The scalar triple product is one of the important concepts of vector algebra in which we take the product of three vectors. This can be performed by taking the dot product of one vector with the cross product of the other two vectors, and results in some scalar quantity, as the dot product always gives some particular value.
A scalar triple product is also known as a box product, a triple scalar product and a mixed product. In this section, aspirants will learn about the geometrical interpretation of scalar triple product, formula and its expansion, properties and much more.
Definition
\(\begin{array}{l}\text{If}\ \vec{a}, \vec{b}\ and\ \vec{c}\ \text{are three vectors,}\end{array} \)
then the scalar triple product is defined as the dot product of \(\begin{array}{l}\vec{a}\ \text{with the cross product of }\ \vec{b}\ \text{and}\ \vec{c},\ \text{i.e. dot product of}\ \vec{a}\ \text{with}\ \vec{b} \times \vec{c}.\end{array} \)
It is generally denoted by
\(\begin{array}{l}\vec{a} . ( \vec{b} \times \vec{c})\end{array} \)
or \(\begin{array}{l}[\vec{a}\ \; \vec{b}\ \; \vec{c}]\end{array} \)
\(\begin{array}{l}\text{Scalar triple product is read as box product of }\ \vec{a}, \vec{b}, \vec{c}.\end{array} \)
Similarly, we can define the other scalar triple product as
\(\begin{array}{l}\vec{b}. (\vec{c} \times \vec{a})\ or\ \vec{c} . (\vec{a} \times \vec{b})\end{array} \)
Geometrical Interpretation of Scalar Triple Product
We can find the volume of a parallelopiped using the scalar triple product.
\(\begin{array}{l}\text{Area of parallelogram is cross product of vectors}\ \vec{a}\ \text{and}\ \vec{b},\end{array} \)
i.e. \(\begin{array}{l}\vec{a} \times \vec{b}= |\vec{a}||\vec{b}|\sin(\theta)\end{array} \)
\(\begin{array}{l}\text{Height of parallelopiped is }\ |\vec{c}|\ cos(\phi).\end{array} \)
Then, the volume of parallelopiped is
\(\begin{array}{l}|(|\vec{a}|\vec{b}|sin(\theta)) \times (|\vec{c}|cos(\phi))| = |(|\vec{a}||\vec{a}|sin(\theta)\hat{n} . \vec{c})| = |(\vec{a} \times \vec{b}).\vec{c}|\end{array} \)
So, we can say that
\(\begin{array}{l}|(\vec{a} \times \vec{b}).\vec{c}| = |[\vec{a}\;\ \vec{b}\;\ \vec{c}]|\end{array} \)
\(\begin{array}{l}\text{is the volume of parallelopiped with coterminous edges}\ \vec{a}, \vec{b}\ \text{and}\ \vec{c}.\end{array} \)
Formula
The scalar triple product equation is given as
\(\begin{array}{l}(\vec{a} \times \vec{b}).\vec{c}| = [\vec{a}\;\ \vec{b}\;\ \vec{c}] = \begin{vmatrix} a_1 &a_2 &a_3 \\ b_1 & b_2 &b_3 \\ c_1& c_2 &c_3 \end{vmatrix}\end{array} \)
Where
\(\begin{array}{l}\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\\ \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\\ \vec{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}\end{array} \)
Properties
Below are some of the important properties of the scalar triple product.
- If we interchange the position of (.) and (x), the result will be the same, i.e.,
\(\begin{array}{l}\vec{a}. (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b} ). \vec{c}\end{array} \)
- Value of the scalar triple product doesn’t change if we don’t change the cyclic order of
\(\begin{array}{l}\vec{a}, \vec{b}\ \text{and}\ \vec{c}\ \text{is}\end{array} \)
\(\begin{array}{l}[\vec{a}\; \ \vec{b}\; \ \vec{c}] = [\vec{b}\; \ \vec{c}\; \ \vec{a}] = [\vec{c}\; \ \vec{a}\; \ \vec{b}]\end{array} \)
- If we change the cyclic order of the vectors, then the sign of the scalar triple product is changed.
i.e., \(\begin{array}{l}[\vec{a}\; \ \vec{b}\; \ \vec{c}] = -[\vec{a}\; \ \vec{c}\; \ \vec{b}]\ or\ \vec{a}. (\vec{b} \times \vec{c}) = -\vec{a}. (\vec{c} \times \vec{b})\end{array} \)
- If any two of three vectors are equal or parallel, then the scalar triple product is zero.
\(\begin{array}{l}[\vec{a}\; \ \vec{b}\; \ \vec{a}] = [\vec{a}\; \ \vec{c}\; \ \vec{c}]=0 \end{array} \)
- If three vectors are mutually perpendicular, then the scalar triple product is ±1.
\(\begin{array}{l}[\hat{i}\; \ \hat{j}\; \ \hat{k}] = 1, [\hat{j}\; \ \hat{i}\; \ \hat{k}] = -1 \end{array} \)
- If three vectors are coplanar, then
\(\begin{array}{l}[\vec{a}\; \ \vec{b}\; \ \vec{c}] = 0,\end{array} \)
This is the necessary and sufficient condition for three non zero and non-collinear vectors to be coplanar.
\(\begin{array}{l}\text{For any four vectors }\ \vec{a}, \vec{b}\ \vec{c}\ \text{and}\ \vec{d},\end{array} \)
\(\begin{array}{l}[\vec{a}+ \vec{d}\; \ \vec{b}\; \ \vec{c}] = [\vec{a}\; \ \vec{b}\; \ \vec{c}] + [\vec{d}\; \ \vec{b}\; \ \vec{c}] \end{array} \)
\(\begin{array}{l}[\vec{a}+ \vec{b}\; \ \vec{b}+ \vec{c}\; \ \vec{c}+ \vec{a}] = 2[\vec{a}\; \ \vec{b}\; \ \vec{c}] \end{array} \)
\(\begin{array}{l}[\vec{a}- \vec{b}\; \ \vec{b}- \vec{c}\; \ \vec{c}- \vec{a}]\ \text{is always zero.}\end{array} \)
\(\begin{array}{l}[\vec{l}\; \ \vec{m}\; \ \vec{n}][\vec{a}\; \ \vec{b}\; \ \vec{a}] = \begin{vmatrix} \vec{l}.\vec{a} & \vec{l}.\vec{b} & \vec{l}.\vec{c}\\ \vec{m}.\vec{a} & \vec{m}.\vec{b} &\vec{m}.\vec{c} \\ \vec{n}.\vec{a}&\vec{n}.\vec{b} & \vec{n}.\vec{c} \end{vmatrix}\end{array} \)
Where \(\begin{array}{l}\vec{l}, \vec{m}, \vec{n}\ \text{and}\ \vec{a}, \vec{b}\ \text{and}\ \vec{c}\ \text{are non-coplanar vectors.}\end{array} \)
\(\begin{array}{l}[\vec{l}\; \ \vec{m}\; \ \vec{n}](\vec{a} \times \vec{b}) = \begin{vmatrix} \vec{l}.\vec{a} & \vec{l}.\vec{b} & \vec{l}\\ \vec{m}.\vec{a} & \vec{m}.\vec{b} &\vec{m} \\ \vec{n}.\vec{a}&\vec{n}.\vec{b} & \vec{n} \end{vmatrix}\end{array} \)
Scalar Triple Product Proof
If
\(\begin{array}{l}\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\\ \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\\ \vec{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}\end{array} \)
\(\begin{array}{l}\vec{a}.(\vec{b} \times \vec{c}) = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) . \begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ b_1& b_2& b_3\\ c_1&c_2 &c_3 \end{vmatrix}\end{array} \)
\(\begin{array}{l}=\begin{vmatrix} a_1\hat{i} . \hat{i}& a_2\hat{j} .\hat{j}& a_3 \hat{k} .\hat{k}\\ b_1& b_2& b_3\\ c_1&c_2 &c_3 \end{vmatrix}\end{array} \)
\(\begin{array}{l}= [\vec{a}\; \ \vec{b}\; \ \vec{c}] \end{array} \)
Expansion:
\(\begin{array}{l}\vec{a}.(\vec{b} \times \vec{c}) = [\vec{a}\; \ \vec{b}\; \ \vec{c}] = \begin{vmatrix} a_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1&c_2 &c_3 \end{vmatrix}\end{array} \)
\(\begin{array}{l}=a_1(b_2c_3 – b_3c_2) – a_2(b_1c_3 – b_3c_1) + a_3(b_1c_2 – b_2c_1) \end{array} \)
Finding Volume of Tetrahedron
\(\begin{array}{l}\text{Let}\ \vec{a}, \vec{b}\ \text{and}\ \vec{c}\ \text{are the position vectors of vertices}\end{array} \)
A, B, C with respect to O, then the volume of tetrahedron OABC is given by
Volume = ⅓ (area of base) x height
\(\begin{array}{l}\text{Area of base}=\frac{1}{2} |(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})|\end{array} \)
Let
\(\begin{array}{l}(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a}) = \vec{n}\end{array} \)
\(\begin{array}{l}\text{Area of base} = \frac{1}{2}|\vec{n}|\end{array} \)
\(\begin{array}{l}\text{Height} = \text{projection of}\ \vec{a}\ \text{on}\ \vec{n} \end{array} \)
\(\begin{array}{l}\text{Volume} = \frac{1}{3}\times \frac{1}{2} |\vec{n}| \times \frac{|[\vec{a}\;\ \vec{b}\;\ \vec{c} ]|}{|\vec{n}|} \end{array} \)
\(\begin{array}{l}\text{Volume} = \frac{1}{6}\times |[\vec{a}\;\ \vec{b}\;\ \vec{c}]| \end{array} \)
Note:
\(\begin{array}{l}\text{If}\ \vec{a}, \vec{b}, \vec{c}\ \text{and}\ \vec{d},\ \text{are position vectors of vertices A, B ,C, and D of tetrahedron ABCD, then}\end{array} \)
Volume of tetrahedron
\(\begin{array}{l}=\frac{1}{6} \times |[\vec{AB}\; \ \vec{AC}\; \ \vec{AD} ]|\end{array} \)
\(\begin{array}{l}=\frac{1}{6} \times |[\vec{b} – \vec{a}\; \ \vec{c} – \vec{a}\; \ \vec{d}-\vec{a}]|\end{array} \)
Scalar Triple Product Examples
Example 1: If
\(\begin{array}{l}\vec{a} = (\hat{i} + 2 \hat{j} + \hat{k}), \vec{b} = (4\hat{i} – \hat{j} + 2\hat{k})\ \text{and}\ \vec{c} = (3\hat{i} + \hat{j}\end{array} \)
represent three coterminous edges of the parallelopiped, then find its volume.
Solution:
Using the volume formula, we get
\(\begin{array}{l}\text{Volume} = \begin{vmatrix} 1& 2& 1\\ 4& -1& 2\\ 3&1 &0 \end{vmatrix}\end{array} \)
= |3(4 + 1) – 1(2 – 4)| = |15 + 2| = 17
Example 2: If vertices of a tetrahedron are
\(\begin{array}{l}\vec{a} = (\hat{i} + 2 \hat{j} + \hat{k}), \vec{b} = (4\hat{i} – 2 \hat{j} + 3\hat{k}), \vec{c} = 2\hat{i} + 2\hat{j}\ \text{and}\ \vec{d} = \hat{j} + 2\hat{k}.\end{array} \)
Find the volume of the tetrahedron.
Solution:
Let A, B, C, and D be the vertices, and their position vectors with respect to O are
\(\begin{array}{l}\vec{a}, \vec{b}, \vec{c}\ \text{and}\ \vec{d},\ \text{then}\end{array} \)
\(\begin{array}{l}\vec{AB} = \vec{b} – \vec{a} = 4 \hat{i} – 2\hat{j} + 3\hat{k} – (\hat{i} + 2\hat{j} + \hat{k}) = 3\hat{i} – 4\hat{j} + 2\hat{k}\end{array} \)
\(\begin{array}{l}\vec{AC} = \vec{c} – \vec{a} = 2 \hat{i} + 2\hat{j} – (\hat{i} + 2\hat{j} + \hat{k}) = \hat{i} – \hat{k}\end{array} \)
\(\begin{array}{l}\vec{AD} = \vec{d} – \vec{a} = \hat{j} + 2\hat{k} – (\hat{i} + 2\hat{j} + \hat{k}) = -\hat{i} – \hat{j} + \hat{k}\end{array} \)
Now,
Example 3: For any three vectors,
\(\begin{array}{l}\vec{a}, \vec{b}\ \text{and}\ \vec{c},\ \text{prove that}\end{array} \)
\(\begin{array}{l}[\vec{a}+ \vec{b}\; \ \vec{b}+ \vec{c}\; \ \vec{c}+ \vec{a}] = 2[\vec{a}\; \ \vec{b}\; \ \vec{a}]\end{array} \)
Solution:
\(\begin{array}{l}[\vec{a}+ \vec{b}\; \ \vec{b}+ \vec{c}\; \ \vec{c}+ \vec{a}] = (\vec{a}+ \vec{b}) . [(\vec{b} + \vec{c}) \times (\vec{c} + \vec{a})]\end{array} \)
\(\begin{array}{l}\Rightarrow (\vec{a} + \vec{b}) . [(\vec{b} \times \vec{c}) + (\vec{b} \times \vec{a}) + (\vec{c} \times \vec{c}) + (\vec{c} \times \vec{a})]\end{array} \)
\(\begin{array}{l}\Rightarrow (\vec{a} + \vec{b}) . [(\vec{b} \times \vec{c}) + (\vec{b} \times \vec{a}) + (\vec{c} \times \vec{a})]\ [\text{As}\ (\vec{c} \times \vec{c}) = 0]\end{array} \)
\(\begin{array}{l}\Rightarrow [\vec{a}\; \ \vec{b}\; \ \vec{c}] + [\vec{a}\; \ \vec{b}\; \ \vec{a}] + [\vec{a}\; \ \vec{c}\; \ \vec{a}] + [\vec{b}\; \ \vec{b}\; \ \vec{c}] + [\vec{b}\; \ \vec{b}\; \ \vec{a}] + [\vec{b}\; \ \vec{c}\; \ \vec{a}]\end{array} \)
\(\begin{array}{l}\Rightarrow [\vec{a}\; \ \vec{b}\; \ \vec{c}] + [\vec{b}\; \ \vec{c}\; \ \vec{a}] = 2[\vec{a}\; \ \vec{b}\; \ \vec{c}] \end{array} \)
. Hence Proved.
Example 4: If a, b, c are three non-coplanar vectors, then
\(\begin{array}{l}\frac{\mathbf{a}\,.\,\mathbf{b}\times \mathbf{c}}{\mathbf{c}\times \mathbf{a}\,.\,\mathbf{b}}+\frac{\mathbf{b}\,.\,\mathbf{a}\times \mathbf{c}}{\mathbf{c}\,.\,\mathbf{a}\times \mathbf{b}}=\end{array} \)
Solution:
\(\begin{array}{l}\frac{\mathbf{a}.\mathbf{b}\times \mathbf{c}}{\mathbf{c}\times \mathbf{a}.\mathbf{b}}+\frac{\mathbf{b}.\mathbf{a}\times \mathbf{c}}{\mathbf{c}.\mathbf{a}\times \mathbf{b}}\\=\frac{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}{[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]}+\frac{[\mathbf{b}\,\mathbf{a}\,\mathbf{c}]}{[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]}\\=\frac{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}{[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]}-\frac{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}{[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]}\\=0\\\end{array} \)
Example 5: If a, b, c be any three non-coplanar vectors, then
\(\begin{array}{l}[\mathbf{a}+\mathbf{b}\,\,\,\mathbf{b}+\mathbf{c}\,\,\,\mathbf{c}+\mathbf{a}]=\end{array} \)
Solution:
\(\begin{array}{l}[\mathbf{a}+\mathbf{b}\,\,\mathbf{b}+\mathbf{c}\,\,\mathbf{c}+\mathbf{a}]=(\mathbf{a}+\mathbf{b}).\{(\mathbf{b}+\mathbf{c})\times (\mathbf{c}+\mathbf{a})\}\\ =(\mathbf{a}+\mathbf{b}).(\mathbf{b}\times \mathbf{c}+\mathbf{b}\times \mathbf{a}+\mathbf{c}\times \mathbf{c}+\mathbf{c}\times \mathbf{a})\\ =(\mathbf{a}+\mathbf{b}).(\mathbf{b}\times \mathbf{c}+\mathbf{b}\times \mathbf{a}+\mathbf{c}\times \mathbf{a}), \left\{ \text because \,\mathbf{c}\times \mathbf{c}=0 \right\}\\ =\mathbf{a}.\mathbf{b}\times \mathbf{c}+\mathbf{a}.\mathbf{b}\times \mathbf{a}+\mathbf{a}.\mathbf{c}\times \mathbf{a}+\mathbf{b}.\mathbf{b}\times \mathbf{c}+\mathbf{b}.\mathbf{b}\times \mathbf{a}+\mathbf{b}.\mathbf{c}\times \mathbf{a} \\=[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]+[\mathbf{b}\,\mathbf{c}\,\mathbf{a}]\\=2[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]\\\end{array} \)
Example 6: If the points whose position, vectors are
\(\begin{array}{l}3\mathbf{i}-2\mathbf{j}-\mathbf{k}, 2\mathbf{i}+3\mathbf{j}-4\mathbf{k}, -\mathbf{i}+\mathbf{j}+2\mathbf{k}\ \text{and}\ 4\mathbf{i}+5\mathbf{j}+\lambda \mathbf{k}\end{array} \)
then find the value of λ.
Solution:
Let
\(\begin{array}{l}\mathbf{a}=3\mathbf{i}-2\mathbf{j}-\mathbf{k},\mathbf{b}=2\mathbf{i}+3\mathbf{j}-4\mathbf{k},\mathbf{c}=-\mathbf{i}+\mathbf{j}+2\mathbf{k}\ \text{and}\ \mathbf{d}=4\mathbf{i}+5\mathbf{j}+\lambda \mathbf{k}\end{array} \)
Since the points are coplanar,
\(\begin{array}{l}[\mathbf{d}\,\mathbf{b}\,\mathbf{c}]+[\mathbf{d}\,\mathbf{c}\,\mathbf{a}]+[\mathbf{d}\,\mathbf{a}\,\mathbf{b}]=[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]\\ \Rightarrow \left| \begin{matrix} 4 & 5 & \lambda \\ 2 & 3 & -4 \\ -1 & 1 & 2 \\ \end{matrix} \right|+\left| \begin{matrix} 4 & 5 & \lambda \\ -1 & 1 & 2 \\ 3 & -2 & -1 \\ \end{matrix} \right|+\left| \,\begin{matrix} 4 & 5 & \lambda \\ 3 & -2 & -1 \\ 2 & 3 & -4 \\ \end{matrix}\, \right| =\left| \begin{matrix} 3 & -2 & -1 \\ 2 & 3 & -4 \\ -1 & 1 & 2 \\ \end{matrix} \right|\\ \Rightarrow 40+5\lambda +37-\lambda +94+13\lambda =25\\\Rightarrow \lambda =\frac{-146}{17}\end{array} \)
Example 7: Find the altitude of a parallelopiped determined by the vectors \(\begin{array}{l}\vec{a},\vec{b}\ \text{and}\ \vec{c}\end{array} \)
\(\begin{array}{l}\text{if the base is taken as a parallelogram determined by}\ \vec{a}\ \text{and}\ \vec{b}\ \text{and if}\end{array} \)
\(\begin{array}{l}\vec{a} = \hat{i}+\hat{j}+\hat{k},\ \vec{b} = 2\hat{i}+4\hat{j}-\hat{k},\ \text{and}\ \vec{c} = \hat{i}+\hat{j}+3\hat{k}.\end{array} \)
Solution:
Let V be the volume of the parallelopiped, determined by the vectors \(\begin{array}{l}\vec{a} , \vec{b}\ \text{and}\ \vec{c}.\end{array} \)
then \(\begin{array}{l}V = \begin{bmatrix} \vec{a} & \vec{b} & \vec{c} \end{bmatrix}\end{array} \)
\(\begin{array}{l}V = \begin{vmatrix} 1 & 1& 1\\ 2 & 4 & -1\\ 1 &1 & 3 \end{vmatrix}\end{array} \)
= 1(12 + 1) – (6 + 1) + 1(2 – 4)
= (12 + 1) – (6 + 1) + (2 – 4)
= 13 – 7 – 2
= 4 cubic units
Let A be the area of the base of the parallelopiped. Then, \(\begin{array}{l}A = \left | \vec{a}\times \vec{b} \right |\end{array} \)
\(\begin{array}{l}=\begin{vmatrix} \hat{i}& \hat{j} & \hat{k}\\ 1 & 1 & 1\\ 2& 4 & -1 \end{vmatrix}\end{array} \)
\(\begin{array}{l}=-5\hat{i}+3\hat{j}+2\hat{k}\end{array} \)
\(\begin{array}{l}A = \left | \vec{a} \times \vec{b}\right |\end{array} \)
\(\begin{array}{l}=\sqrt{5^{2}+3^{2}+2^{2}}\end{array} \)
\(\begin{array}{l}=\sqrt{25+9+4}\end{array} \)
\(\begin{array}{l}=\sqrt{38}\end{array} \)
The volume of parallelopiped = Area of base × Altitude
So, Altitude = V/A = 4/√38 units
Example 8: \(\begin{array}{l}\text{Find}\ [\vec{a}\; \vec{b}\; \vec{c}]\ \text{when}\end{array} \)
\(\begin{array}{l}\vec{a}= 2\hat{i}-3\hat{j}+4\hat{k},\ \vec{b}= \hat{i}+2\hat{j}-\hat{k},\ \text{and}\ \vec{c}= 3\hat{i}-\hat{j}+2\hat{k}.\end{array} \)
Solution:
Given \(\begin{array}{l}\begin{bmatrix} \vec{a} & \vec{b} & \vec{c} \end{bmatrix}=\begin{vmatrix} 2 & -3 &4 \\ 1&2 &-1 \\ 3& -1 &2 \end{vmatrix}\end{array} \)
= 2(4 – 1) – (-3)(2 + 3) + 4(-1 – 6)
= 2(3) + 3(5) + 4(-7)
= 6 + 15 – 28
= -7
Also read
Vector Dot Product
FAQs on Vector Operations
Video Lessons
Vector Algebra and 3D Geometry Important Topics
Vector Algebra and 3D Geometry Important Questions
Frequently Asked Questions
Q1
What do you mean by a scalar triple product?
Let vector a, vector b, and vector c be three vectors. Then, the scalar triple product is the dot product of vector a with the cross product of vector b and vector c. It is denoted by
a.(b×c) or [a b c].
Q2
List two properties of the scalar vector product.
Let a, b and c be three vectors, then a.(b×c) = (b×c).a. i.e., the scalar product is commutative.
If three vectors are coplanar, then [a b c] =0.
Q3
Can the scalar triple product be negative?
Yes. The scalar triple product can be positive, negative or zero.
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