Example 3 of Triple Product (Volume of a Parallelpiped 3D Shape)

In linear algebra courses you may be asked to find the volume of a parallelpiped, and you'll be given either points of vertices or vectors that make up the parallelpiped. Here's an example of finding the volume of a parallelpiped, which can be done either by the triple product or a 3x3 determinant.

Find the volume of the parallelpiped formed from the vectors \(\left\langle {2,1,4} \right\rangle \), \(\left\langle {1,2,3} \right\rangle \) and \(\left\langle { - 1,1,2} \right\rangle \).

This can be done either by the triple product \(\left| {a \cdot (b \times c)} \right|\) or by just putting the three vectors into a 3x3 matrix and finding the determinant.

Let's use the triple product method first. Note that those are absolute signs around the triple product, it's NOT the norm. This is because the way you choose the a b and c vectors may cause you to get a negative of the answer, just take the positive of it!

Let b and c be \(\left\langle {1,2,3} \right\rangle \) and \(\left\langle { - 1,1,2} \right\rangle \), respectively.

To find the cross product of b and c, write them into a matrix, and find each component by covering up the column you want to find with you hand and finding the determinant of the remaining 2x2 matrix, and make sure the middle term is multiplied by negative 1.

\[\begin{array}{*{20}{c}}i&j&k\\1&2&3\\{ - 1}&1&2\end{array}\]

For the first component for example, you put your hand over the first column, and find the determinant of what remains. Same with the second component, but after finding the determinant multiply it by negative one because of the cofactor matrix, and repeat for the third component but don't multiply that by a negative. See our worked through calculations:

\[(b \times c) = \left\langle {(2)(2) - (3)(1), - [(1)(2) - ( - 1)(3)],(1)(1) - ( - 1)(2)} \right\rangle \]

\[(b \times c) = \left\langle {1, - 5,3} \right\rangle \]

And now just find the dot product part of the triple product, which is just multiplying both components by each other and summing it up.

\[\left| {a \cdot (b \times c)} \right| = \left\langle {2,1,4} \right\rangle \cdot \left\langle {1, - 5,3} \right\rangle \]

\[\left| {a \cdot (b \times c)} \right| = (2)(1) + (1)( - 5) + (4)(3)\]

\[\left| {a \cdot (b \times c)} \right| = 9\]

And we found the volume, 9 units cubed!

What about the other way, by putting all the vectors into a matrix and finding the determinant? Let's do that:

\[V = \left| {\begin{array}{*{20}{c}}2&1&4\\1&2&3\\{ - 1}&1&2\end{array}} \right|\]

And just find the determinant of the 3x3 matrix, which just involves making multiple 2x2 determinants by covering the row and column of the piece you are working with. Note that the middle part is negative.

\[V = 2[(2)(2) - (1)(3)] - 1[(1)(2) - ( - 1)(3)] + 4[(1)(1) - (2)( - 1)]\]

\[V = 2(1) - 1(5) + 4(3)\]

\[V = 9\]


Please see our other linear algebra examples below:

Triple product examples (dot product of a cross product), including finding the volume of a parallelpiped:

Triple Product (Dot Product and Cross Product) Example 1

Triple Product (Dot Product and Cross Product) Example 2

Volume of Parallelpiped Triple Product Example 3

Volume of Parallelpiped Triple Product Example 4

Alternatively return to the linear algebra hub page:

Linear Algebra Examples

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