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:
where $\theta \in [0, \pi]$ is the interior angle between them. In orthonormal Cartesian components:
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:
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:
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 $\vec{a}, \vec{b} \in \mathbb{R}^3$:
Proof:
From geometric definitions:
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:
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:
This is precisely the expansion of the $3 \times 3$ determinant:
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:
- Volume of a Tetrahedron:
A tetrahedron with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ has volume equal to one-sixth of the parallelepiped:
3. Cyclic Symmetries and Coplanarity
From the determinant properties under row swaps:
Interchange of Dot and Cross:
- Condition of Coplanarity:
Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if their scalar triple product vanishes:
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):
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:
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:
Now evaluate $\vec{a} \times (\vec{b} \times \vec{c})$:
4. Compute RHS:
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:
Summing all three equations:
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:
2. Vector Product of Four Vectors:
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.
Given three vectors:
(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 $[\vec{a}, \vec{b}, \vec{c}]$
Step 2: Expand the determinant along the first row
Evaluate each $2 \times 2$ minor:
Substitute back:
Step 3: Parallelepiped and Tetrahedral Volumes
- Parallelepiped Volume:
- Tetrahedron Volume:
For the vectors:
compute explicitly both sides of the BAC-CAB theorem:
and verify that they yield identically equal vectors.
Step 1: Compute LHS by direct cross products First, compute $\vec{b} \times \vec{c}$:
Next, evaluate $\vec{a} \times (\vec{b} \times \vec{c})$:
Thus:
Step 2: Compute RHS using dot products Evaluate the dot products:
Now evaluate $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$:
Step 3: Conclusion
The identity is fully verified.
Prove analytically for any four vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d} \in \mathbb{R}^3$ that:
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:
Using the cyclic property of the scalar triple product:
Substitute back $\vec{v} = \vec{c} \times \vec{d}$:
Step 2: Apply the BAC-CAB Rule to $\vec{b} \times (\vec{c} \times \vec{d})$
Step 3: Take the dot product with $\vec{a}$
Step 4: Express as a $2 \times 2$ determinant Notice that:
This precisely equals the expression derived in Step 3. $\blacksquare$
Step 5: Deduction of Lagrange's Identity Set $\vec{c} = \vec{a}$ and $\vec{d} = \vec{b}$:
This yields Lagrange's Identity immediately as a special case. $\blacksquare$