# Scalar Triple Product

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## Scalar Triple Product Definition

Scalar triple product is one of the primary concepts of vector algebra where we consider the product of three vectors. This can be carried out by taking the dot products of any one of the vectors with the cross product of the remaining two vectors and results in some scalar quantity as the dot product always gives some particular value.

The scalar triple product is also referred to as some other names which are namely, triple scalar product, box product, and mixed product.

If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are three vectors, then the scalar triple product is defined as the dot product of $\vec{a}$ with the cross product of $\vec{b}$ and $\vec{c}$, i.e. dot product of $\vec{a}$ with $\vec{b}$ х $\vec{c}$

It is generally denoted by $\vec{a}$ . ($\vec{b}$ х $\vec{c}$) or [$\vec{a}$ $\vec{b}$ $\vec{c}$]

Therefore, we can define a different scalar triple product by,

$\vec{b}$ . ($\vec{c}$ х $\vec{a}$) or $\vec{c}$ . ($\vec{a}$ х $\vec{b}$)

### Geometrical Interpretation of the Scalar Triple Product

Geometrically, the absolute value of the scalar triple product ($\vec{a}$ х $\vec{b}$) . $\vec{c}$ is the volume of the parallelepiped formed by using the three vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ as coterminous edges. Indeed, the magnitude of the vector ($\vec{a}$ х $\vec{b}$) is the area of the parallelogram formed by using $\vec{a}$ and $\vec{b}$; and the direction of the vector ($\vec{a}$ х $\vec{b}$) is perpendicular to the plane parallel to both $\vec{a}$ and $\vec{b}$

Therefore, 丨($\vec{a}$ х $\vec{b}$) . $\vec{c}$丨 is 丨($\vec{a}$ х $\vec{b}$)丨।$\vec{c}$।丨cos θ, where θ is the angle between ($\vec{a}$ х $\vec{b}$) and丨($\vec{a}$ х $\vec{b}$) . $\vec{c}$丨.

From the above figure, we can observe that 丨$\vec{c}$丨丨cos θ丨 is the height of the parallelepiped formed by using the three vectors as adjacent vectors. Thus, 丨($\vec{a}$ х $\vec{b}$) . $\vec{c}$丨is the volume of the parallelepiped.

So we can say that 丨($\vec{a}$ х $\vec{b}$) . $\vec{c}$丨 = 丨[$\vec{a}$ $\vec{b}$ $\vec{c}$ ]丨 is the volume of parallelepiped with coterminous edges [$\vec{a}$, [$\vec{b}$ and [$\vec{c}$

### Formula

Scalar triple product equation is given as

($\vec{a}$ х $\vec{b}$) . $\vec{c}$ = [$\vec{a}$ $\vec{b}$ $\vec{c}$] = $\begin{vmatrix}a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{vmatrix}$

Where,

$\vec{a}$ = a$_{1}$$\hat{i}$ + a$_{2}$$\hat{j}$ + a$_{3}$$\hat{k}$

$\vec{b}$ = b$_{1}$$\hat{i}$ + b$_{2}$$\hat{j}$ + b$_{3}$$\hat{k}$

$\vec{c}$ = c$_{1}$$\hat{i}$ + c$_{2}$$\hat{j}$ + c$_{3}$$\hat{k}$

### Properties

Below are some of the important properties of the scalar triple product:

1. If we interchange the position of (.) and (х), the result will be the same i.e. $\vec{a}$ . ($\vec{b}$ х $\vec{c}$) = ($\vec{a}$ х $\vec{b}$) . $\vec{c}$

2. Value of scalar triple product remains the same when we do not change the cyclic order of $\vec{a}$, $\vec{b}$ and $\vec{c}$, [$\vec{a}$ $\vec{b}$ $\vec{c}$] = [$\vec{b}$ $\vec{c}$ $\vec{a}$] = [$\vec{c}$ $\vec{a}$ $\vec{b}$]

3. If we change the cyclic order of the vectors then the sign of the scalar triple product is changed.

i.e. [$\vec{a}$ $\vec{b}$ $\vec{c}$] = -[$\vec{a}$ $\vec{c}$ $\vec{b}$] or $\vec{a}$ . ($\vec{b}$ х $\vec{c}$) = -$\vec{a}$ . ($\vec{c}$ х $\vec{b}$)

4. If any two of three vectors are equal or parallel, then the scalar triple product is zero.

[$\vec{a}$ $\vec{b}$ $\vec{a}$] = [$\vec{a}$ $\vec{c}$ $\vec{c}$] = 0

5. If three vectors are mutually perpendicular, then the scalar triple product is ±1.

[$\hat{i}$ $\hat{j}$ $\hat{k}$] = 1, [$\hat{j}$ $\hat{i}$ $\hat{k}$] = -1

6. If three vectors are coplanar then [$\vec{a}$ $\vec{b}$ $\vec{c}$][abc]=0. this is the necessary and sufficient condition for three non-zero and non-collinear vectors to be coplanar.

7. For any four-vectors $\vec{a}$ $\vec{b}$ $\vec{c}$ and $\vec{d}$,

[$\vec{a}$ + $\vec{d}$ $\vec{b}$ $\vec{c}$] = [$\vec{a}$ $\vec{b}$ $\vec{c}$] + [$\vec{d}$ $\vec{b}$ $\vec{c}$]

8. [$\vec{a}$ + $\vec{b}$ $\vec{b}$ + $\vec{c}$ $\vec{c}$ + $\vec{a}$] = 2[$\vec{a}$ $\vec{b}$ $\vec{c}$]

9.  [$\vec{a}$ - $\vec{b}$ $\vec{b}$ - $\vec{c}$ $\vec{c}$ - $\vec{a}$] = 2[$\vec{a}$ $\vec{b}$ $\vec{c}$]  is always zero.

10. [$\vec{l}$ $\vec{m}$ $\vec{n}$][$\vec{a}$ $\vec{b}$ $\vec{a}$] =  $\begin{vmatrix} \vec{l} \vec{a} & \vec{l} \vec{b} & \vec{l} \vec{c} \\ \vec{m} \vec{a} & \vec{m} \vec{b} & \vec{m} \vec{c} \\ \vec{n} \vec{a} & \vec{n} \vec{b} & \vec{n} \vec{c} \end{vmatrix}$ Where $\vec{l}$, $\vec{m}$, $\vec{n}$ and $\vec{a}$ $\vec{b}$ and $\vec{c}$ are non-coplanar vectors.

11. [$\vec{l}$ $\vec{m}$ $\vec{n}$] ($\vec{a}$ x $\vec{b}$) = $\begin{vmatrix} \vec{l} \vec{a} & \vec{l} \vec{b} & \vec{l} \\ \vec{m} \vec{a} & \vec{m} \vec{b} & \vec{m} \\ \vec{n} \vec{a} & \vec{n} \vec{b} & \vec{n} \end{vmatrix}$

## Scalar Triple Product Proof

If,

$\vec{a}$ = a$_{1}$ $\hat{i}$ + a$_{2}$ $\hat{j}$ + a$_{3}$ $\hat{k}$

$\vec{b}$ = b$_{1}$ $\hat{i}$ + b$_{2}$ $\hat{j}$ + b$_{3}$ $\hat{k}$

$\vec{c}$ = c$_{1}$ $\hat{i}$ + c$_{2}$ $\hat{j}$ + c$_{3}$ $\hat{k}$

Then,

$\vec{a}$ . ($\vec{b}$ x $\vec{c}$) = (a$_{1}$ $\hat{i}$ + a$_{2}$ $\hat{j}$ + a$_{3}$ $\hat{k}$) . $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{vmatrix}$

= $\begin{vmatrix} a_{1} \hat{i} . \hat{i} & a_{2} \hat{j} . \hat{j} & a_{3} \hat{k} . \hat{k} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{vmatrix}$

= [$\vec{a}$ b $\vec{c}$]

### Expansion:

$\vec{a}$ . ($\vec{b}$ x $\vec{c}$) = [$\vec{a}$ ($\vec{b}$ $\vec{c}$)] = $\begin{vmatrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{vmatrix}$

= a$_{1}$(b$_{2}$c$_{3}$ - b$_{3}$c$_{2}$) - a$_{2}$(b$_{1}$c$_{3}$ - b$_{3}$c$_{1}$) + a$_{3}$(b$_{1}$c$_{2}$ - b$_{2}$c$_{1}$)

## Finding Volume of Tetrahedron

Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ are the position vectors of vertices A, B, C concerning O, then the volume of tetrahedron OABC is given by:

[Image will be Uploaded Soon]

Volume = $\frac{1}{3}$(area of base) x height

Area of base = $\frac{1}{2}$ |($\vec{a}$ x $\vec{b}$) + ($\vec{b}$ x $\vec{c}$) + ($\vec{c}$ x $\vec{a}$)|

Let, ($\vec{a}$ x $\vec{b}$) + ($\vec{b}$ x $\vec{c}$) + ($\vec{c}$ x $\vec{a}$) = $\vec{n}$

Area of base = $\frac{1}{2}$|$\vec{n}$|

Height = projection of $\vec{a}$ on $\vec{n}$

= $\frac{\mid \vec{a} . \vec{n} \mid}{\mid \vec{n} \mid}$ = $\frac{\mid \vec{a}.((\vec{a} \times \vec{b})+(\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})) \mid}{\mid \vec{a} \times \vec{b}+\vec{b} \times \vec{c} + \vec{c} \times \vec{a} \mid}$ = $\frac{\mid [\vec{a} \vec{b} \vec{c}] \mid}{\mid \vec{n} \mid}$

Volume = $\frac{1}{3}$ x $\frac{1}{2}$ $\mid$ $\vec{n}$ $\mid$ x $\frac{\mid [\vec{a} b \vec{c}] \mid }{\mid \vec{n} \mid}$

Volume = $\frac{1}{6}$ x $\mid$[$\vec{a}$ $\vec{b}$ $\vec{c}$ ]$\mid$

Note: If $\vec{a}$, $\vec{b}$ $\vec{c}$ and $\vec{d}$, are position vectors of vertices A, B, C, D of tetrahedron ABCD, then

The volume of a tetrahedron,

= $\frac{1}{6}$ x $\mid$[$\vec{AB}$ $\vec{AC}$ $\vec{AD}$] $\mid$

=  $\frac{1}{6}$ x $\mid$[$\vec{b}$ - $\vec{a}$ $\vec{c}$ - $\vec{a}$ $\vec{d}$ - $\vec{a}$]$\mid$

### Scalar Triple Product Examples

1. If $\vec{a}$ = ($\hat{i}$ + 2$\hat{j}$ + $\hat{k}$), $\vec{b}$ = (4$\hat{i}$ - $\hat{j}$ + 2$\hat{k}$) and $\vec{c}$ = (3$\hat{i}$ + $\hat{j}$)​ represent three coterminous edges of a parallelepiped, then find its volume.

Solution:

Using the Volume formula, we get

Volume = $\begin{vmatrix} 1 & 2 & 1\\ 4 & -1 & 2 \\ 3 & 1 & 0 \end{vmatrix}$

= $\mid$ 3(4+1)-1(2-4)$\mid$

= $\mid$ 15 + 2 $\mid$

= 17

2. For any three vectors, $\vec{a}$, $\vec{b}$ and $\vec{c}$, prove that [$\vec{a}$ + $\vec{b}$ $\vec{b}$ + $\vec{c}$ $\vec{c}$ + $\vec{a}$] = 2[$\vec{a}$ $\vec{b}$ $\vec{a}$]

Solution:

[$\vec{a}$ + $\vec{b}$ $\vec{b}$ + $\vec{c}$ $\vec{c}$ + $\vec{a}$] = ($\vec{a}$ + $\vec{b}$).[($\vec{b}$ + $\vec{c}$) x ($\vec{c}$ + $\vec{a}$)]

⇒ ($\vec{a}$ + $\vec{b}$) . [($\vec{b}$ x $\vec{c}$) + ($\vec{b}$ x $\vec{a}$) + ($\vec{c}$ x $\vec{c}$) + ($\vec{c}$ x  $\vec{a}$)]

⇒ ($\vec{a}$ + $\vec{b}$).[($\vec{b}$ x $\vec{c}$) + ($\vec{b}$ x $\vec{a}$) +  ($\vec{c}$ x  $\vec{a}$)] (a + b) [As, ($\vec{c}$ x $\vec{c}$) = 0 ( c x c) = 0]

⇒ [$\vec{a}$ $\vec{b}$ $\vec{c}$] + [$\vec{a}$ $\vec{b}$ $\vec{a}$] + [$\vec{a}$ $\vec{c}$ $\vec{a}$] + [$\vec{b}$ $\vec{b}$ $\vec{c}$] + [$\vec{b}$ $\vec{b}$ $\vec{a}$] + [$\vec{b}$ $\vec{c}$ $\vec{a}$]

⇒ [$\vec{a}$ $\vec{b}$ $\vec{c}$] + [$\vec{b}$ $\vec{c}$ $\vec{a}$] = 2[$\vec{a}$ $\vec{b}$ $\vec{c}$

Hence Proved.

1. Find [$\vec{a}$ $\vec{b}$ $\vec{c}$] when $\vec{a}$ = 2$\hat{i}$-3$\hat{j}$+4k, $\vec{b}$ = $\hat{i}$ + 2$\hat{j}$ - k and $\vec{c}$] = 3$\hat{i}$ - $\hat{j}$ + 2

Solution:

Given, [$\vec{a}$ = 2$\hat{i}$ - 3$\hat{j}$ + 4k, $\vec{b}$ = $\hat{i}$ + 2$\hat{j}$ - k and $\vec{c}$ = 3$\hat{i}$ - $\hat{j}$ + 2

Therefore,

[$\vec{a}$ $\vec{b}$ $\vec{c}$] = $\begin{vmatrix} 2 & -3 & 4\\ 1 & 2 & -1\\ 3 & -1 & 2 \end{vmatrix}$

[$\vec{a}$ $\vec{b}$ $\vec{c}$] = 2(4-1)-(-3)(2+3)+4(-1-6)

[$\vec{a}$ $\vec{b}$ $\vec{c}$] = 2(3)+3(5)+4(-7)

[$\vec{a}$ $\vec{b}$ $\vec{c}$] = 6+15-28

[$\vec{a}$ $\vec{b}$ $\vec{c}$] = -7

### Note:

[$\vec{a}$ $\vec{b}$ $\vec{c}$] = ($\vec{a}$ x $\vec{b}$) . $\vec{c}$ = $\vec{a}$ . ($\vec{b}$ x $\vec{c}$) = ($\vec{b}$ x $\vec{c}$) . $\vec{a}$ = $\vec{b}$ . ($\vec{c}$ x $\vec{a}$) = [ [$\vec{b}$ $\vec{c}$ $\vec{a}$] ]

[$\vec{b}$ $\vec{c}$ $\vec{a}$] = ($\vec{b}$ x $\vec{c}$) . $\vec{a}$ = $\vec{b}$ . ($\vec{c}$ x $\vec{a}$) = ($\vec{c}$ x $\vec{a}$) . $\vec{b}$ = $\vec{c}$ . ($\vec{a}$ x $\vec{b}$) = [$\vec{c}$ $\vec{a}$ $\vec{b}$]

In other words, [$\vec{a}$ $\vec{b}$ $\vec{c}$] = [$\vec{b}$ $\vec{c}$ $\vec{a}$] = [$\vec{c}$ $\vec{a}$ $\vec{b}$]; that is, if the three vectors are permuted in the same cyclic order, the value of the scalar triple product remains the same.

If 2 vectors are swapped by each other in their position in a scalar triple product, in that case, the result of the scalar triple product is times of the original result. More explicitly, [$\vec{a}$ $\vec{b}$ $\vec{c}$] = [$\vec{b}$ $\vec{c}$ $\vec{a}$] = [$\vec{c}$ $\vec{a}$ $\vec{b}$] = -[$\vec{a}$ $\vec{b}$ $\vec{c}$] = - [$\vec{c}$ $\vec{a}$ $\vec{b}$] = -[$\vec{b}$ $\vec{c}$ $\vec{a}$]

### Did You Know

From the formula of the scalar triple product, that is [$\vec{a}$ $\vec{b}$ $\vec{c}$] = ($\vec{a}$ x $\vec{b}$) . $\vec{c}$,  the following conclusions can be drawn:

• The resultant product is always a scalar quantity.

• At first, the Cross product of the vectors is calculated and then with the dot product which yields the scalar triple product.

• Talking about the physical significance of the scalar triple product formula, it represents the volume of the parallelepiped whose three co-terminus edges represent the three vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$. The following figure will make this point clearer.

FAQ (Frequently Asked Questions)

1. What is Meant By Equality of Vectors?

Two vectors are said to be identical if their magnitudes and directions are identical. Here we are discussing two values of the same physical quantity, i.e. we cannot talk about equality of two vectors if they don’t represent the same physical quantity. For example, one can’t say that the velocity vector of 5 m/s in the positive x-axis and force vector of 5 N also in the positive x-axis is identical.

2. What is the Scalar Triple Product of the Vector?

The scalar triple output of three vectors a,b and c is (a x b) . c. It is a scalar product because, just like the dot product, it calculates to a single number. (In this manner, it is different from the cross product, which is a vector.) The scalar triple product is important because its absolute value 丨(a x b) . c丨 is the quantity of the parallelepiped spanned by a, b and c (i.e., the parallelepiped whose neighboring sides are the vectors (a, b and c).

3. What is the Triple Scalar Product Used For?

The triple scalar product is equivalent to multiplying the area of the base times the height. This is the recipe for finding the volume. The absolute value of the triple scalar product is the volume of the three-dimensional figure defined by the vectors a, b and c.

4. Why Does Dot Product Give Scalar?

The easy answer to this question is that the dot product is scalar and the cross product is vector since they are defined that way. The dot product is defining the elements of a vector in the path of another when the second vector is normalized. As such, it is a scalar multiplier.