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Chapter 8 โ€ข Theory & Derivations

Unit 8: Applications of Vectors in Spatial Geometry

Advanced geometric applications of vector calculus: vector representations of lines and planes, intersection piercing points, distance theorems, and the theory of reciprocal vector triads.

ยง8.1 8.1 Vector Formulation of Lines and Planes

The vector formalism condenses three-dimensional analytical geometry into coordinate-free, coordinate-independent equations.


1. Vector Equations of a Straight Line

1. Line through a Point with a Given Direction:

Passing through $\vec{a}$ parallel to $\vec{b}$:

$$\mathbf{\vec{r} = \vec{a} + t\vec{b}} \quad (t \in \mathbb{R})$$

In non-parametric form (since $\vec{r} - \vec{a} \parallel \vec{b}$):

$$\mathbf{(\vec{r} - \vec{a}) \times \vec{b} = \vec{0}}$$

2. Line Passing Through Two Points:

Passing through $\vec{a}$ and $\vec{b}$:

$$\mathbf{\vec{r} = (1 - t)\vec{a} + t\vec{b}} \quad \iff \quad (\vec{r} - \vec{a}) \times (\vec{b} - \vec{a}) = \vec{0}$$

2. Vector Equations of a Plane

1. Point-Normal Form:

Passing through $\vec{a}$ with normal vector $\vec{n}$:

$$\mathbf{(\vec{r} - \vec{a}) \cdot \vec{n} = 0 \iff \vec{r} \cdot \vec{n} = d} \quad (d = \vec{a} \cdot \vec{n})$$

2. Hesse Normal Form:

Using unit normal $\hat{n}$ pointing away from the origin, $p$ is the perpendicular distance from origin:

$$\mathbf{\vec{r} \cdot \hat{n} = p} \quad (p \ge 0)$$

3. Plane Through Three Non-Collinear Points:

Passing through $\vec{a}, \vec{b}, \vec{c}$. The vectors $\vec{r} - \vec{a}$, $\vec{b} - \vec{a}$, and $\vec{c} - \vec{a}$ are coplanar:

$$(\vec{r} - \vec{a}) \cdot [(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})] = 0$$

Expanding:

$$\mathbf{[\vec{r}, \vec{b}, \vec{c}] + [\vec{r}, \vec{c}, \vec{a}] + [\vec{r}, \vec{a}, \vec{b}] = [\vec{a}, \vec{b}, \vec{c}]}$$

3. Intersection of a Vector Line and a Vector Plane

Let the line be $\vec{r} = \vec{a} + t\vec{d}$ and the plane be $\vec{r} \cdot \vec{n} = q$.

Substitute the line parametrization into the plane equation:

$$(\vec{a} + t\vec{d}) \cdot \vec{n} = q \implies \vec{a} \cdot \vec{n} + t(\vec{d} \cdot \vec{n}) = q$$

If $\vec{d} \cdot \vec{n} \neq 0$ (line is not parallel to the plane):

$$\mathbf{t = \frac{q - \vec{a} \cdot \vec{n}}{\vec{d} \cdot \vec{n}}}$$

Substituting this unique parameter $t$ into $\vec{r}(t)$ yields the exact coordinates of the piercing point.

ยง8.2 8.2 Spatial Distance Formulas via Vector Projection

Vector cross and dot products provide concise derivations of all spatial distance formulas.


1. Distance from a Point to a Straight Line

Let $P(\vec{p})$ be a point in space, and let the line be $\mathcal{L}: \vec{r} = \vec{a} + t\vec{d}$.

Consider the vector $\vec{aP} = \vec{p} - \vec{a}$. The area of the parallelogram formed by $\vec{p} - \vec{a}$ and the direction vector $\vec{d}$ is:

$$\text{Area} = |(\vec{p} - \vec{a}) \times \vec{d}|$$

On the other hand, $\text{Area} = \text{base} \times \text{height} = |\vec{d}| \cdot D$, where $D$ is the perpendicular distance from $P$ to the line. Equating:

$$\mathbf{D = \frac{|(\vec{p} - \vec{a}) \times \vec{d}|}{|\vec{d}|}}$$

2. Distance from a Point to a Plane

Let $P(\vec{p})$ be a point and let the plane be $\Pi: \vec{r} \cdot \vec{n} = d$.

Let $A(\vec{a})$ be any point on the plane, so that $\vec{a} \cdot \vec{n} = d$. The perpendicular distance $D$ is the absolute value of the scalar projection of $\vec{p} - \vec{a}$ onto the normal vector $\vec{n}$:

$$D = \left| (\vec{p} - \vec{a}) \cdot \frac{\vec{n}}{|\vec{n}|} \right| = \frac{|\vec{p} \cdot \vec{n} - \vec{a} \cdot \vec{n}|}{|\vec{n}|}$$
$$\mathbf{D = \frac{|\vec{p} \cdot \vec{n} - d|}{|\vec{n}|}}$$

3. Angle Between a Line and a Plane

Let the line have direction $\vec{d}$ and the plane have normal $\vec{n}$. The angle $\theta$ between the line and the plane is the complement of the angle between $\vec{d}$ and $\vec{n}$:

$$\sin\theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|}$$

ยง8.3 8.3 Reciprocal Triad of Vectors and Polyhedral Geometry

1. Definition of the Reciprocal System

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors in $\mathbb{R}^3$, so that their scalar triple product is non-zero:

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

The reciprocal triad (or dual basis) of vectors, denoted by $\vec{a}', \vec{b}', \vec{c}'$, is defined as:

$$\mathbf{\vec{a}' = \frac{\vec{b} \times \vec{c}}{[\vec{a}, \vec{b}, \vec{c}]}, \qquad \vec{b}' = \frac{\vec{c} \times \vec{a}}{[\vec{a}, \vec{b}, \vec{c}]}, \qquad \vec{c}' = \frac{\vec{a} \times \vec{b}}{[\vec{a}, \vec{b}, \vec{c}]}}$$

2. Fundamental Orthogonality and Normalization Relations

Theorem: For the direct triad $(\vec{a}_1, \vec{a}_2, \vec{a}_3) = (\vec{a}, \vec{b}, \vec{c})$ and the reciprocal triad $(\vec{a}'_1, \vec{a}'_2, \vec{a}'_3) = (\vec{a}', \vec{b}', \vec{c}')$:

$$\mathbf{\vec{a}_i \cdot \vec{a}'_j = \delta_{ij} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}}$$
Proof:

For $i = j = 1$:

$$\vec{a} \cdot \vec{a}' = \vec{a} \cdot \left( \frac{\vec{b} \times \vec{c}}{[\vec{a}, \vec{b}, \vec{c}]} \right) = \frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{[\vec{a}, \vec{b}, \vec{c}]} = \frac{[\vec{a}, \vec{b}, \vec{c}]}{[\vec{a}, \vec{b}, \vec{c}]} = 1$$

For $i \neq j$, say $\vec{a} \cdot \vec{b}'$:

$$\vec{a} \cdot \vec{b}' = \vec{a} \cdot \left( \frac{\vec{c} \times \vec{a}}{[\vec{a}, \vec{b}, \vec{c}]} \right) = \frac{[\vec{a}, \vec{c}, \vec{a}]}{[\vec{a}, \vec{b}, \vec{c}]} = \frac{0}{[\vec{a}, \vec{b}, \vec{c}]} = 0$$

since any determinant with two identical vectors vanishes identically. $\blacksquare$


3. Reciprocal Volume Theorem

Theorem: The scalar triple product of the reciprocal vectors is the reciprocal of the scalar triple product of the original vectors:

$$\mathbf{[\vec{a}', \vec{b}', \vec{c}'] = \frac{1}{[\vec{a}, \vec{b}, \vec{c}]}}$$
Proof:

By definition:

$$[\vec{a}', \vec{b}', \vec{c}'] = \vec{a}' \cdot (\vec{b}' \times \vec{c}') = \frac{\vec{b} \times \vec{c}}{V} \cdot \left( \frac{\vec{c} \times \vec{a}}{V} \times \frac{\vec{a} \times \vec{b}}{V} \right)$$
$$= \frac{1}{V^3} (\vec{b} \times \vec{c}) \cdot \left[ (\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b}) \right]$$

Using the product of four vectors identity $(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$ with $\vec{w} = \vec{a} \times \vec{b}$:

$$[(\vec{c} \times \vec{a}) \times \vec{w}] = [\vec{c}, \vec{a}, \vec{b}]\vec{a} - [\vec{a}, \vec{a}, \vec{b}]\vec{c} = V\vec{a} - \vec{0} = V\vec{a}$$

Substitute this back:

$$[\vec{a}', \vec{b}', \vec{c}'] = \frac{1}{V^3} (\vec{b} \times \vec{c}) \cdot (V\vec{a}) = \frac{V}{V^3} (\vec{b} \times \vec{c}) \cdot \vec{a} = \frac{V \cdot V}{V^3} = \frac{1}{V} = \frac{1}{[\vec{a}, \vec{b}, \vec{c}]} \quad \blacksquare$$

4. Expansion of an Arbitrary Vector

Any vector $\vec{r} \in \mathbb{R}^3$ can be decomposed immediately in terms of either basis without inverting matrices:

$$\mathbf{\vec{r} = (\vec{r} \cdot \vec{a}')\vec{a} + (\vec{r} \cdot \vec{b}')\vec{b} + (\vec{r} \cdot \vec{c}')\vec{c}}$$
$$\mathbf{\vec{r} = (\vec{r} \cdot \vec{a})\vec{a}' + (\vec{r} \cdot \vec{b})\vec{b}' + (\vec{r} \cdot \vec{c})\vec{c}'}$$

This property makes reciprocal triads indispensable in solid-state physics and crystallography for defining reciprocal lattices.

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 8.1: Perpendicular Distance from a Point to a Vector Line

Find the perpendicular distance from the point $P(1, 2, 3)$ to the straight line given by:

$$\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + t(\hat{i} + 2\hat{j} - 2\hat{k})$$

Step 1: Identify reference vectors

  • Line reference point $\vec{a} = (2, -1, 4)$.
  • Line direction vector $\vec{d} = (1, 2, -2)$.
  • Given point $\vec{p} = (1, 2, 3)$.

The connecting displacement vector is:

$$\vec{p} - \vec{a} = (1 - 2)\hat{i} + (2 - (-1))\hat{j} + (3 - 4)\hat{k} = (-1, 3, -1)$$

Step 2: Compute the cross product $(\vec{p} - \vec{a}) \times \vec{d}$

$$(\vec{p} - \vec{a}) \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & -1 \\ 1 & 2 & -2 \end{vmatrix}$$
$$= \hat{i}(3(-2) - (-1)(2)) - \hat{j}((-1)(-2) - (-1)(1)) + \hat{k}((-1)(2) - 3(1))$$
$$= \hat{i}(-6 + 2) - \hat{j}(2 + 1) + \hat{k}(-2 - 3) = -4\hat{i} - 3\hat{j} - 5\hat{k}$$

Step 3: Evaluate magnitudes

$$|(\vec{p} - \vec{a}) \times \vec{d}| = \sqrt{(-4)^2 + (-3)^2 + (-5)^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2}$$
$$|\vec{d}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$

Step 4: Compute perpendicular distance

$$D = \frac{|(\vec{p} - \vec{a}) \times \vec{d}|}{|\vec{d}|} = \frac{5\sqrt{2}}{3} \approx 2.357 \text{ units}$$
Solved Problem Example 8.2: Piercing Point and Angle of Line-Plane Intersection

Find the point of intersection (piercing point) of the line:

$$\vec{r} = (3\hat{i} - \hat{j} + 2\hat{k}) + t(2\hat{i} + \hat{j} - \hat{k})$$

with the plane:

$$\vec{r} \cdot (2\hat{i} - 3\hat{j} + 4\hat{k}) = 7$$

Also compute the angle $\theta$ between the line and the plane.

Step 1: Determine the intersection parameter $t$ The line is $\vec{r}(t) = (3 + 2t)\hat{i} + (-1 + t)\hat{j} + (2 - t)\hat{k}$. Substitute into the plane equation $\vec{r} \cdot \vec{n} = 7$ where $\vec{n} = (2, -3, 4)$:

$$(3 + 2t)(2) + (-1 + t)(-3) + (2 - t)(4) = 7$$
$$(6 + 4t) + (3 - 3t) + (8 - 4t) = 7$$
$$(6 + 3 + 8) + (4t - 3t - 4t) = 7$$
$$17 - 3t = 7$$
$$-3t = 7 - 17 = -10 \implies t = \frac{10}{3}$$

Step 2: Compute piercing point coordinates Substitute $t = \frac{10}{3}$ into $\vec{r}(t)$:

$$x = 3 + 2\left(\frac{10}{3}\right) = 3 + \frac{20}{3} = \frac{29}{3}$$
$$y = -1 + \left(\frac{10}{3}\right) = \frac{7}{3}$$
$$z = 2 - \left(\frac{10}{3}\right) = -\frac{4}{3}$$

Thus, the piercing point is $P\left(\frac{29}{3}, \frac{7}{3}, -\frac{4}{3}\right)$.


Step 3: Angle between line and plane Line direction $\vec{d} = (2, 1, -1)$, plane normal $\vec{n} = (2, -3, 4)$.

$$\vec{d} \cdot \vec{n} = 2(2) + 1(-3) + (-1)(4) = 4 - 3 - 4 = -3$$
$$|\vec{d}| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$$
$$|\vec{n}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$$
$$\sin\theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|} = \frac{|-3|}{\sqrt{6}\sqrt{29}} = \frac{3}{\sqrt{174}}$$
$$\theta = \arcsin\left(\frac{3}{\sqrt{174}}\right) \approx \arcsin(0.2274) \approx 13.14^\circ$$
Solved Problem Example 8.3: Construction and Verification of Reciprocal Vector Triad

Given the non-coplanar triad of vectors:

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

(a) Compute the scalar triple product $[\vec{a}, \vec{b}, \vec{c}]$. (b) Construct the reciprocal triad of vectors $(\vec{a}', \vec{b}', \vec{c}')$. (c) Verify explicitly that $\vec{a} \cdot \vec{a}' = 1$, $\vec{a} \cdot \vec{b}' = 0$, and $[\vec{a}', \vec{b}', \vec{c}'] = \frac{1}{[\vec{a}, \vec{b}, \vec{c}]}$.

Part (a): Evaluate $[\vec{a}, \vec{b}, \vec{c}]$

$$\vec{a} = (1, 1, 0), \quad \vec{b} = (0, 1, 1), \quad \vec{c} = (1, 0, 1)$$
$$V = [\vec{a}, \vec{b}, \vec{c}] = \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix}$$

Expanding along the first row:

$$V = 1(1 - 0) - 1(0 - 1) + 0 = 1 + 1 = 2$$

Since $V = 2 \neq 0$, the vectors form a valid non-coplanar triad.


Part (b): Construct the reciprocal triad

  1. Compute $\vec{b} \times \vec{c}$:
$$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} = \hat{i}(1 - 0) - \hat{j}(0 - 1) + \hat{k}(0 - 1) = \hat{i} + \hat{j} - \hat{k} = (1, 1, -1)$$
$$\vec{a}' = \frac{\vec{b} \times \vec{c}}{V} = \frac{1}{2}(\hat{i} + \hat{j} - \hat{k})$$
  1. Compute $\vec{c} \times \vec{a}$:
$$\vec{c} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - 1) - \hat{j}(0 - 1) + \hat{k}(1 - 0) = -\hat{i} + \hat{j} + \hat{k} = (-1, 1, 1)$$
$$\vec{b}' = \frac{\vec{c} \times \vec{a}}{V} = \frac{1}{2}(-\hat{i} + \hat{j} + \hat{k})$$
  1. Compute $\vec{a} \times \vec{b}$:
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1 - 0) - \hat{j}(1 - 0) + \hat{k}(1 - 0) = \hat{i} - \hat{j} + \hat{k} = (1, -1, 1)$$
$$\vec{c}' = \frac{\vec{a} \times \vec{b}}{V} = \frac{1}{2}(\hat{i} - \hat{j} + \hat{k})$$

Part (c): Verification of Properties

1. Normalization:

$$\vec{a} \cdot \vec{a}' = (1, 1, 0) \cdot \frac{1}{2}(1, 1, -1) = \frac{1}{2}(1(1) + 1(1) + 0(-1)) = \frac{1}{2}(2) = 1 \quad \checkmark$$

2. Orthogonality:

$$\vec{a} \cdot \vec{b}' = (1, 1, 0) \cdot \frac{1}{2}(-1, 1, 1) = \frac{1}{2}(1(-1) + 1(1) + 0(1)) = \frac{1}{2}(0) = 0 \quad \checkmark$$

3. Reciprocal Volume:

$$[\vec{a}', \vec{b}', \vec{c}'] = \begin{vmatrix} 1/2 & 1/2 & -1/2 \\ -1/2 & 1/2 & 1/2 \\ 1/2 & -1/2 & 1/2 \end{vmatrix} = \left(\frac{1}{2}\right)^3 \begin{vmatrix} 1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$$

Evaluate the integer determinant:

$$\begin{vmatrix} 1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} = 1(1 - (-1)) - 1(-1 - 1) + (-1)(1 - 1)$$
$$= 1(2) - 1(-2) - 1(0) = 2 + 2 = 4$$

Thus:

$$[\vec{a}', \vec{b}', \vec{c}'] = \frac{1}{8} \times 4 = \frac{4}{8} = \frac{1}{2}$$

Notice that:

$$\frac{1}{[\vec{a}, \vec{b}, \vec{c}]} = \frac{1}{2}$$

Therefore, $[\vec{a}', \vec{b}', \vec{c}'] = \frac{1}{[\vec{a}, \vec{b}, \vec{c}]} = \frac{1}{2}$ holds with absolute mathematical precision. $\blacksquare$