)} (the unit vector in the direction of the positive {y} axis)
{{\LARGE{\hat{\normalsize \mathbf{k}}}} = (0, 0, 1)} (the unit vector in the direction of the positive {z} axis)
Here, {{\LARGE{\hat{\normalsize \mathbf{i}}}}, \, } {{\LARGE{\hat{\normalsize \mathbf{j}}}},} and {{\LARGE{\hat{\normalsize \mathbf{k}}}}} are the {3}-dimensional
unit vectors that point in the direction of the positive {x,} {y,} and {z} axes,
respectively. These three vectors are called the standard basis vectors
of
The advantage of learning the cross product using the {3 \mathord{\times} 3} determinant is that you only need to learn a single procedure that works for both the determinant and the cross product.
Here’s an example of a cross product calculated using the {3 \mathord{\times} 3} determinant:
{ \mathbf{a} = (1, 2, 3) }
{ \mathbf{b} = (4, 5, 6) } { \mathbf{a} \times \mathbf{b} \ = \left| \hspace+1pt \begin{array}{ccc} {\LARGE{\hat{\normalsize \mathbf{i}}}} &\hspace-3pt {\LARGE{\hat{\normalsize \mathbf{j}}}} &\hspace-3pt {\LARGE{\hat{\normalsize \mathbf{k}}}} \cr 1 &\hspace-3pt 2 &\hspace-3pt 3 \cr 4 &\hspace-3pt 5 &\hspace-3pt 6 \cr \end{array} \hspace+1pt \right| } { \mathbf{a} \times \mathbf{b} \ = \ { \left| \hspace+1pt \begin{array}{rr} \hfil 2 &\hfil\hspace-3pt 3 \cr \hfil 5 &\hfil\hspace-3pt 6 \cr \end{array} \hspace+1pt \right| } \ {\LARGE{\hat{\normalsize \mathbf{i}}}} \ - \ { \left| \hspace+1pt \begin{array}{rr} \hfil 1 &\hfil\hspace-3pt 3 \cr \hfil 4 &\hfil\hspace-3pt 6 \cr \end{array} \hspace+1pt \right| } \ {\LARGE{\hat{\normalsize \mathbf{j}}}} \ + \ { \left| \hspace+1pt \begin{array}{rr} \hfil 1 &\hfil\hspace-3pt 2 \cr \hfil 4 &\hfil\hspace-3pt 5 \cr \end{array} \hspace+1pt \right| } \ {\LARGE{\hat{\normalsize \mathbf{k}}}} } { \mathbf{a} \times \mathbf{b} \ = \ (-3) \ {\LARGE{\hat{\normalsize \mathbf{i}}}} \ - \ (-6) \ {\LARGE{\hat{\normalsize \mathbf{j}}}} \ + \ (-3) \ {\LARGE{\hat{\normalsize \mathbf{k}}}} } { \mathbf{a} \times \mathbf{b} \ = \ (-3)(1, 0, 0) \ - \ (-6)(0, 1, 0) \ + \ (-3)(0, 0, 1) } { \mathbf{a} \times \mathbf{b} \ = \ (-3, 0, 0) \ - \ (0, -6, 0) \ + \ (0, 0, -3) } { \mathbf{a} \times \mathbf{b} \ = \ (-3, 6, -3) }
Given the definition of the cross product:
{ \mathbf{a} \times \mathbf{b} \ = \left| \hspace+1pt \begin{array}{ccc} {\LARGE{\hat{\normalsize \mathbf{i}}}} &\hspace-5pt {\LARGE{\hat{\normalsize \mathbf{j}}}} &\hspace-5pt {\LARGE{\hat{\normalsize \mathbf{k}}}} \cr \mathbf{a} _1 &\hspace-5pt \mathbf{a} _2 &\hspace-5pt \mathbf{a} _3 \cr \mathbf{b} _1 &\hspace-5pt \mathbf{b} _2 &\hspace-5pt \mathbf{b} _3 \cr \end{array} \hspace+1pt \right| \ = \ { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{a} _2 &\hfil\hspace-5pt \mathbf{a} _3 \cr \hfil \mathbf{b} _2 &\hfil\hspace-5pt \mathbf{b} _3 \cr \end{array} \hspace+1pt \right| } \ {\LARGE{\hat{\normalsize \mathbf{i}}}} \ - \ { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _3 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _3 \cr \end{array} \hspace+1pt \right| } \ {\LARGE{\hat{\normalsize \mathbf{j}}}} \ + \ { \left| \hspace+1pt \begin{array}{rr} \hfil \mathbf{a} _1 &\hfil\hspace-5pt \mathbf{a} _2 \cr \hfil \mathbf{b} _1 &\hfil\hspace-5pt \mathbf{b} _2 \cr \end{array} \hspace+1pt \right| } \ {\LARGE{\hat{\normalsize \mathbf{k}}}} }
Calculate the cross product:
{(-2, 1, -1) \times (1, 3, 2)}
In this section, we’ll see how the vectors {\mathbf{a} \times \mathbf{b}} and {\mathbf{b} \times \mathbf{a}} are positioned in relation to the vectors {\mathbf{a}} and {\mathbf{b}}.
Here’s a diagram showing all four vectors:
Observe the following:
The value {|\mathbf{a} \times \mathbf{b}|} is the area of the parallelogram that’s defined by the vertex points {\mathbf{a},} {\mathbf{b},} {\mathbf{a} \mathord{+} \mathbf{b},} and the origin.
The area of the parallelogram can also be calculated as {|\mathbf{a}| \, |\mathbf{b}| \, \sin \theta,} so we have:
{|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| \, |\mathbf{b}| \, \sin \theta} {(}for {0° \le \theta \le 180°)}
From the above identity, we can see that if {\mathbf{a}} and {\mathbf{b}} are parallel, then {|\mathbf{a} \times \mathbf{b}| = 0.}
The parallelogram can be located anywhere in 3D space, positioned at any angle.
The vector {\mathbf{a} \times \mathbf{b}} points in a direction that’s perpendicular to both {\mathbf{a}} and {\mathbf{b}.}
To understand how the vectors {\mathbf{b}} and {\mathbf{a} \times \mathbf{b}} are perpendicular, you’ll need to view {\mathbf{b}} as pointing into the plane of the page, into the third dimension.
The vector {\mathbf{b} \times \mathbf{a}} also points in a direction that’s perpendicular to both {\mathbf{a}} and {\mathbf{b},} but in the opposite direction of {\mathbf{a} \times \mathbf{b}:}
{\mathbf{b} \times \mathbf{a} \, = \, -(\mathbf{a} \times \mathbf{b})}
This identity shows the anticommutative property of the cross product.
{|\mathbf{a} \times \mathbf{b}| = |\mathbf{b} \times \mathbf{a}|}
It’s sometimes convenient to define a 3D unit vector {{\LARGE{\hat{\normalsize \mathbf{n}}}}} that points in the direction of {\mathbf{a} \times \mathbf{b}.} This allows us to define the geometric form of the cross product as:
{\mathbf{a} \times \mathbf{b} = (|\mathbf{a}| \, |\mathbf{b}| \, \sin \theta) \ {\LARGE{\hat{\normalsize \mathbf{n}}}}} {(}for {0° \le \theta \le 180°)}
where:
[[ {\LARGE{\hat{\normalsize \mathbf{n}}}} = {{\mathbf{a} \times \mathbf{b}} \over {|\mathbf{a} \times \mathbf{b}|}} ]] {(|\mathbf{a} \times \mathbf{b}| \ne 0)}
The letter {{\LARGE{\hat{\normalsize \mathbf{n}}}}} is an abbreviation of the word “normal”. A normal vector is a vector that’s perpendicular to every vector that lies on a given plane.
In the previous section, we saw that the vectors {\mathbf{a} \times \mathbf{b}} and {\mathbf{b} \times \mathbf{a}} point in opposite directions.
But how do we tell which vector is located where?
To answer that question, we use the right hand rule:
The right hand rule for the cross product:
- Point the fingers of your right hand in the same direction as the first vector, facing your palm toward the second vector.
- Point your right thumb out, and it will point in the direction of the cross product.
This rule explains the orientation of {\mathbf{a} \times \mathbf{b}} in relation
to {\mathbf{a}} and
The right hand rule applies only when the {x,} {y,} and {z} axes are in their standard orientation:
Each of these cross products illustrates the right hand rule for the above diagram:
{{\LARGE{\hat{\normalsize \mathbf{i}}}} \ \times \ {\LARGE{\hat{\normalsize \mathbf{j}}}} \ = \ {\LARGE{\hat{\normalsize \mathbf{k}}}}}{{\LARGE{\hat{\normalsize \mathbf{j}}}} \ \times \ {\LARGE{\hat{\normalsize \mathbf{k}}}} \ = \ {\LARGE{\hat{\normalsize \mathbf{i}}}}}{{\LARGE{\hat{\normalsize \mathbf{k}}}} \ \times \ {\LARGE{\hat{\normalsize \mathbf{i}}}} \ = \ {\LARGE{\hat{\normalsize \mathbf{j}}}}}
