Understanding the Vector Triple Product in Vectors

The vector triple product is a fundamental result in vector algebra that describes the cross product of one vector with the cross product of two other vectors. This concept appears frequently in advanced mathematics and physics, especially in the context of vector spaces and spatial relationships. Understanding its algebraic structure and geometric implications is essential for progressing through topics such as Vector Algebra Concepts and related applications in physical sciences.

Meaning and Computational Purpose of the Vector Triple Product

The vector triple product calculates a new vector by evaluating the cross product of one vector with the cross product of two others. Algebraically, for three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$, the expression is denoted as $\vec{a} \times (\vec{b} \times \vec{c})$.

This result is significant because it transforms a combination of cross products into a linear combination of the two inner vectors, streamlining many vector calculations in mechanics and geometry.

Conditions and Applicability of the Vector Triple Product

The vector triple product operates in three-dimensional Euclidean space. All participating vectors may be arbitrary; however, special cases such as coplanarity or collinearity have additional implications explained in subsequent sections.

It is important to note that the vector triple product is not associative. In symbols, $(\vec{a} \times \vec{b}) \times \vec{c} \neq \vec{a} \times (\vec{b} \times \vec{c})$ in general.

Formal Statement: Vector Triple Product Identity (BAC–CAB Rule)

For any vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ in $\mathbb{R}^3$, the following identity holds:

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

This expression rewrites the complicated cross product into a difference involving only dot products and scalar multiples of vectors $\vec{b}$ and $\vec{c}$.

Stepwise Proof of the Vector Triple Product Identity

Let $\vec{a} = (a_1, a_2, a_3)$, $\vec{b} = (b_1, b_2, b_3)$, $\vec{c} = (c_1, c_2, c_3)$. First, compute $\vec{b} \times \vec{c}$.

$\vec{b} \times \vec{c} = (b_2 c_3 - b_3 c_2, \, b_3 c_1 - b_1 c_3, \, b_1 c_2 - b_2 c_1)$

Now, compute $\vec{a} \times (\vec{b} \times \vec{c})$, component-wise. The $i$-th component (using Levi-Civita symbol) is:

$[\vec{a} \times (\vec{b} \times \vec{c})]_i = \varepsilon_{ijk} a_j (\vec{b} \times \vec{c})_k$

Expanding using the property $\varepsilon_{ijk} \varepsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$, regroup terms:

$(\vec{a} \times (\vec{b} \times \vec{c}))_i = b_i (\vec{a} \cdot \vec{c}) - c_i (\vec{a} \cdot \vec{b})$

Therefore, returning to vector notation:

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

This derivation validates the stated BAC–CAB identity.

Special Cases and Geometric Interpretations

If $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$, then $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$. Thus, $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{0}$.

If $\vec{b}$ and $\vec{c}$ are linearly dependent, then $\vec{b} \times \vec{c} = \vec{0}$, yielding $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{0}$.

Key Properties and Exam Patterns

• Result lies in $\mathrm{span}\{\vec{b}, \vec{c}\}$
• Identity applies in $\mathbb{R}^3$ only
• Not associative in general
• Formula simplifies multi-cross products
• Useful for proofs in mechanics or geometry

Worked Examples on the Vector Triple Product

Example 1: Direct Computation

Given $\vec{a} = \hat{i}$, $\vec{b} = \hat{j}$, $\vec{c} = \hat{k}$, compute $\vec{a} \times (\vec{b} \times \vec{c})$.

First, $\vec{b} \times \vec{c} = \hat{j} \times \hat{k} = \hat{i}$.

Then, $\vec{a} \times (\vec{b} \times \vec{c}) = \hat{i} \times \hat{i} = \vec{0}$.

Alternatively, the identity gives $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$:

$\vec{a} \cdot \vec{c} = \hat{i} \cdot \hat{k} = 0$

$\vec{a} \cdot \vec{b} = \hat{i} \cdot \hat{j} = 0$

Thus, the result is $0 \cdot \hat{j} - 0 \cdot \hat{k} = \vec{0}$.

Example 2: Application with Parameter Identification

Given $|\vec{a}| = 1$, $|\vec{b}| = 2$, $|\vec{c}| = 1$, solve:

$\vec{a} \times (\vec{a} \times \vec{b}) + \vec{c} = \vec{0}$

From the identity:

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

Substitute into the given equation:

$(\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b} + \vec{c} = \vec{0}$

Since $|\vec{a}| = 1$, $|\vec{b}| = 2$, $|\vec{c}| = 1$, let the angle between $\vec{a}$ and $\vec{b}$ be $\theta$; then $\vec{a} \cdot \vec{b} = 1 \cdot 2 \cos\theta = 2 \cos\theta$.

$\vec{a} \cdot \vec{a} = 1$.

From the equation:

$2\cos\theta \vec{a} - \vec{b} + \vec{c} = \vec{0}$

$2\cos\theta \vec{a} - \vec{b} = -\vec{c}$

Square both sides:

$|2\cos\theta \vec{a} - \vec{b}|^2 = |\vec{c}|^2$

$4\cos^2\theta |\vec{a}|^2 - 4\cos\theta (\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = 1$

With $|\vec{a}| = 1$, $|\vec{b}| = 2$, $\vec{a} \cdot \vec{b} = 2\cos\theta$:

$4\cos^2\theta - 8\cos^2\theta + 4 = 1$

$-4\cos^2\theta + 4 = 1$

$-4\cos^2\theta = -3$

$\cos^2\theta = \dfrac{3}{4}$

$\cos\theta = \dfrac{\sqrt{3}}{2}$

$\theta = \dfrac{\pi}{6}$

Therefore, the angle between $\vec{a}$ and $\vec{b}$ is $\dfrac{\pi}{6}$ radians.

Example 3: Collinear Vectors

Let $\vec{a}$ and $\vec{c}$ be collinear. Show that $(\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c})$.

Since $\vec{a} = \lambda \vec{c}$ for some scalar $\lambda$, note:

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

Similarly, $(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$, per the alternate BAC–CAB rule.

But if $\vec{a} = \lambda \vec{c}$, then $\vec{a} \cdot \vec{b} = \lambda (\vec{c} \cdot \vec{b})$ and $(\vec{b} \cdot \vec{c})\vec{a} = \lambda (\vec{b} \cdot \vec{c})\vec{c}$, ensuring equality.

Therefore, both sides are equal when $\vec{a}$ and $\vec{c}$ are collinear.

Exam Identification and Error Patterns

• Non-associativity must be recognised
• Formula selection is context-dependent
• Components must be kept separate
• Zero vectors cause trivial results

Further Concepts and Extensions

The scalar triple product, defined as $\vec{a} \cdot (\vec{b} \times \vec{c})$, is distinct from the vector triple product. It yields the (signed) volume of the parallelepiped formed by the three vectors. For more details, refer to Understanding Scalar Triple Product.

Knowledge of these products allows the simplification of advanced vector expressions in rotational dynamics and electromagnetism. For deeper exploration, reviewing the Triangle Law of Vector Addition may be beneficial to fortify the geometric intuitions underpinning such cross-product relationships.