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Chapter 7 โ€ข Theory & Derivations

Unit 7: Vector Algebra in Space & Multi-Vector Products

Exhaustive treatment of vector products in Euclidean 3-space: scalar product, vector cross product, scalar triple product, vector triple product (BAC-CAB theorem), and higher-order 4-vector identities.

ยง7.1 7.1 Inner and Outer Products in R^3

Vector algebra in three-dimensional Euclidean space $\mathbb{R}^3$ forms the foundational language of mathematical physics, mechanics, and geometric analysis.


1. The Dot (Scalar Inner) Product

Given two vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, their scalar product is defined geometrically as:

$$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$$

where $\theta \in [0, \pi]$ is the interior angle between them. In orthonormal Cartesian components:

$$\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$$
Key Properties:
  • Commutativity: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$.
  • Magnitude: $|\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}$.
  • Orthogonality: $\vec{a} \perp \vec{b} \iff \vec{a} \cdot \vec{b} = 0$ (for non-zero vectors).
  • Cauchy-Schwarz Inequality: $|\vec{a} \cdot \vec{b}| \le |\vec{a}| |\vec{b}|$, with equality if and only if $\vec{a} \parallel \vec{b}$.

2. The Cross (Vector Outer) Product

The vector product $\vec{a} \times \vec{b}$ produces a vector perpendicular to both $\vec{a}$ and $\vec{b}$, oriented according to the right-hand rule:

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

where $\hat{n}$ is the unit normal vector such that $(\vec{a}, \vec{b}, \hat{n})$ forms a right-handed orthogonal triad.

In Cartesian component form:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\hat{i} + (a_3 b_1 - a_1 b_3)\hat{j} + (a_1 b_2 - a_2 b_1)\hat{k}$$
Key Properties:
  • Anti-commutativity: $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$.
  • Collinearity: $\vec{a} \parallel \vec{b} \iff \vec{a} \times \vec{b} = \vec{0}$.
  • Geometric Area: The magnitude $|\vec{a} \times \vec{b}|$ equals the area of the parallelogram spanned by $\vec{a}$ and $\vec{b}$.

3. Lagrange's Identity

Theorem (Lagrange's Identity): For any vectors $\vec{a}, \vec{b} \in \mathbb{R}^3$:

$$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$$
Proof:

From geometric definitions:

$$\text{LHS} = (|\vec{a}| |\vec{b}| \sin\theta)^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2\theta$$
$$= |\vec{a}|^2 |\vec{b}|^2 (1 - \cos^2\theta) = |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}|^2 |\vec{b}|^2 \cos^2\theta$$
$$= |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 = \text{RHS} \quad \blacksquare$$

ยง7.2 7.2 The Scalar Triple Product (Box Product)

1. Definition and Determinant Representation

The scalar triple product (or box product) of three vectors $\vec{a}, \vec{b}, \vec{c} \in \mathbb{R}^3$ is defined as:

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

Let $\vec{a} = (a_1, a_2, a_3)$, $\vec{b} = (b_1, b_2, b_3)$, and $\vec{c} = (c_1, c_2, c_3)$. Expanding via the dot product:

$$\vec{a} \cdot (\vec{b} \times \vec{c}) = a_1 (b_2 c_3 - b_3 c_2) + a_2 (b_3 c_1 - b_1 c_3) + a_3 (b_1 c_2 - b_2 c_1)$$

This is precisely the expansion of the $3 \times 3$ determinant:

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

2. Geometric Interpretation: Parallelepiped and Tetrahedral Volume

  • Volume of a Parallelepiped:

The base of the parallelepiped formed by $\vec{b}$ and $\vec{c}$ has area $A = |\vec{b} \times \vec{c}|$. The unit normal to the base is $\hat{n} = \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|}$. The height $h$ is the projection of $\vec{a}$ onto $\hat{n}$: $h = |\vec{a} \cdot \hat{n}|$. Hence, the volume is:

$$V_{para} = A \cdot h = |\vec{b} \times \vec{c}| \frac{|\vec{a} \cdot (\vec{b} \times \vec{c})|}{|\vec{b} \times \vec{c}|} = |[\vec{a}, \vec{b}, \vec{c}]|$$
  • Volume of a Tetrahedron:

A tetrahedron with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ has volume equal to one-sixth of the parallelepiped:

$$V_{tet} = \frac{1}{6} |[\vec{a}, \vec{b}, \vec{c}]|$$

3. Cyclic Symmetries and Coplanarity

From the determinant properties under row swaps:

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

Interchange of Dot and Cross:

$$\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}$$
  • Condition of Coplanarity:

Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if their scalar triple product vanishes:

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

ยง7.3 7.3 The Vector Triple Product & Higher-Order Identities

1. The Vector Triple Product (BAC-CAB Theorem)

Given three vectors $\vec{a}, \vec{b}, \vec{c}$, their vector triple product is $\vec{a} \times (\vec{b} \times \vec{c})$.

Theorem (The BAC-CAB Rule):

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

1. Geometric Orientation: The vector $\vec{b} \times \vec{c}$ is perpendicular to the plane spanned by $\vec{b}$ and $\vec{c}$. Therefore, $\vec{a} \times (\vec{b} \times \vec{c})$ is perpendicular to $\vec{b} \times \vec{c}$, which implies it must lie in the plane of $\vec{b}$ and $\vec{c}$.

Hence, there exist scalars $\lambda, \mu$ such that:

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

2. Coordinate Setup: Without loss of generality, choose an orthonormal coordinate system such that:

  • The $x$-axis lies along $\vec{b}$: $\vec{b} = (b_1, 0, 0)$.
  • The $xy$-plane contains $\vec{c}$: $\vec{c} = (c_1, c_2, 0)$.
  • $\vec{a} = (a_1, a_2, a_3)$ is general.

3. Compute the Cross Products:

$$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ b_1 & 0 & 0 \\ c_1 & c_2 & 0 \end{vmatrix} = (0)\hat{i} + (0)\hat{j} + (b_1 c_2)\hat{k} = (0, 0, b_1 c_2)$$

Now evaluate $\vec{a} \times (\vec{b} \times \vec{c})$:

$$\vec{a} \times (\vec{b} \times \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 0 & 0 & b_1 c_2 \end{vmatrix} = (a_2 b_1 c_2)\hat{i} - (a_1 b_1 c_2)\hat{j} + (0)\hat{k}$$

4. Compute RHS:

$$\vec{a} \cdot \vec{c} = a_1 c_1 + a_2 c_2$$
$$\vec{a} \cdot \vec{b} = a_1 b_1$$
$$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = (a_1 c_1 + a_2 c_2)(b_1 \hat{i}) - (a_1 b_1)(c_1 \hat{i} + c_2 \hat{j})$$
$$= (a_1 b_1 c_1 + a_2 b_1 c_2 - a_1 b_1 c_1)\hat{i} - (a_1 b_1 c_2)\hat{j}$$
$$= (a_2 b_1 c_2)\hat{i} - (a_1 b_1 c_2)\hat{j}$$

The LHS and RHS are identical in every component. Since the choice of coordinate axes is arbitrary, the result holds universally for all vectors in $\mathbb{R}^3$. $\blacksquare$


2. Jacobi's Identity

Summing cyclic permutations of the vector triple product:

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

Summing all three equations:

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

This is Jacobi's Identity, proving that the Lie algebra of 3D rotations $(\mathbb{R}^3, \times)$ satisfies the Jacobi relation.


3. Products of Four Vectors

1. Scalar Product of Four Vectors:

$$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c}) = \begin{vmatrix} \vec{a} \cdot \vec{c} & \vec{a} \cdot \vec{d} \\ \vec{b} \cdot \vec{c} & \vec{b} \cdot \vec{d} \end{vmatrix}$$

2. Vector Product of Four Vectors:

$$(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = [\vec{a}, \vec{b}, \vec{d}]\vec{c} - [\vec{a}, \vec{b}, \vec{c}]\vec{d} = [\vec{a}, \vec{c}, \vec{d}]\vec{b} - [\vec{b}, \vec{c}, \vec{d}]\vec{a}$$
TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Solved Problem Example 7.1: Computation of Box Product and Parallelepiped Volume

Given three vectors:

$$\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}, \qquad \vec{b} = \hat{i} + \hat{j} - 2\hat{k}, \qquad \vec{c} = 3\hat{i} - \hat{j} - \hat{k}$$

(a) Compute the scalar triple product $[\vec{a}, \vec{b}, \vec{c}]$. (b) Find the volume of the parallelepiped spanned by them. (c) Find the volume of the tetrahedron with coterminous edges $\vec{a}, \vec{b}, \vec{c}$.

Step 1: Set up the determinant for $[\vec{a}, \vec{b}, \vec{c}]$

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

Step 2: Expand the determinant along the first row

$$[\vec{a}, \vec{b}, \vec{c}] = 2\begin{vmatrix} 1 & -2 \\ -1 & -1 \end{vmatrix} - (-3)\begin{vmatrix} 1 & -2 \\ 3 & -1 \end{vmatrix} + 1\begin{vmatrix} 1 & 1 \\ 3 & -1 \end{vmatrix}$$

Evaluate each $2 \times 2$ minor:

$$\begin{vmatrix} 1 & -2 \\ -1 & -1 \end{vmatrix} = 1(-1) - (-2)(-1) = -1 - 2 = -3$$
$$\begin{vmatrix} 1 & -2 \\ 3 & -1 \end{vmatrix} = 1(-1) - (-2)(3) = -1 + 6 = 5$$
$$\begin{vmatrix} 1 & 1 \\ 3 & -1 \end{vmatrix} = 1(-1) - 1(3) = -1 - 3 = -4$$

Substitute back:

$$[\vec{a}, \vec{b}, \vec{c}] = 2(-3) + 3(5) + 1(-4) = -6 + 15 - 4 = 5$$

Step 3: Parallelepiped and Tetrahedral Volumes

  • Parallelepiped Volume:
$$V_{para} = |[\vec{a}, \vec{b}, \vec{c}]| = |5| = 5 \text{ cubic units}$$
  • Tetrahedron Volume:
$$V_{tet} = \frac{1}{6} |[\vec{a}, \vec{b}, \vec{c}]| = \frac{5}{6} \text{ cubic units}$$
Solved Problem Example 7.2: Verification of the Vector Triple Product BAC-CAB Identity

For the vectors:

$$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \qquad \vec{b} = 2\hat{i} - \hat{j} + \hat{k}, \qquad \vec{c} = 3\hat{i} + \hat{j} - \hat{k}$$

compute explicitly both sides of the BAC-CAB theorem:

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

and verify that they yield identically equal vectors.

Step 1: Compute LHS by direct cross products First, compute $\vec{b} \times \vec{c}$:

$$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & 1 & -1 \end{vmatrix}$$
$$= \hat{i}((-1)(-1) - (1)(1)) - \hat{j}((2)(-1) - (1)(3)) + \hat{k}((2)(1) - (-1)(3))$$
$$= \hat{i}(1 - 1) - \hat{j}(-2 - 3) + \hat{k}(2 + 3) = 0\hat{i} + 5\hat{j} + 5\hat{k} = (0, 5, 5)$$

Next, evaluate $\vec{a} \times (\vec{b} \times \vec{c})$:

$$\vec{a} \times (\vec{b} \times \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 5 & 5 \end{vmatrix}$$
$$= \hat{i}(2 \cdot 5 - 3 \cdot 5) - \hat{j}(1 \cdot 5 - 3 \cdot 0) + \hat{k}(1 \cdot 5 - 2 \cdot 0)$$
$$= \hat{i}(10 - 15) - \hat{j}(5 - 0) + \hat{k}(5 - 0) = -5\hat{i} - 5\hat{j} + 5\hat{k}$$

Thus:

$$\text{LHS} = (-5, -5, 5)$$

Step 2: Compute RHS using dot products Evaluate the dot products:

$$\vec{a} \cdot \vec{c} = 1(3) + 2(1) + 3(-1) = 3 + 2 - 3 = 2$$
$$\vec{a} \cdot \vec{b} = 1(2) + 2(-1) + 3(1) = 2 - 2 + 3 = 3$$

Now evaluate $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$:

$$\text{RHS} = 2(2\hat{i} - \hat{j} + \hat{k}) - 3(3\hat{i} + \hat{j} - \hat{k})$$
$$= (4\hat{i} - 2\hat{j} + 2\hat{k}) - (9\hat{i} + 3\hat{j} - 3\hat{k})$$
$$= (4 - 9)\hat{i} + (-2 - 3)\hat{j} + (2 - (-3))\hat{k}$$
$$= -5\hat{i} - 5\hat{j} + 5\hat{k}$$

Step 3: Conclusion

$$\text{LHS} = -5\hat{i} - 5\hat{j} + 5\hat{k} = \text{RHS}$$

The identity is fully verified.

Solved Problem Example 7.3: Proof of the Four-Vector Inner Product Identity

Prove analytically for any four vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d} \in \mathbb{R}^3$ that:

$$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \begin{vmatrix} \vec{a} \cdot \vec{c} & \vec{a} \cdot \vec{d} \\ \vec{b} \cdot \vec{c} & \vec{b} \cdot \vec{d} \end{vmatrix}$$

Hence, deduce Lagrange's Identity as a special corollary when $\vec{c} = \vec{a}$ and $\vec{d} = \vec{b}$.

Step 1: Treat $\vec{c} \times \vec{d}$ as a single vector Let $\vec{v} = \vec{c} \times \vec{d}$. The left-hand side is:

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

Using the cyclic property of the scalar triple product:

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

Substitute back $\vec{v} = \vec{c} \times \vec{d}$:

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

Step 2: Apply the BAC-CAB Rule to $\vec{b} \times (\vec{c} \times \vec{d})$

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

Step 3: Take the dot product with $\vec{a}$

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

Step 4: Express as a $2 \times 2$ determinant Notice that:

$$\begin{vmatrix} \vec{a} \cdot \vec{c} & \vec{a} \cdot \vec{d} \\ \vec{b} \cdot \vec{c} & \vec{b} \cdot \vec{d} \end{vmatrix} = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})$$

This precisely equals the expression derived in Step 3. $\blacksquare$


Step 5: Deduction of Lagrange's Identity Set $\vec{c} = \vec{a}$ and $\vec{d} = \vec{b}$:

$$(\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b}) = \begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} \end{vmatrix}$$
$$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$$

This yields Lagrange's Identity immediately as a special case. $\blacksquare$