Vector Triple Product

Last Updated : 26 Jun, 2026

The Vector Triple Product is the cross product of one vector with the cross product of two other vectors.

If A, B, and C are three vectors, then the vector triple product is:

\mathbf{A}\times(\mathbf{B}\times\mathbf{C})

Since the operation contains two cross products and three vectors, it is called a vector triple product. Unlike the scalar triple product, whose result is a scalar quantity, the result of a vector triple product is always a vector.

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Interpretation of the Formula

  • (A⋅C) is a scalar.
  • (A⋅B) is a scalar.
  • The result is a linear combination of vectors B and C.

Therefore, the result is a vector lying in the plane formed by B and C or coplaner to BC.

Proof

Let 3 vectors be: \mathbf{A}=(a_1,a_2,a_3),\quad \mathbf{B}=(b_1,b_2,b_3),\quad \mathbf{C}=(c_1,c_2,c_3)

We want to prove that \mathbf{A}\times(\mathbf{B}\times\mathbf{C})=(\mathbf{A}\cdot\mathbf{C})\mathbf{B}-(\mathbf{A}\cdot\mathbf{B})\mathbf{C}

Step 1: Find \mathbf{B}\times\mathbf{C}

Using the cross product formula, \mathbf{B}\times\mathbf{C}=(b_2c_3-b_3c_2, b_3c_1-b_1c_3, b_1c_2-b_2c_1)

Step 2: Compute (\mathbf{A}\times(\mathbf{B}\times\mathbf{C}))

Let, \mathbf{D}=\mathbf{B}\times\mathbf{C}

Then , \mathbf{A}\times\mathbf{D}

\mathbf{A}\times(\mathbf{B}\times\mathbf{C}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ a_1 & a_2 & a_3\\ D_1 & D_2 & D_3 \end{vmatrix}

Expanding and simplifying gives

\mathbf{A}\times(\mathbf{B}\times\mathbf{C}) = \Big( b_1(\mathbf{A}\cdot\mathbf{C}) - c_1(\mathbf{A}\cdot\mathbf{B}),b_2(\mathbf{A}\cdot\mathbf{C}) - c_2(\mathbf{A}\cdot\mathbf{B}),b_3(\mathbf{A}\cdot\mathbf{C}) - c_3(\mathbf{A}\cdot\mathbf{B}) \Big)

Step 3: Factor the Common Terms

Taking common factors from each component, (\mathbf{A}\cdot\mathbf{C}) (b_1,b_2,b_3)- (\mathbf{A}\cdot\mathbf{B}) (c_1,c_2,c_3)

Step 4: Write in Vector Form

Since \mathbf{B}=(b_1,b_2,b_3) , \mathbf{C}= (c_1,c_2,c_3) ,we obtain
\mathbf{A}\times(\mathbf{B}\times\mathbf{C})=(\mathbf{A}\cdot\mathbf{C})\mathbf{B}-(\mathbf{A}\cdot\mathbf{B})\mathbf{C}

This is known as the BAC–CAB identity or the Vector Triple Product Formula.

Properties

These property are useful for simplifying complex vector expressions.

1. Resultant Vector: The vector triple product always produces a vector quantity.

2. The Result Lies in the Plane of B and C: Since A×(B×C) = (A⋅C)B − (A⋅B)C, the resulting vector is a linear combination of B and C. Therefore, it always lies in the plane formed by B and C.

3. Cross Product is Not Associative: The order of operations cannot be changed i.e., A × (B×C) ≠ (A×B) × C

4. Becomes Zero When B and C are Parallel: If B and C are parallel, then B×C = 0

Therefore, A×(B×C) = A×0=0

5. Distributive Property: The vector triple product is distributive over vector addition. A×[(B+D)×C] = A×(B×C)+A×(D×C)

Solved Examples

Example 1: Given three vectors \vec{a} = \hat{i} + 2\hat{j} - \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k}, and \vec{c} = \hat{i} + \hat{j} + \hat{k}, calculate the vector triple product \vec{a} \times (\vec{b} \times \vec{c})

Solution: Find \vec{b} \times \vec{c}

\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 2\hat{i} - 2\hat{j}

Multiply by \vec{a}

\vec{a} \times (\vec{b} \times \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & -2 & 0 \end{vmatrix} = -2\hat{i} + 4\hat{j} + 6\hat{k}

Therefore, \vec{a} \times (\vec{b} \times \vec{c}) = -2\hat{i} + 4\hat{j} + 6\hat{k}

Example 2: Verify whether the equation \vec{p} = \vec{q} \times (\vec{r} \times \vec{s}) holds true, where \vec{p} = \hat{i} + \hat{j}, \vec{q} = 2\hat{i} - \hat{j} , \vec{r} = \hat{i} + 3\hat{j} + 2\hat{k} , and \vec{s} = \hat{k}.

Solution: Calculate \vec{r} \times \vec{s}:

\vec{r} \times \vec{s} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 2 \\ 0 & 0 & 1 \end{vmatrix} = -\hat{j} + \hat{k}

Multiply by \vec{q}

\vec{q} \times (\vec{r} \times \vec{s}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 0 \\ -1 & 1 & 0 \end{vmatrix} = \hat{k}

Compare with \vec{p}

\vec{p} = \hat{i} + \hat{j} and \vec{q} \times (\vec{r} \times \vec{s}) = \hat{k}

Since \vec{p} and \vec{q} \times (\vec{r} \times \vec{s}) are not equal, the equation \vec{p} = \vec{q} \times (\vec{r} \times \vec{s}) does not hold true.

Practice Problems

Problem 1. Given three vectors \vec{a}, \vec{b}, \vec{c}, calculate the vector triple product \vec{a} \times (\vec{b} \times \vec{c}).

Problem 2. Determine the unit vector that is coplanar with \vec{u} = 3\hat{i} + 2\hat{j} - \hat{k} and \vec{v} = \hat{i} - \hat{j} + 2\hat{k}, and perpendicular to \vec{w} = 2\hat{i} + \hat{j} + 3\hat{k} using the vector triple product.

Problem 3. Verify whether the equation \vec{p} = \vec{q} \times (\vec{r} \times \vec{s}) holds true, where \vec{p} = \hat{i} + \hat{j}, \vec{q} = 2\hat{i} - \hat{j}, \vec{r} = \hat{i} + 3\hat{j} + 2\hat{k}, and \vec{s} = \hat{k}.

Problem 4. If \vec{a} = 2\hat{i} - \hat{j}, \vec{b} = \hat{i} + \hat{j} + \hat{k}, and \vec{c} = -\hat{j} + 3\hat{k}, find the angle between \vec{a} \times (\vec{b} \times \vec{c}) and \vec{c}.

Problem 5. Given non-coplanar vectors \vec{u} = \hat{i} - 2\hat{j} + 3\hat{k}, \vec{v} = 2\hat{i} + \hat{j} - \hat{k}, and \vec{w} = -\hat{i} + \hat{j} + 2\hat{k}, prove that \vec{u} \times \vec{v}, \vec{v} \times \vec{w}, and \vec{w} \times \vec{u} are also non-coplanar.

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