Cross[a,b]
gives the vector cross product of a and b.
 
     
   Cross 
Cross[a,b]
gives the vector cross product of a and b.
Details
 
   - If a and b are lists of length 3, corresponding to vectors in three dimensions, then Cross[a,b] is also a list of length 3.
- Cross[a,b] can be entered in StandardForm and InputForm as ab, a  cross cross b or a\[Cross]b. Note the difference between \[Cross] and \[Times]. b or a\[Cross]b. Note the difference between \[Cross] and \[Times].
- Cross is antisymmetric, so that Cross[b,a] is -Cross[a,b]. »
- Cross[{x,y}] gives the perpendicular vector {-y,x}.
- In general, Cross[v1,v2,…,vn-1] is a totally antisymmetric product which takes vectors of length n and yields a vector of length n that is orthogonal to all of the vi.
- Cross[v1,v2,…] gives the dual (Hodge star) of the wedge product of the vi, viewed as one‐forms in n dimensions.
Examples
open all close allBasic Examples (3)
Scope (9)
Find the cross product of machine-precision vectors:
Cross product of complex vectors:
Cross product of exact vectors:
The cross product of arbitrary-precision vectors:
Cross product of symbolic vectors:
Compute the cross product of QuantityArray vectors:
The QuantityArray structure is preserved:
The cross product of a single vector in two dimensions:
The result is perpendicular to the original vector:
Define two vectors in three dimensions:
Verify that Cross is antisymmetric:
Define three vectors in four dimensions:
Compute the cross product of the vectors:
Verify that the product is orthogonal to all three vectors:
Compute all possible product orders; each swap of two vectors merely changes the overall sign:
Applications (10)
Geometric Applications (5)
Find the normal to the plane spanned by two vectors:
Verify that the result is perpendicular to both inputs:
Find a vector perpendicular to a vector in the plane:
Verify that u and v are perpendicular:
Find a vector orthogonal to n-1 vectors in n dimensions:
Find the area of the parallelogram defined by two vectors:
Compare with a direct computation using Area:
This can also be computed as ![TemplateBox[{a}, Norm] TemplateBox[{b}, Norm] sin(theta) TemplateBox[{a}, Norm] TemplateBox[{b}, Norm] sin(theta)](Files/Cross.en/13.png) , with
, with  the angle between the vectors:
 the angle between the vectors:
The Frenet–Serret system encodes every space curve's properties in a vector basis and scalar functions. Consider the following curve:
Define the tangent, normal and binormal vectors in terms of cross products of the first two derivatives:
These three vectors define a right-handed, orthonormal basis for  :
:
Compute the curvature,  , and torsion,
, and torsion,  , which quantify how the curve bends:
, which quantify how the curve bends:
Verify the answers using FrenetSerretSystem:
Visualize the curve and the associated moving basis, also called a frame:
Physical Applications (5)
Find the torque about the origin of a force ![TemplateBox[{10, "N", newtons, "Newtons"}, QuantityTF] TemplateBox[{10, "N", newtons, "Newtons"}, QuantityTF]](Files/Cross.en/18.png) straight down applied at the point
 straight down applied at the point ![TemplateBox[{{{, {1.5`, ,, 3.2`, ,, 1.25`}, }}, "m", meters, "Meters"}, QuantityTF] TemplateBox[{{{, {1.5`, ,, 3.2`, ,, 1.25`}, }}, "m", meters, "Meters"}, QuantityTF]](Files/Cross.en/19.png) :
:
Torque is given by the formula  :
:
Find the angular momentum of a particle of mass ![TemplateBox[{3, "kg", kilograms, "Kilograms"}, QuantityTF] TemplateBox[{3, "kg", kilograms, "Kilograms"}, QuantityTF]](Files/Cross.en/21.png) , velocity
, velocity ![TemplateBox[{{{, {1.5`, ,, 2.3`, ,, {-, 3.4`}}, }}, {"m", , "/", , "s"}, meters per second, {{(, "Meters", )}, /, {(, "Seconds", )}}}, QuantityTF] TemplateBox[{{{, {1.5`, ,, 2.3`, ,, {-, 3.4`}}, }}, {"m", , "/", , "s"}, meters per second, {{(, "Meters", )}, /, {(, "Seconds", )}}}, QuantityTF]](Files/Cross.en/22.png) and position
 and position ![TemplateBox[{{{, {2.5`, ,, {-, 3.3`}, ,, 1.4`}, }}, "m", meters, "Meters"}, QuantityTF] TemplateBox[{{{, {2.5`, ,, {-, 3.3`}, ,, 1.4`}, }}, "m", meters, "Meters"}, QuantityTF]](Files/Cross.en/23.png) about the origin:
 about the origin:
Angular momentum is given by  , with linear momentum
, with linear momentum  equal to
 equal to  :
:
Find the magnetic force on the particle of charge ![TemplateBox[{2, "C", coulombs, "Coulombs"}, QuantityTF] TemplateBox[{2, "C", coulombs, "Coulombs"}, QuantityTF]](Files/Cross.en/27.png) and velocity
 and velocity ![TemplateBox[{{{, {3, ,, {-, 4}, ,, 5}, }}, {"m", , "/", , "s"}, meters per second, {{(, "Meters", )}, /, {(, "Seconds", )}}}, QuantityTF] TemplateBox[{{{, {3, ,, {-, 4}, ,, 5}, }}, {"m", , "/", , "s"}, meters per second, {{(, "Meters", )}, /, {(, "Seconds", )}}}, QuantityTF]](Files/Cross.en/28.png) moving through a magnetic field of
 moving through a magnetic field of ![TemplateBox[{0.075`, "T", teslas, "Teslas"}, QuantityTF] TemplateBox[{0.075`, "T", teslas, "Teslas"}, QuantityTF]](Files/Cross.en/29.png) in the positive
 in the positive  direction:
 direction:
Use UnitSimplify get the expected unit Newtons, and MatrixForm to format the vector:
Consider a particle constrained to rotate at a fixed distance from the  axis:
 axis:
Define the angular velocity by means of a cross product:
Many properties can be expressed in terms of  . The linear velocity
. The linear velocity  equals
 equals  :
:
The perpendicular or centripetal acceleration equals  :
:
Since  and
 and  are orthogonal, it is immediate that
 are orthogonal, it is immediate that ![TemplateBox[{{a, _, {(, perp, )}}}, Norm]=TemplateBox[{omega}, Norm] TemplateBox[{{r, ^, {(, ', )}}}, Norm] TemplateBox[{{a, _, {(, perp, )}}}, Norm]=TemplateBox[{omega}, Norm] TemplateBox[{{r, ^, {(, ', )}}}, Norm]](Files/Cross.en/39.png) :
:
The well-known formula for centripetal acceleration, ![TemplateBox[{{a, _, {(, perp, )}}}, Norm]=(TemplateBox[{{r, ^, {(, ', )}}}, Norm]^2)/(TemplateBox[{r}, Norm])=(TemplateBox[{v}, Norm]^2)/(TemplateBox[{r}, Norm]) TemplateBox[{{a, _, {(, perp, )}}}, Norm]=(TemplateBox[{{r, ^, {(, ', )}}}, Norm]^2)/(TemplateBox[{r}, Norm])=(TemplateBox[{v}, Norm]^2)/(TemplateBox[{r}, Norm])](Files/Cross.en/40.png) , also holds:
, also holds:
The derivative of  is the angular velocity
 is the angular velocity  :
:
The acceleration parallel to the direction of motion,  , equals
, equals  :
:
Note that the linear acceleration  equals the sum
 equals the sum  :
:
Cross products with respect to fixed three-dimensional vectors can be represented by matrix multiplication, which is useful in studying rotational motion. Construct the antisymmetric matrix representing the linear operator  , where
, where  is an angular velocity about the
 is an angular velocity about the  axis:
 axis:
Verify that the action of  is the same as doing a cross product with
 is the same as doing a cross product with  :
:
The rotation matrix at time  is the matrix exponential of
 is the matrix exponential of  times the previous matrix:
 times the previous matrix:
Verify  using RotationMatrix:
 using RotationMatrix:
The point  at time zero will be
 at time zero will be  at time
 at time  :
:
The velocity of  will be given by
 will be given by  :
:
Properties & Relations (10)
If u and v are linearly independent, u×v is nonzero and orthogonal to u and v:
If u and v are linearly dependent, u×v is zero:
For three-dimensional vectors, ![TemplateBox[{{{u, _, 1}, x, {u, _, 2}}}, Norm]=TemplateBox[{{u, _, 1}}, Norm] TemplateBox[{{u, _, 2}}, Norm]sin(theta)  TemplateBox[{{{u, _, 1}, x, {u, _, 2}}}, Norm]=TemplateBox[{{u, _, 1}}, Norm] TemplateBox[{{u, _, 2}}, Norm]sin(theta)](Files/Cross.en/62.png) , with
, with  the angle between
 the angle between  and
 and  :
:
The norm of Cross[u1,…,uk] is the measure of the k-dimensional parallelopiped spanned by ui:
Cross is antisymmetric:
Cross is linear in each argument:
Since Cross is linear, the operator  can be represented by matrix multiplication:
 can be represented by matrix multiplication:
Multiplying a vector  by by the antisymmetric matrix
 by by the antisymmetric matrix  is equivalent to
 is equivalent to  :
:
There is a corresponding operator,  , for computing the product in the opposite order:
, for computing the product in the opposite order:
These two matrices are transposes or—equivalently, due to antisymmetry—negations of each other:
Cross in dimension  is the contraction of
 is the contraction of  vectors into the Levi-Civita tensor:
 vectors into the Levi-Civita tensor:
Cross of  vectors in dimension
 vectors in dimension  is (
 is ( times the Hodge dual of their tensor product:
 times the Hodge dual of their tensor product:
The Hodge dual of the TensorWedge of  
  -vectors coincides with the Cross of those vectors:
-vectors coincides with the Cross of those vectors:
TensorWedge can treat higher-rank forms:
See Also
Dot Signature Outer Curl TensorWedge LeviCivitaTensor HodgeDual KroneckerProduct Orthogonalize VectorAngle
Characters: \[Cross]
Tech Notes
Related Guides
History
Introduced in 1996 (3.0)
Text
Wolfram Research (1996), Cross, Wolfram Language function, https://reference.wolfram.com/language/ref/Cross.html.
CMS
Wolfram Language. 1996. "Cross." Wolfram Language & System Documentation Center. Wolfram Research. https://reference.wolfram.com/language/ref/Cross.html.
APA
Wolfram Language. (1996). Cross. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/Cross.html
BibTeX
@misc{reference.wolfram_2025_cross, author="Wolfram Research", title="{Cross}", year="1996", howpublished="\url{https://reference.wolfram.com/language/ref/Cross.html}", note=[Accessed: 30-October-2025]}
BibLaTeX
@online{reference.wolfram_2025_cross, organization={Wolfram Research}, title={Cross}, year={1996}, url={https://reference.wolfram.com/language/ref/Cross.html}, note=[Accessed: 30-October-2025]}

 :
:

![(v(t)xomega)/(TemplateBox[{omega}, Norm]^2) (v(t)xomega)/(TemplateBox[{omega}, Norm]^2)](Files/Cross.en/61.png)


