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The triple scalar product

Because the cross product of two vectors is a vector, it is possible to combine the dot product and the cross product. The dot product of a vector with the cross product of two other vectors is called the triple scalar product because the result is a scalar.

Definition

The triple scalar product    of vectors u , v , and w is u · ( v × w ) .

Calculating a triple scalar product

The triple scalar product of vectors u = u 1 i + u 2 j + u 3 k , v = v 1 i + v 2 j + v 3 k , and w = w 1 i + w 2 j + w 3 k is the determinant of the 3 × 3 matrix formed by the components of the vectors:

u · ( v × w ) = | u 1 u 2 u 3 v 1 v 2 v 3 w 1 w 2 w 3 | .

Proof

The calculation is straightforward.

u · ( v × w ) = u 1 , u 2 , u 3 · v 2 w 3 v 3 w 2 , v 1 w 3 + v 3 w 1 , v 1 w 2 v 2 w 1 = 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 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 1 u 2 u 3 v 1 v 2 v 3 w 1 w 2 w 3 |

Calculating the triple scalar product

Let u = 1 , 3 , 5 , v = 2 , −1 , 0 and w = −3 , 0 , −1 . Calculate the triple scalar product u · ( v × w ) .

Apply [link] directly:

u · ( v × w ) = | 1 3 5 2 −1 0 −3 0 −1 | = 1 | −1 0 0 −1 | 3 | 2 0 −3 −1 | + 5 | 2 −1 −3 0 | = ( 1 0 ) 3 ( −2 0 ) + 5 ( 0 3 ) = 1 + 6 15 = −8 .
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Calculate the triple scalar product a · ( b × c ) , where a = 2 , −4 , 1 , b = 0 , 3 , −1 , and c = 5 , −3 , 3 .

17

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When we create a matrix from three vectors, we must be careful about the order in which we list the vectors. If we list them in a matrix in one order and then rearrange the rows, the absolute value of the determinant remains unchanged. However, each time two rows switch places, the determinant changes sign:

| a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 | = d | b 1 b 2 b 3 a 1 a 2 a 3 c 1 c 2 c 3 | = d | b 1 b 2 b 3 c 1 c 2 c 3 a 1 a 2 a 3 | = d | c 1 c 2 c 3 b 1 b 2 b 3 a 1 a 2 a 3 | = d .

Verifying this fact is straightforward, but rather messy. Let’s take a look at this with an example:

| 1 2 1 −2 0 3 4 1 −1 | = | 0 3 1 −1 | 2 | −2 3 4 −1 | + | −2 0 4 1 | = ( 0 3 ) 2 ( 2 12 ) + ( −2 0 ) = −3 + 20 2 = 15.

Switching the top two rows we have

| −2 0 3 1 2 1 4 1 −1 | = −2 | 2 1 1 −1 | + 3 | 1 2 4 1 | = −2 ( −2 1 ) + 3 ( 1 8 ) = 6 21 = −15 .

Rearranging vectors in the triple products is equivalent to reordering the rows in the matrix of the determinant. Let u = u 1 i + u 2 j + u 3 k , v = v 1 i + v 2 j + v 3 k , and w = w 1 i + w 2 j + w 3 k . Applying [link] , we have

u · ( v × w ) = | u 1 u 2 u 3 v 1 v 2 v 3 w 1 w 2 w 3 | and u · ( w × v ) = | u 1 u 2 u 3 w 1 w 2 w 3 v 1 v 2 v 3 | .

We can obtain the determinant for calculating u · ( w × v ) by switching the bottom two rows of u · ( v × w ) . Therefore, u · ( v × w ) = u · ( w × v ) .

Following this reasoning and exploring the different ways we can interchange variables in the triple scalar product lead to the following identities:

u · ( v × w ) = u · ( w × v ) u · ( v × w ) = v · ( w × u ) = w · ( u × v ) .

Let u and v be two vectors in standard position. If u and v are not scalar multiples of each other, then these vectors form adjacent sides of a parallelogram. We saw in [link] that the area of this parallelogram is u × v . Now suppose we add a third vector w that does not lie in the same plane as u and v but still shares the same initial point. Then these vectors form three edges of a parallelepiped    , a three-dimensional prism with six faces that are each parallelograms, as shown in [link] . The volume of this prism is the product of the figure’s height and the area of its base. The triple scalar product of u , v , and w provides a simple method for calculating the volume of the parallelepiped defined by these vectors.

Volume of a parallelepiped

The volume of a parallelepiped with adjacent edges given by the vectors u , v , and w is the absolute value of the triple scalar product:

V = | u · ( v × w ) | .

See [link] .

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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