Core Definition
Given three 3D vectors u β = β¨ u 1 , u 2 , u 3 β© \vec{u} = \langle u_1, u_2, u_3 \rangle u = β¨ u 1 β , u 2 β , u 3 β β© v β = β¨ v 1 , v 2 , v 3 β© \vec{v} = \langle v_1, v_2, v_3 \rangle v = β¨ v 1 β , v 2 β , v 3 β β© w β = β¨ w 1 , w 2 , w 3 β© \vec{w} = \langle w_1, w_2, w_3 \rangle w = β¨ w 1 β , w 2 β , w 3 β β© Triple Scalar Product (or box product) is defined as:
u β β
( v β Γ w β ) = β£ u 1 u 2 u 3 v 1 v 2 v 3 w 1 w 2 w 3 β£ \vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix} u β
( v Γ w ) = β u 1 β v 1 β w 1 β β u 2 β v 2 β w 2 β β u 3 β v 3 β w 3 β β β
The absolute value
β£ u β β
( v β Γ w β ) β£ |\vec{u} \cdot (\vec{v} \times \vec{w})| β£ u β
( v Γ w ) β£ represents the exact geometric
volume of the parallelepiped spanned by
u β \vec{u} u ,
v β \vec{v} v , and
w β \vec{w} w in
R 3 \mathbb{R}^3 R 3 .
Concept Explanation
Geometric Intuition
To understand why the triple scalar product computes volume, consider the geometric definition of a parallelepiped:
Volume = ( BaseΒ Area ) Γ ( PerpendicularΒ Height ) \text{Volume} = (\text{Base Area}) \times (\text{Perpendicular Height}) Volume = ( BaseΒ Area ) Γ ( PerpendicularΒ Height )
Base Area: The cross product v β Γ w β \vec{v} \times \vec{w} v Γ w Area base = β₯ v β Γ w β β₯ \text{Area}_{\text{base}} = \|\vec{v} \times \vec{w}\| Area base β = β₯ v Γ w β₯ Perpendicular Height: The height h h h u β \vec{u} u n β = v β Γ w β \vec{n} = \vec{v} \times \vec{w} n = v Γ w h = β₯ u β β₯ β£ cos β‘ Ο β£ h = \|\vec{u}\| |\cos\phi| h = β₯ u β₯β£ cos Ο β£ Ο \phi Ο u β \vec{u} u v β Γ w β \vec{v} \times \vec{w} v Γ w Volume Synthesis:
Volume = β₯ v β Γ w β β₯ ( β₯ u β β₯ β£ cos β‘ Ο β£ ) = β£ u β β
( v β Γ w β ) β£ \text{Volume} = \|\vec{v} \times \vec{w}\| (\|\vec{u}\| |\cos\phi|) = |\vec{u} \cdot (\vec{v} \times \vec{w})| Volume = β₯ v Γ w β₯ ( β₯ u β₯β£ cos Ο β£ ) = β£ u β
( v Γ w ) β£
u /_________________
/ /|
/ / |
/_________________/ |
| | | h = ||u|| |cos Ο|
| | /
| | / Area = ||v x w||
|_________________|/
w v
Determinant Formula
Evaluating the cross product and then the dot product collapses into the determinant of a 3 Γ 3 3 \times 3 3 Γ 3
u β β
( v β Γ w β ) = u 1 ( v 2 w 3 β v 3 w 2 ) β u 2 ( v 1 w 3 β v 3 w 1 ) + u 3 ( v 1 w 2 β v 2 w 1 ) \vec{u} \cdot (\vec{v} \times \vec{w}) = u_1 (v_2 w_3 - v_3 w_2) - u_2 (v_1 w_3 - v_3 w_1) + u_3 (v_1 w_2 - v_2 w_1) u β
( v Γ w ) = u 1 β ( v 2 β w 3 β β v 3 β w 2 β ) β u 2 β ( v 1 β w 3 β β v 3 β w 1 β ) + u 3 β ( v 1 β w 2 β β v 2 β w 1 β ) Coplanarity Test
Three vectors u β , v β , w β \vec{u}, \vec{v}, \vec{w} u , v , w R 3 \mathbb{R}^3 R 3 same geometric plane (coplanar) if and only if the volume of the spanned parallelepiped collapses to zero:
u β , v β , w β Β areΒ coplanar β
β βΊ β
β u β β
( v β Γ w β ) = 0 \vec{u}, \vec{v}, \vec{w} \text{ are coplanar} \iff \vec{u} \cdot (\vec{v} \times \vec{w}) = 0 u , v , w Β areΒ coplanar βΊ u β
( v Γ w ) = 0 Volume of a Tetrahedron
A tetrahedron formed by four vertices in space spanned by vectors u β , v β , w β \vec{u}, \vec{v}, \vec{w} u , v , w 1 6 \frac{1}{6} 6 1 β
V tetrahedron = 1 6 β£ u β β
( v β Γ w β ) β£ V_{\text{tetrahedron}} = \frac{1}{6}|\vec{u} \cdot (\vec{v} \times \vec{w})| V tetrahedron β = 6 1 β β£ u β
( v Γ w ) β£ Engineering Applications
Civil & Structural Engineering: Calculating the volume, mass, and center of gravity of irregular concrete foundations, rock excavation blocks, and 3D truss polyhedra.Aerospace & Mechanical Engineering: Modeling fluid flow flux across slanted 3D control volumes and computing moment of inertia tensors for non-symmetric satellite bodies.Computer Graphics & CAD Modeling: Rapid coplanarity and degeneracy checks in 3D tetrahedral finite element meshes (FEM) and 3D geometric intersection algorithms.
Example Problems
Example 1: Volume of a Parallelepiped
Find the volume of the parallelepiped with adjacent edges u β = β¨ 1 , 4 , β 7 β© \vec{u} = \langle 1, 4, -7 \rangle u = β¨ 1 , 4 , β 7 β© v β = β¨ 2 , β 1 , 4 β© \vec{v} = \langle 2, -1, 4 \rangle v = β¨ 2 , β 1 , 4 β© w β = β¨ 0 , β 9 , 18 β© \vec{w} = \langle 0, -9, 18 \rangle w = β¨ 0 , β 9 , 18 β©
Solution
Set up the triple scalar product determinant:
u β β
( v β Γ w β ) = β£ 1 4 β 7 2 β 1 4 0 β 9 18 β£ \vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} 1 & 4 & -7 \\ 2 & -1 & 4 \\ 0 & -9 & 18 \end{vmatrix} u β
( v Γ w ) = β 1 2 0 β 4 β 1 β 9 β β 7 4 18 β β
Expand along Row 1:
u β β
( v β Γ w β ) = 1 [ ( β 1 ) ( 18 ) β ( 4 ) ( β 9 ) ] β 4 [ ( 2 ) ( 18 ) β ( 4 ) ( 0 ) ] + ( β 7 ) [ ( 2 ) ( β 9 ) β ( β 1 ) ( 0 ) ] \vec{u} \cdot (\vec{v} \times \vec{w}) = 1[(-1)(18) - (4)(-9)] - 4[(2)(18) - (4)(0)] + (-7)[(2)(-9) - (-1)(0)] u β
( v Γ w ) = 1 [( β 1 ) ( 18 ) β ( 4 ) ( β 9 )] β 4 [( 2 ) ( 18 ) β ( 4 ) ( 0 )] + ( β 7 ) [( 2 ) ( β 9 ) β ( β 1 ) ( 0 )] = 1 [ β 18 + 36 ] β 4 [ 36 β 0 ] β 7 [ β 18 β 0 ] = 1[-18 + 36] - 4[36 - 0] - 7[-18 - 0] = 1 [ β 18 + 36 ] β 4 [ 36 β 0 ] β 7 [ β 18 β 0 ] = 1 ( 18 ) β 4 ( 36 ) β 7 ( β 18 ) = 18 β 144 + 126 = 0 = 1(18) - 4(36) - 7(-18) = 18 - 144 + 126 = 0 = 1 ( 18 ) β 4 ( 36 ) β 7 ( β 18 ) = 18 β 144 + 126 = 0
Geometric Conclusion: Because the triple scalar product is 0 0 0 0 Β units 3 0\text{ units}^3 0 Β units 3 u β , v β , w β \vec{u}, \vec{v}, \vec{w} u , v , w coplanar .
Example 2: Volume of a Structural Tetrahedron
A foundation footing has tetrahedral vertices at A ( 0 , 1 , 2 ) A(0, 1, 2) A ( 0 , 1 , 2 ) B ( 2 , 4 , 5 ) B(2, 4, 5) B ( 2 , 4 , 5 ) C ( β 1 , 0 , 1 ) C(-1, 0, 1) C ( β 1 , 0 , 1 ) D ( 3 , β 1 , 4 ) D(3, -1, 4) D ( 3 , β 1 , 4 )
Solution
Find the three edge displacement vectors originating from vertex A A A u β = A B β = β¨ 2 β 0 , β
β 4 β 1 , β
β 5 β 2 β© = β¨ 2 , 3 , 3 β© \vec{u} = \vec{AB} = \langle 2-0, \; 4-1, \; 5-2 \rangle = \langle 2, 3, 3 \rangle u = A B = β¨ 2 β 0 , 4 β 1 , 5 β 2 β© = β¨ 2 , 3 , 3 β© v β = A C β = β¨ β 1 β 0 , β
β 0 β 1 , β
β 1 β 2 β© = β¨ β 1 , β 1 , β 1 β© \vec{v} = \vec{AC} = \langle -1-0, \; 0-1, \; 1-2 \rangle = \langle -1, -1, -1 \rangle v = A C = β¨ β 1 β 0 , 0 β 1 , 1 β 2 β© = β¨ β 1 , β 1 , β 1 β© w β = A D β = β¨ 3 β 0 , β
β β 1 β 1 , β
β 4 β 2 β© = β¨ 3 , β 2 , 2 β© \vec{w} = \vec{AD} = \langle 3-0, \; -1-1, \; 4-2 \rangle = \langle 3, -2, 2 \rangle w = A D = β¨ 3 β 0 , β 1 β 1 , 4 β 2 β© = β¨ 3 , β 2 , 2 β©
Evaluate the triple scalar product:
u β β
( v β Γ w β ) = β£ 2 3 3 β 1 β 1 β 1 3 β 2 2 β£ \vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} 2 & 3 & 3 \\ -1 & -1 & -1 \\ 3 & -2 & 2 \end{vmatrix} u β
( v Γ w ) = β 2 β 1 3 β 3 β 1 β 2 β 3 β 1 2 β β = 2 [ ( β 1 ) ( 2 ) β ( β 1 ) ( β 2 ) ] β 3 [ ( β 1 ) ( 2 ) β ( β 1 ) ( 3 ) ] + 3 [ ( β 1 ) ( β 2 ) β ( β 1 ) ( 3 ) ] = 2[(-1)(2) - (-1)(-2)] - 3[(-1)(2) - (-1)(3)] + 3[(-1)(-2) - (-1)(3)] = 2 [( β 1 ) ( 2 ) β ( β 1 ) ( β 2 )] β 3 [( β 1 ) ( 2 ) β ( β 1 ) ( 3 )] + 3 [( β 1 ) ( β 2 ) β ( β 1 ) ( 3 )] = 2 [ β 2 β 2 ] β 3 [ β 2 + 3 ] + 3 [ 2 + 3 ] = 2 ( β 4 ) β 3 ( 1 ) + 3 ( 5 ) = β 8 β 3 + 15 = 4 = 2[-2 - 2] - 3[-2 + 3] + 3[2 + 3] = 2(-4) - 3(1) + 3(5) = -8 - 3 + 15 = 4 = 2 [ β 2 β 2 ] β 3 [ β 2 + 3 ] + 3 [ 2 + 3 ] = 2 ( β 4 ) β 3 ( 1 ) + 3 ( 5 ) = β 8 β 3 + 15 = 4
Compute tetrahedron volume:
V = 1 6 β£ u β β
( v β Γ w β ) β£ = 1 6 ( 4 ) = 2 3 Β units 3 V = \frac{1}{6}|\vec{u} \cdot (\vec{v} \times \vec{w})| = \frac{1}{6}(4) = \frac{2}{3}\text{ units}^3 V = 6 1 β β£ u β
( v Γ w ) β£ = 6 1 β ( 4 ) = 3 2 β Β units 3
Connections & References