By now, you've probably encountered a lot of vector material and wondered how theoretical principles like dot multiplication relate to real life. That's where the Scalar Triple Product comes in. This product provides an easy way to find the volumes of some complex shapes via a combination of previously seen theoretical principles.
This article will show how we can take vectors and apply them to a physical context.
Scalar Triple Product Meaning
The scalar triple product is a principle that we use to find the volume of a parallelepiped - a \(6-\)sided shape where each side is a parallelogram or a tetrahedron.
The scalar triple product actually involves two previously seen vector operations - dot and cross multiplication.
The cross multiplication of two vectors will result in a vector quantity, but the subsequent dot multiplication to find the scalar product will reduce the vectors down into a scalar value.
This is how we can calculate the volume of the shapes named above from three vectors- we get a single number at the end of the process.
You can recall the definition of vector quantity as follows.
A vector quantity is represented in terms of \(x,y,z\) and, as such, has three components to it. Vectors also have a definite magnitude and direction.
The definition of a scalar quantity is as follows.
A scalar quantity is a singular value that only has only magnitude. It has no direction.
Scalar Triple Product of Vectors
We know that vectors can be used to describe motion and are usually in the form of movement in the \(x,\, y,\, z\) directions. In vector form, these become \(\vec{i},\, \vec{j},\, \vec{k}\) respectively and with this notation, we can perform many operations on vectors.
To find the scalar triple product of three vectors you should be familiar with the principle of dot products and cross products and how they work. In case you aren't, you can check our articles on Scalar Products and Vector Product respectively for refreshers.
The scalar triple product is finding the dot product of a vector with the cross product of two vectors. This is a more complex methodology than the dot product of two, but the methodology is useful in finding volumes of certain shapes.
We find the vector product of the first two vectors first. This will lead to one vector that will be used in the dot product with the third vector. And this will lead to a scalar value.
Scalar Triple Product Formula
Consider three vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\), where,\[\vec{a}=a_1\vec{i}+a_2\vec{j}+a_3\vec{k},\]\[\vec{b}=b_1\vec{i}+b_2\vec{j}+b_3\vec{k},\] and \[\vec{c}=c_1\vec{i}+c_2\vec{j}+c_3\vec{k}.\] To find the scalar triple product of these vectors we must find the cross product of two of them and find the dot product of this result with the third vector. In mathematical notation, this looks like,\[\vec{a}\cdot (\vec{b}\times\vec{c}).\]This absolute value of this formula gives us the volume of a parallelepiped.
For the volume of a tetrahedron, the formula you would apply is \(\frac{1}{6}\left[|\vec{a}\cdot (\vec{b}\times\vec{c})|\right]\) when the vectors describe three non-coplanar sides of the shape.
From Vector Product we know that for the cross product of \(\vec{b}\times \vec{c}\) is given as,\[\vec{b}\times \vec{c}=(b_2c_3-b_3c_2)\vec{i}-(b_1c_3-b_3c_1)\vec{j}+(b_1c_2-b_2c_1)\vec{k}.\]If we then consider the scalar product of the result of the vector product and vector \(\vec{a}\) we get the formula for the scalar triple product,\[\vec{a}\cdot (\vec{b}\times\vec{c})=a_1(b_2c_3-b_3c_2)+a_2(b_3c_1-b_1c_3)+a_3(b_1c_2-b_2c_1).\]
Scalar Triple Product Properties
As stated earlier, the scalar triple product is used to find the volume of a parallelepiped, but what does this actually mean?
If we consider the vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) to be three non-parallel sides of a parallelepiped, we can conduct the scalar triple product formula to get a result for the volume of the shape.
When we are seeking to find the volume of shapes, the order we apply these vectors doesn't matter as long as the process is cyclic. This means:\[\vec{a}\cdot (\vec{b}\times\vec{c})=\vec{b}\cdot (\vec{c}\times\vec{a})=\vec{c}\cdot (\vec{a}\times\vec{b}).\]
Let's look into an example of this.
Show that \(\vec{a}\cdot (\vec{b}\times\vec{c})=\vec{b}\cdot (\vec{c}\times\vec{a})\) using the vectors below,
\[\vec{a}=5\vec{i}+2\vec{j}+6\vec{k},\]\[\vec{b}=-2\vec{i}+17\vec{j}+1\vec{k},\] and \[\vec{c}=8\vec{i}-5\vec{j}+13\vec{k}.\]
Solution
Using our general formula for \(\vec{a}\cdot (\vec{b}\times\vec{c})\),
\[\begin{align}\vec{a}\cdot (\vec{b}\times\vec{c})&=a_1(b_2c_3-b_3c_2)+a_2(b_3c_1-b_1c_3)+a_3(b_1c_2-b_2c_1)\\&=5[(17\cdot13)-(1\cdot-5)]+2[(1\cdot8)-(-2\cdot13)]\\ & \qquad +6[(-2\cdot-5)-(17\cdot8)]\\&=5(226)+2(34)+6(-126)\\&=1130+68-756\\&=442.\end{align}\]
We can then use the general formula again for \(\vec{b}\cdot (\vec{c}\times\vec{a})\), where we shift the letters- where there were \(a's\) there will now be \(b's\), \(b's\) will be replaced by \(c's\) and \(c's\) replaced by \(a's\). This will take the form, \[\begin{align}\vec{b}\cdot (\vec{c}\times\vec{a})&=b_1(c_2a_3-c_3a_2)+b_2(c_3a_1-c_1a_3)+b_3(c_1a_2-c_2a_1)\\&=-2[(-5\cdot6)-(13\cdot2)]+17[(13\cdot5)-(8\cdot6)]\\& \qquad+1[(8\cdot2)-(-5\cdot5)]\\&=-2(-56)+17(17)+1(41)\\&=112+289+41\\&=442.\end{align}\]
As you can see, the numbers through the process change, but as the process is cyclic, the end result is the same.
Thus, \[\vec{a}\cdot (\vec{b}\times\vec{c})=\vec{b}\cdot (\vec{c}\times\vec{a}).\]
There is another property of the scalar triple product that hasn't yet been discussed- let's array our three vectors into a \(3\times 3\) matrix,\[\begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{bmatrix}\]
If you expand the above matrix, you should be getting the scalar triple product. Let's see how!
\[\begin{align} \begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{bmatrix} &=a_1(b_2c_3-b_3c_2)+a_2(b_3c_1-b_1c_3)+a_3(b_1c_2-b_2c_1)\\ &=\vec{a}\cdot (\vec{b}\times\vec{c}) \end{align}\]
The scalar triple product is the same as the determinant of this matrix. For a refresher on why this is, see our article on Matrix Determinants.
The key takeaway here is that the minors and expansion of a \(3\times 3\) matrix determinant reflect the formula of the scalar triple product, so this may be an easier way to remember the process for you.
Let's look into an example to find the scalar triple product by expanding the determinant.
Find the volume of the parallelepiped formed by the coterminous edges given by vectors, \[\vec{a}=3\vec{i}-1\vec{j}-2\vec{k},\]\[\vec{b}=\vec{i}+3\vec{j}-2\vec{k},\] and \[\vec{c}=6\vec{i}-2\vec{j}+\vec{k}.\]
Solution
To find the volume of the parallelepiped, you should find the scalar triple product. Here you will find the scalar triple product by the determinant method.
\[\begin{align} \begin{bmatrix}3&-1&-2\\1&3&-2\\6&-2&1\end{bmatrix} &=3(3\cdot1-(-2)\cdot(-2))+1(1\cdot1-(-2)\cdot6)\\&-2(1\cdot(-2)-3\cdot6)\\ &= 3(3-4)+1(1+12)-2(-2-18)\\ &=3(-1)+1(13)-2(-20)\\ & =50 \mbox{ units}^3 . \end{align}\]
Thus, the volume of the parallelepiped formed by the coterminous edges of the given vectors is \(50 \mbox{ units}^3\).
Please note that even if you get a negative determinant, you need to take the modulus of the scalar triple product to get the volume.
Also, there are many other properties of scalar triple products which are beyond the scope of Further Maths.
- The scalar triple product is unchanged if we swap the positions of the operations without changing the positions of the vectors. \[\vec{a}\cdot (\vec{b}\times\vec{c})=(\vec{a}\times \vec{b})\cdot\vec{c}.\]
- The scalar triple product negates if you swap any two of the three given vectors. \[\vec{a}\cdot (\vec{b}\times\vec{c})=-\vec{a}\cdot (\vec{c}\times\vec{b}).\]
- The scalar triple product is zero if any of all the three given vectors are coplanar and vice versa.
Example of Scalar Triple Product
Let's start by looking at an example where we need to find the volume of a parallelepiped.
Find the volume of the parallelepiped with three non-parallel sides described by the vectors,\[\vec{a}=2\vec{i}+1\vec{j}-1\vec{k},\]\[\vec{b}=-5\vec{i}+14\vec{j}-7\vec{k},\] and \[\vec{c}=16\vec{i}-3\vec{j}+12\vec{k}.\]
Solution
We know that the volume of a parallelepiped is given as \(|\vec{a}\cdot(\vec{b}\times\vec{c})|\). Therefore:\[\begin{align}\mbox{Volume}&=|\vec{a}\cdot(\vec{b}\times\vec{c})|\\&=|a_1(b_2c_3-b_3c_2)+a_2(b_3c_1-b_1c_3)+a_3(b_1c_2-b_2c_1)|\\&=|2[(14\cdot12)-(-7\cdot-3)]+1[(-7\cdot16)-(-5\cdot12)]\\ &\quad \quad +(-1)[(-5\cdot-3)-(14\cdot16)]|\\&=451 \mbox{ units}^3.\end{align}\]
Let's now have a look at an example where we need to find the volume of a tetrahedron.
Find the volume of the tetrahedron with three non-coplanar sides described by the vectors, \[\vec{a}=-4\vec{i}+12\vec{j}+2\vec{k},\]\[\vec{b}=3\vec{i}+1\vec{j}-1\vec{k},\] and \[\vec{c}=4\vec{i}+3\vec{j}+2\vec{k}.\]
Solution
We know that the volume of a tetrahedron is given as \(\frac{1}{6}\left[\left|\vec{a}\cdot(\vec{b}\times\vec{c})\right|\right]\). Therefore,\[\begin{align}\mbox{Volume}&=\frac{1}{6}\left[\left|\vec{a}\cdot(\vec{b}\times\vec{c})\right|\right]\\&=\frac{1}{6}\bigg[\left|a_1(b_2c_3-b_3c_2)+a_2(b_3c_1-b_1c_3)+a_3(b_1c_2-b_2c_1)\right|\bigg]\\&=\frac{1}{6}\bigg[|(-4)[(1\cdot2)-(-1\cdot3)]+12[(-1\cdot4)-(3\cdot2)]\bigg.\\ \bigg. &\quad \quad +2[(3\cdot3)-(1\cdot4)]|\bigg]\\&=\frac{1}{6}\bigg[\left | -4(5)+12(-10)+2(5)\right |\bigg]\\&=\frac{1}{6}\cdot130\\&=\frac{65}{3} \mbox{ units}^3.\end{align}\]
Scalar Triple Product - Key takeaways
- The scalar triple product can be found with the formula below,\[\vec{a}\cdot(\vec{b}\times\vec{c}).\]
- The scalar triple product is cyclic thus,\[\vec{a}\cdot (\vec{b}\times\vec{c})=\vec{b}\cdot (\vec{c}\times\vec{a})=\vec{c}\cdot (\vec{a}\times\vec{b}).\]
- The absolute value of the formula above can be used to find the volume of a parallelepiped if the vectors are three non-parallel sides of this shape.
- The volume of a tetrahedron where the vectors are three non-coplanar sides is given as, \[\frac{1}{6}\left[|\vec{a}\cdot (\vec{b}\times\vec{c})|\right].\]
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