UnsafeNativeMethods.Vector3.Cross Method

Language:

Determines the cross product of two 3-D vectors.

Note: For programming in Microsoft Visual Basic .NET or Microsoft JScript .NET, use the equivalent method in the Microsoft.DirectX structures.

Definition

Visual Basic Public Shared Function Cross( _
    ByVal pOut As Vector3, _
    ByVal pV1 As Vector3, _
    ByVal pV2 As Vector3 _
) As Vector3

C# public static Vector3 Cross(
    Vector3 pOut,
    Vector3 pV1,
    Vector3 pV2
);

C++ public:
static Vector3 Cross(
    Vector3 pOut,
    Vector3 pV1,
    Vector3 pV2
);

JScript public static function Cross(
    pOut : Vector3,
    pV1 : Vector3,
    pV2 : Vector3
) : Vector3;

Visual Basic	Public Shared Function Cross( _ ByVal pOut As Vector3, _ ByVal pV1 As Vector3, _ ByVal pV2 As Vector3 _ ) As Vector3
C#	public static Vector3 Cross( Vector3 pOut, Vector3 pV1, Vector3 pV2 );
C++	public: static Vector3 Cross( Vector3 pOut, Vector3 pV1, Vector3 pV2 );
JScript	public static function Cross( pOut : Vector3, pV1 : Vector3, pV2 : Vector3 ) : Vector3;

Parameters

pOut Microsoft.DirectX.Vector3
A Vector3 structure that is the cross product of two 3-D vectors.

pV1 Microsoft.DirectX.Vector3
Source Vector3 structure.

pV2 Microsoft.DirectX.Vector3
Source Vector3 structure.

Return Value

Microsoft.DirectX.Vector3
A Vector3 structure that is the cross product of two 3-D vectors.

Remarks

The following C# example demonstrates how the Cross method finds the cross-product.
[C#]
Vector3 returnValue;
    returnValue.x = left.y * right.z - left.z * right.y;
    returnValue.y = left.z * right.x - left.x * right.z;
    returnValue.z = left.x * right.y - left.y * right.x;
The return value for this method is the same value returned in the pOut parameter. This allows you to use the Cross method as a parameter for another method.

pOut	Microsoft.DirectX.Vector3 A Vector3 structure that is the cross product of two 3-D vectors.
pV1	Microsoft.DirectX.Vector3 Source Vector3 structure.
pV2	Microsoft.DirectX.Vector3 Source Vector3 structure.