Scalar Triple Product of Vectors

Scalar Triple Product – Definition

The Scalar Triple Product, also known as the mixed product or box product, is defined as the dot product of one vector with the cross product of the other two vectors. It is one of the most important concepts in vector algebra and is mainly used to find the volume of a parallelepiped, test coplanarity of vectors, and understand the spatial orientation of three vectors.

If $\vec{a}, \vec{b}$ and $\vec{c}$ are any three vectors, then their scalar triple product is defined as

$\vec{a}\cdot(\vec{b}\times\vec{c})$

and it is denoted by

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

Scalar Triple Product – Formula

The scalar triple product can be evaluated using any of the following equivalent forms:

$(\vec{a}\times\vec{b})\cdot\vec{c}
=\vec{a}\cdot(\vec{b}\times\vec{c})
=\vec{b}\cdot(\vec{c}\times\vec{a})
=\vec{c}\cdot(\vec{a}\times\vec{b})$

In symbolic notation:

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

This shows that:

• Scalar triple product remains the same under cyclic permutation.

• It changes sign when any two vectors are interchanged.

The brackets are usually written without parentheses because the dot product cannot be evaluated before the cross product. A dot product of a scalar with a vector is not defined, so ambiguity never arises.

Scalar Triple Product – Proof in Component Form

Let $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$
$\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$
$\vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$

Then, $[\vec{a}\ \vec{b}\ \vec{c}] = (\vec{a}\times\vec{b})\cdot\vec{c}$

First find $\vec{a}\times\vec{b}$:

$\vec{a}\times\vec{b}
=\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{vmatrix}$

Now dot it with $\vec{c}$:

$[\vec{a}\ \vec{b}\ \vec{c}]
=\begin{vmatrix}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{vmatrix}$

Hence, the scalar triple product is numerically equal to the determinant formed by the components of $\vec{a}$, $\vec{b}$ and $\vec{c}$.

This determinant form is extremely useful for calculations and competitive exams.

Important Properties of Scalar Triple Product

We have provided below the useful and important properties of scalar triple product:

1. Scalar Multiplication Property

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

where $m$ is a scalar.

2. Product of Multiple Scalars

$[m_1\vec{a}\ m_2\vec{b}\ m_3\vec{c}]
=m_1m_2m_3[\vec{a}\ \vec{b}\ \vec{c}]$

where $m_1, m_2, m_3$ are scalars.

3. Distributive Property

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

This shows linearity of the scalar triple product in each vector.

Coplanarity Condition Using Scalar Triple Product

The necessary and sufficient condition for three non-zero, non-collinear vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ to be coplanar is:

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

If the scalar triple product is zero, the volume of the parallelepiped formed by the three vectors is zero, which means all three vectors lie in the same plane.

Geometrical Interpretation of Scalar Triple Product

The geometrical interpretation of the scalar triple product gives it a powerful physical meaning. It represents the volume of the parallelepiped formed by three vectors in space. This is why the scalar triple product is widely used in 3D geometry, vector algebra, and physics.

Let the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ represent the edges $OA$, $OB$, and $OC$ of a parallelepiped respectively.

Understanding the Geometry Step by Step

The vector $\vec{b}\times\vec{c}$ is perpendicular to the plane formed by $\vec{b}$ and $\vec{c}$.
Let:

• $\theta$ be the angle between $\vec{b}$ and $\vec{c}$

• $\alpha$ be the angle between $\vec{a}$ and $\vec{b}\times\vec{c}$

• $\hat{n}$ be the unit vector along $\vec{b}\times\vec{c}$

Then,

$\vec{b}\times\vec{c}
=|\vec{b}||\vec{c}|\sin\theta \hat{n}$

Now compute the scalar triple product:

$[\vec{a}\ \vec{b}\ \vec{c}]
=\vec{a}\cdot(\vec{b}\times\vec{c})$

$=\vec{a}\cdot(|\vec{b}||\vec{c}|\sin\theta \hat{n})$

$=|\vec{b}||\vec{c}|\sin\theta (\vec{a}\cdot\hat{n})$

But,

$\vec{a}\cdot\hat{n}
=|\vec{a}|\cos\alpha$

So,

$[\vec{a}\ \vec{b}\ \vec{c}]
=|\vec{a}|\cos\alpha |\vec{b}||\vec{c}|\sin\theta$

Physical Meaning

Notice that:

• $|\vec{b}||\vec{c}|\sin\theta$ is the area of the base (parallelogram formed by $\vec{b}$ and $\vec{c}$)

• $|\vec{a}|\cos\alpha$ is the height of the parallelepiped

Therefore,

$[\vec{a}\ \vec{b}\ \vec{c}]
=\text{(Area of Base)} \times \text{(Height)}$

$=\text{Volume of the parallelepiped}$

Volume of Tetrahedron

A tetrahedron is a pyramid having a triangular base. Therefore

$
\therefore \quad \text { Volume }=\frac{1}{6}\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]
$

Properties of Scalar Triple Product

If $\vec{a}, \vec{b}$ and $\vec{c}$ are vectors
1) $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}), \vec{c}=\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})$ i.e. position of the dot and the cross can be interchanged without altering the product.
2) $\vec{a}, \vec{b}$ and $\vec{c}$ in that order form a right-handed system if $[\vec{a} \quad \vec{b} \quad \vec{c}]>0$
$\vec{a}, \vec{b}$ and $\vec{c}$ in that order form a lett-handed system if $[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]<0$
3) $[\vec{a} \vec{a} \vec{b}]=0(\vec{a}$ is perpendicular to ( $\vec{a} \times \vec{b}), \vec{a} \cdot(\vec{a} \times \vec{b})=0)$

Solved Examples Based on Scalar Triple Product of Vectors

Example 1: Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=1,\ |\vec{b}|=4$ and $\vec{a}\cdot\vec{b}=2$.
If $\vec{c}=(2\vec{a}\times\vec{b})-3\vec{b}$, find $\vec{b}\cdot\vec{c}$. [JEE MAINS 2023]

Solution:
$\vec{b}\cdot\vec{c}=\vec{b}\cdot\big((2\vec{a}\times\vec{b})-3\vec{b}\big)$

$=\vec{b}\cdot(2\vec{a}\times\vec{b})-3\vec{b}\cdot\vec{b}$

$=0-3|\vec{b}|^2$

$=-3\times16=-48$

Hence, $\vec{b}\cdot\vec{c}=-48$.

Example 2: If four distinct points with position vectors $\vec{a},\vec{b},\vec{c},\vec{d}$ are coplanar, then find $[\vec{a}\ \vec{b}\ \vec{c}]$. [JEE MAINS 2023]

Solution:
Since $\vec{a},\vec{b},\vec{c},\vec{d}$ are coplanar, the vectors

$\vec{b}-\vec{a},\ \vec{c}-\vec{b},\ \vec{d}-\vec{c}$ are coplanar.

So, $[\vec{b}-\vec{a},\ \vec{c}-\vec{b},\ \vec{d}-\vec{c}]=0$

$(\vec{b}-\vec{a})\cdot\big((\vec{c}-\vec{b})\times(\vec{d}-\vec{c})\big)=0$

On expansion and rearrangement,

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

Hence, the answer is
$[\vec{d}\ \vec{c}\ \vec{a}]+[\vec{b}\ \vec{d}\ \vec{a}]+[\vec{c}\ \vec{d}\ \vec{b}]$.

Example 3: Let $\vec{v}=\alpha\hat{i}+2\hat{j}-3\hat{k}$,
$\vec{w}=2\alpha\hat{i}+\hat{j}-\hat{k}$, and $\vec{u}$ be a vector such that $|\vec{u}|=\alpha>0$.
If the minimum value of $[\vec{u}\ \vec{v}\ \vec{w}]$ is $-\alpha\sqrt{3401}$ and $|\vec{u}\cdot\hat{i}|^2=\dfrac{m}{n}$, find $m+n$. [JEE MAINS 2023]

Solution:

$-\alpha\sqrt{1+34\alpha^2}=-\alpha\sqrt{3401}$

$\Rightarrow 1+34\alpha^2=3401$

$\Rightarrow \alpha^2=100$

$\Rightarrow \alpha=10$

For minimum value, $\vec{u}$ is parallel to $\vec{v}\times\vec{w}$.

$\vec{v}\times\vec{w}=\hat{i}-50\hat{j}-30\hat{k}$

$\vec{u}=\lambda(\hat{i}-50\hat{j}-30\hat{k})$

$|\vec{u}|=10$

$|\lambda|\sqrt{3401}=10$

$|\lambda|=\dfrac{10}{\sqrt{3401}}$

$\vec{u}=\pm\dfrac{10}{\sqrt{3401}}(\hat{i}-50\hat{j}-30\hat{k})$

$\vec{u}\cdot\hat{i}=\pm\dfrac{10}{\sqrt{3401}}$

$|\vec{u}\cdot\hat{i}|^2=\dfrac{100}{3401}$

So, $m=100,\ n=3401$

$m+n=3501$

Hence, the answer is $3501$.

Example 4: If $\vec{a}=2\hat{i}+\hat{j}+3\hat{k}$,
$\vec{b}=3\hat{i}+3\hat{j}+\hat{k}$, $\vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$ are coplanar and $\vec{a}\cdot\vec{c}=5$,
$\vec{b}\perp\vec{c}$, find $122(c_1+c_2+c_3)$. [JEE MAINS 2022]

Solution:

From $\vec{a}\cdot\vec{c}=5$,
$2c_1+c_2+3c_3=5$ ...(1)

From $\vec{b}\perp\vec{c}$,
$3c_1+3c_2+c_3=0$ ...(2)

Coplanarity condition:

$\begin{vmatrix}
c_1 & c_2 & c_3\\
2 & 1 & 3\\
3 & 3 & 1
\end{vmatrix}=0$

$-8c_1+7c_2+3c_3=0$ ...(3)

Solving (1), (2), and (3),

$c_1=\dfrac{5}{61}$,
$c_2=-\dfrac{85}{122}$,
$c_3=-3(c_1+c_2)$

$c_1+c_2+c_3=-2(c_1+c_2)$

$=-2\left(\dfrac{5}{61}-\dfrac{85}{122}\right)$

$=\dfrac{75}{61}$

$122(c_1+c_2+c_3)=122\times\dfrac{75}{61}=150$

Hence, the answer is $150$.

Example 5: The volume of a parallelepiped whose coterminous edges are $\vec{u}=\hat{i}+\hat{j}+\lambda\hat{k}$,
$\vec{v}=\hat{i}+\hat{j}+3\hat{k}$,
$\vec{w}=2\hat{i}+\hat{j}+\hat{k}$ is $1$ cubic unit.
If $\theta$ is the angle between $\vec{u}$ and $\vec{w}$, find $\cos\theta$.

Solution:

$|\vec{u}\ \vec{v}\ \vec{w}|=1$

$\begin{vmatrix}
1 & 1 & \lambda\\
1 & 1 & 3\\
2 & 1 & 1
\end{vmatrix}=\pm1$

$-\lambda+3=\pm1$

$\Rightarrow \lambda=2\ \text{or}\ 4$

Taking $\lambda=4$,

$\vec{u}=\hat{i}+\hat{j}+4\hat{k}$

$\vec{w}=2\hat{i}+\hat{j}+\hat{k}$

$\vec{u}\cdot\vec{w}=2+1+4=7$

$|\vec{u}|=\sqrt{6}$,
$|\vec{w}|=\sqrt{6}$

$\cos\theta=\dfrac{7}{\sqrt{6}\sqrt{6}}=\dfrac{7}{6\sqrt{3}}$

Hence, the answer is
$\dfrac{7}{6\sqrt{3}}$.

List of Topics Related to Vector Algebra

This section presents a carefully organized list of important topics from Vector Algebra that are closely connected to the vector triple product.

Types of Vectors

Vectors and Scalars

Addition of Vectors and Subtraction of Vectors

Multiplication Of Vectors by a Scalar Quantity

Components Of A Vector Along And Perpendicular To Another Vector

NCERT Resources

This section provides a complete collection of NCERT-based resources to help you study Vector Algebra in a structured and stress-free way. It includes clear notes, detailed solutions, and exemplar problems that follow the NCERT syllabus strictly, making it easier to build strong fundamentals, improve conceptual clarity, and prepare confidently for board exams as well as competitive exams.

NCERT Maths Class 12th Notes for Chapter 10 - Vector Algebra

NCERT Maths Class 12th Solutions for Chapter 10 - Vector Algebra

NCERT Maths Class 12th Exemplar Solutions for Chapter 10 - Vector Algebra

Practice Questions based on Scalar Triple Product

This section brings you a focused set of practice questions based on the Scalar Triple Product, carefully designed to help you understand its geometric meaning, master its formulas, and apply its properties effectively. The problems range from basic applications to exam-level questions, so you can build confidence and sharpen your accuracy with regular practice.

Scalar Triple Product Of Vectors- Practice Question MCQ

We have provided below the practice questions based on the topics related to Scalar Triple Product: