How to Calculate the Cross Product of Two Vectors

Last Updated: January 31, 2023 Fact Checked

The cross product is a type of vector multiplication only defined in three and seven dimensions that outputs another vector. This operation, used in almost exclusively three dimensions, is useful for applications in physics and engineering. In this article, we will calculate the cross product of two three-dimensional vectors defined in Cartesian coordinates.

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Cross Product of Vectors Diagram

Method 1

Method 1 of 2:

Calculating the Cross Product

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1
Consider two general three-dimensional vectors defined in Cartesian coordinates. ^{[1]
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- ${\begin{aligned}{\mathbf {a}}&=A{\mathbf {i}}+B{\mathbf {j}}+C{\mathbf {k}}\\{\mathbf {b}}&=D{\mathbf {i}}+E{\mathbf {j}}+F{\mathbf {k}}\end{aligned}}$
- Here, ${\mathbf {i}},{\mathbf {j}},{\mathbf {k}}$ are unit vectors, and $A,B,C,D,E,F$ are constants.
2
Set up the matrix. One of the easiest ways to compute a cross product is to set up the unit vectors with the two vectors in a matrix. ^{[2]
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- ${\mathbf {a}}\times {\mathbf {b}}={\begin{vmatrix}{\mathbf {i}}&{\mathbf {j}}&{\mathbf {k}}\\A&B&C\\D&E&F\end{vmatrix}}$
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3
Calculate the determinant of the matrix. Below, we use cofactor expansion (expansion by minors). ^{[3]
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- ${\mathbf {a}}\times {\mathbf {b}}=(BF-EC){\mathbf {i}}-(AF-DC){\mathbf {j}}+(AE-DB){\mathbf {k}}$
- This vector is orthogonal to both ${\mathbf {a}}$ and ${\mathbf {b}}.$

Method 2

Method 2 of 2:

Example

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1
Consider the two vectors below.
- ${\begin{aligned}{\mathbf {u}}&=2{\mathbf {i}}-{\mathbf {j}}+3{\mathbf {k}}\\{\mathbf {v}}&=5{\mathbf {i}}+7{\mathbf {j}}-4{\mathbf {k}}\end{aligned}}$
2
Set up the matrix.
- ${\mathbf {u}}\times {\mathbf {v}}={\begin{vmatrix}{\mathbf {i}}&{\mathbf {j}}&{\mathbf {k}}\\2&-1&3\\5&7&-4\end{vmatrix}}$
3
Calculate the determinant of the matrix.
- ${\begin{aligned}{\mathbf {u}}\times {\mathbf {v}}&=(4-21){\mathbf {i}}-(-8-15){\mathbf {j}}+(14+5){\mathbf {k}}\\&=-17{\mathbf {i}}+23{\mathbf {j}}+19{\mathbf {k}}\end{aligned}}$

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How do I calculate the vector triple product?

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Given vectors u, v, and w, the scalar triple product is u*(vXw). So by order of operations, first find the cross product of v and w. Set up a 3X3 determinant with the unit coordinate vectors (i, j, k) in the first row, v in the second row, and w in the third row. Evaluate the determinant (you'll get a 3 dimensional vector). Then dot that with u (to get a scalar). Inner products are abelian, so u*(vXw)=(vXw)*u. Interestingly, the absolute value of the T.S.P. yields the volume of a parallelpiped with 3 edges given by vectors u, v, and w.

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What is the vector analog of distance?

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Norm, sometimes also called magnitude, generalizes distance to vectors. Norm is denoted with vertical bars like absolute values. For example, |(3,-4)| = 5, and |(1,1,1,1)| = 2.

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The cross product of a vector with any multiple of itself is 0. This is easier shown when setting up the matrix. The second and third rows are linearly dependent, since you can write one as a multiple of the other. Then, the determinant of the matrix and therefore the cross product is 0.

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One can show that the vector produced by a cross product of two vectors ${\mathbf {a}}\times {\mathbf {b}}$ is orthogonal to both ${\mathbf {a}}$ and ${\mathbf {b}}.$ To do so, compute the dot products. These products are called triple products - since the operation on the outside is a dot product, these are the scalar triple products.
- ${\begin{aligned}{\mathbf {a}}&\cdot ({\mathbf {a}}\times {\mathbf {b}})\\{\mathbf {b}}&\cdot ({\mathbf {a}}\times {\mathbf {b}})\end{aligned}}$
- These triple products follow something known as cyclic permutation - that is, if you swap the positions of the vectors without reordering them, the expressions are equivalent. Then, we can rewrite them such that a vector is crossing with itself.
- ${\begin{aligned}{\mathbf {a}}&\cdot ({\mathbf {b}}\times {\mathbf {b}})\\{\mathbf {b}}&\cdot ({\mathbf {a}}\times {\mathbf {a}})\end{aligned}}$
- However, we know that the cross product of a vector with itself is 0. Since a dot product of the two vectors ends up being 0 as well, they are orthogonal.
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