# Vector3D.Division(Vector3D, Double)Vector3D.Division(Vector3D, Double)Vector3D.Division(Vector3D, Double)Vector3D.Division(Vector3D, Double) Operator

## Definition

Divides the specified Vector3D structure by the specified scalar and returns the result as a Vector3D.

``````public:
static System::Windows::Media::Media3D::Vector3D operator /(System::Windows::Media::Media3D::Vector3D vector, double scalar);``````
``public static System.Windows.Media.Media3D.Vector3D operator / (System.Windows.Media.Media3D.Vector3D vector, double scalar);``
``static member ( / ) : System.Windows.Media.Media3D.Vector3D * double -> System.Windows.Media.Media3D.Vector3D``
``Public Shared Operator / (vector As Vector3D, scalar As Double) As Vector3D``

#### Parameters

vector
Vector3D Vector3D Vector3D Vector3D

The Vector3D structure to be divided.

scalar
Double Double Double Double

The scalar to divide `vector` by.

#### Returns

The result of dividing `vector` by `scalar`.

## Examples

The following example shows how to use the overloaded addition operator to divide a Vector3D structure by a scalar.

``````private Vector overloadedDivisionOperatorExample()
{
Vector vector1 = new Vector(20, 30);
Vector vectorResult = new Vector();
Double scalar1 = 75;

// Divide vector by scalar.
// vectorResult is approximately equal to (0.26667,0.4)
vectorResult = vector1 / scalar1;

return vectorResult;

}
``````
``````Private Function overloadedDivisionOperatorExample() As Vector
Dim vector1 As New Vector(20, 30)
Dim vectorResult As New Vector()
Dim scalar1 As Double = 75

' Divide vector by scalar.
' vectorResult is approximately equal to (0.26667,0.4)
vectorResult = vector1 / scalar1

Return vectorResult

End Function
``````