Chapter 13 - Vectors and Surfaces

Vectors

Vector space: addition and scaling

Vector space - addition and scaling

A vector is an element in a vector space, which is a collection of objects supporting addition and scalar coefficients.

Vector space operations

• $\mathbf{v}+\mathbf{w}$ is a vector when $\mathbf{v},\mathbf{w}=\text{vectors}$
• $\lambda \mathbf{v}$ is a vector when $\mathbf{v}=\text{vector},\lambda =\text{scalar}$

Properties of vector operations

• Commutativity: $\mathbf{v}+\mathbf{w}=\mathbf{w}+\mathbf{v}$
• Associativity: $\mathbf{u}+(\mathbf{v}+\mathbf{w})=(\mathbf{u}+\mathbf{v})+\mathbf{w}$
• Distributivity: $\lambda (\mathbf{v}+\mathbf{w})=\lambda \mathbf{v}+\lambda \mathbf{w}$
• Zero vector: $\mathbf{v}+0=\mathbf{v}$

The elements $\lambda \mathbf{v}$ also satisfy obvious properties like:
$1\mathbf{v}=\mathbf{v}$ and $(\lambda +\mu )\mathbf{v}=\lambda \mathbf{v}+\mu \mathbf{v}$ and $\lambda (\mu \mathbf{v})=(\lambda \mu )\mathbf{v}$

Scalars
Scalars (“scal-ers”) could be taken in any field, but for this class, they are just real numbers $\lambda \in \mathbb{R}$.

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Components of vectors

Components of vectors

In this class we consider vectors in the plane ${\mathbb{R}}^2$ and vectors in space ${\mathbb{R}}^3$.

Vectors in the plane are ordered pairs $\mathbf{v}=(a_1,a_2)$ with $a_1,a_2\in \mathbb{R}$.
Vectors in space are ordered triples $\mathbf{v}=(a_1,a_2,a_3)$ with $a_1,a_2,a_3\in \mathbb{R}$.

Addition is componentwise: $(a_1,a_2)+(b_1,b_2)=(a_1+b_1,a_2+b_2)$
Scalars act by components too: $\lambda (a_1,a_2)=(\lambda a_1,\lambda a_2)$
(Similarly for vectors in space.)

Standard basis vectors
These special vectors have their own symbols:
${\mathbf{e}}_{\mathbf{i}}=(1,0,0),{\mathbf{e}}_{\mathbf{j}}=(0,1,0),{\mathbf{e}}_{\mathbf{k}}=(0,0,1)$

Components
Most often the word ‘component’ of a vector refers to one of the numbers $a_i$ in the ordered sequence $(a_1,a_2)$ or $(a_1,a_2,a_3)$.

Sometimes ‘component’ refers to the basis vector scaled by this number, e.g. $a_1{\mathbf{e}}_{\mathbf{i}}$.
Notice that we always have $(a_1,a_2,a_3)=a_1{\mathbf{e}}_{\mathbf{i}}+a_2{\mathbf{e}}_{\mathbf{j}}+a_3{\mathbf{e}}_{\mathbf{k}}$, which is the reason ‘component’ is sometimes used in this second way.

Other bases
It is possible to select a different set of special vectors, e.g. $\mathbf{B}=\{{\mathbf{u}}_1,{\mathbf{u}}_2,{\mathbf{u}}_3\}$, having the property that any vector $\mathbf{v}$ can be written uniquely as a linear combination in the same way: $\mathbf{v}=a_1{\mathbf{u}}_1+a_2{\mathbf{u}}_2+a_3{\mathbf{u}}_3$. Any such set is called a basis, and the $a_i$ in this case are called components in the basis $\mathbf{B}$. This topic is considered further in Linear Algebra.

Coordinates
Mathematicians say ‘components’ for the parts of a vector, and ‘coordinates’ for the quantities that are used to identify points of space. Since the spaces ${\mathbb{R}}^2$ and ${\mathbb{R}}^3$ are vector spaces, the tuples $(a_1,a_2)$ or $(a_1,a_2,a_3)$ actually give coordinates and vector components!

In more general curved spaces, points cannot be “added together,” so this relation between coordinates and components breaks down.

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Geometric vectors

Geometric vectors

It is possible to consider vectors purely geometrically as the combined data of direction and magnitude in space.
(Provided one has defined space purely geometrically, which nobody does.)

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Products of vectors

Introduction

Vector spaces often come with additional product operations between vectors.

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Dot product

Dot product

The vector space ${\mathbb{R}}^n$ (any $n$) comes with an inner product, also called the dot product or scalar product. Every inner product satisfies basic rules:
Algebraic properties

• $\mathbf{u}\cdot \mathbf{v}=\mathbf{v}\cdot \mathbf{u}\in \mathbb{R}$
• $(\mathbf{u}+\mathbf{v})\cdot \mathbf{w}=\mathbf{u}\cdot \mathbf{w}+\mathbf{v}\cdot \mathbf{w}$
• $(\lambda \mathbf{u})\cdot \mathbf{v}=\lambda (\mathbf{u}\cdot \mathbf{v})$
• $0\cdot \mathbf{u}=0$

Component definition
The dot products for ${\mathbb{R}}^2$ and ${\mathbb{R}}^3$ are defined using components:

• $(a_1,a_2)\cdot (b_1,b_2)=a_1{b}_1+a_2{b}_2$
• $(a_1,a_2,a_3)\cdot (b_1,b_2,b_3)=a_1{b}_1+a_2{b}_2+a_3{b}_3$

$$\begin{array}{c}\mathbf{a}\cdot \mathbf{b}=\\ a_1{b}_1+a_2{b}_2+a_3{b}_3\end{array}$$

Norm
The dot product of a vector with itself gives a number called the (norm or length or magnitude) squared:

• $\mathbf{u}\cdot \mathbf{u}=|\mathbf{u}{|}^2=||\mathbf{u}|{|}^2$

The terminology is justified by the Pythagorean Theorem in ${\mathbb{R}}^3$:

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Exercise

01

Parallelogram identity

Let $\mathbf{u}=(4,0,3)$ and $\mathbf{v}=(-2,1,5)$. Compute $|\mathbf{u}-\mathbf{v}|$ and $|\mathbf{u}+\mathbf{v}|$ and $|\mathbf{u}|$ and $|\mathbf{v}|$. Show that:

$$|\mathbf{u}+\mathbf{v}{|}^2+|\mathbf{u}-\mathbf{v}{|}^2=2|\mathbf{u}{|}^2+2|\mathbf{v}{|}^2$$

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Dot product (part 2)

Geometry
The dot product scalar provides information about the geometric relationship of two vectors.

$$\mathbf{u}\cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos (\theta ),$$

where $\theta$ is the angle between $\mathbf{u}$ and $\mathbf{v}$.
Proof: The rules for dot product imply that:

$$|\mathbf{u}-\mathbf{v}{|}^2=(\mathbf{u}-\mathbf{v})\cdot (\mathbf{u}-\mathbf{v})=\mathbf{u}\cdot \mathbf{u}+\mathbf{v}\cdot \mathbf{v}-2\mathbf{u}\cdot \mathbf{v}.$$

On the other hand, the Law of Cosines for the triangle with sides $\mathbf{u},\mathbf{v},\mathbf{u}-\mathbf{v}$ gives:

$$|\mathbf{u}-\mathbf{v}{|}^2=|\mathbf{u}{|}^2+|\mathbf{v}{|}^2-2\cos (\theta )|\mathbf{u}||\mathbf{v}|.$$

It is worth noticing that the norm determines the dot product:

$$\mathbf{u}\cdot \mathbf{v}=\frac{1}{2}\left(|\mathbf{u}{|}^2+|\mathbf{v}{|}^2-|\mathbf{u}-\mathbf{v}{|}^2\right)$$

The formula for $\theta$ has qualitative geometrical consequences:

• If $\mathbf{u}\cdot \mathbf{v}=0$ then $\mathbf{u}$ and $\mathbf{v}$ are orthogonal (perpendicular)
• If $\mathbf{u}\cdot \mathbf{v}>0$ then $\mathbf{u}$ and $\mathbf{v}$ form an acute angle
• If $\mathbf{u}\cdot \mathbf{v}<0$ then $\mathbf{u}$ and $\mathbf{v}$ form an obtuse angle
• If $\mathbf{u}\cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}|$ then $\mathbf{u}$ and $\mathbf{v}$ are parallel

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Exercise

02

Angle between vectors

Find the angle between $(3,6,2)$ and $(4,2,4)$.

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Exercise

03

Orthogonality by hand

Determine which pairs of $\mathbf{u}=(2,-1,1)$ and $\mathbf{v}=(2,6,1)$ and $\mathbf{w}=(-4,1,2)$ are orthogonal.

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Exercise

04

Acute, orthogonal, or obtuse?

Are the angles between $\mathbf{u}=(1/2,1/2,5)$ and $\mathbf{v}=(3,1,-2)$ and $\mathbf{w}=(4,-3,0)$ obtuse or acute?

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Exercise

05

Dot product via unit vectors

Compute the dot product of $4{\mathbf{e}}_{\mathbf{i}}-3{\mathbf{e}}_{\mathbf{j}}$ and ${\mathbf{e}}_{\mathbf{i}}+2{\mathbf{e}}_{\mathbf{j}}+{\mathbf{e}}_{\mathbf{k}}$.

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Dot product (part 3)

Unit vectors
A vector $\mathbf{u}$ with $|\mathbf{u}|=1$ is called a unit vector.
For example, ${\mathbf{e}}_{\mathbf{i}},{\mathbf{e}}_{\mathbf{j}},{\mathbf{e}}_{\mathbf{k}}$ are unit vectors.

Thinking geometrically, a unit vector carries the information of direction. One could call them ‘direction vectors’ instead.

Any vector can be renormalized to create a new unit vector by dividing out its length:
${\mathbf{e}}_{\mathbf{u}}=\frac{1}{|\mathbf{u}|}\mathbf{u}$
Some texts, especially in physics, use the notation $\hat{\mathbf{u}}$ for this unit vector ${\mathbf{e}}_{\mathbf{u}}$.

Projection
The dot product is a quantity telling how much one vector aligns with another.
The dot product of two unit vectors is $\cos (\theta )$ where $\theta$ is the angle between them.

One can use the dot product to find the components of a vector in the standard basis. Say $\mathbf{u}=(a_1,a_2,a_3)$ is any vector. Then:

$$\mathbf{u}=(\mathbf{u}\cdot {\mathbf{e}}_{\mathbf{i}}){\mathbf{e}}_{\mathbf{i}}+(\mathbf{u}\cdot {\mathbf{e}}_{\mathbf{j}}){\mathbf{e}}_{\mathbf{j}}+(\mathbf{u}\cdot {\mathbf{e}}_{\mathbf{k}}){\mathbf{e}}_{\mathbf{k}}$$

This is because $\mathbf{u}\cdot {\mathbf{e}}_{\mathbf{i}}=a_1$.

More generally, one defines the projection of $\mathbf{u}$ along $\mathbf{v}$ (projection along the direction of $\mathbf{v}$) using the dot product:

$${\mathbf{u}}_{||}={\text{proj}}_{\mathbf{v}}(\mathbf{u})=(\mathbf{u}\cdot {\mathbf{e}}_{\mathbf{v}}){\mathbf{e}}_{\mathbf{v}}=\left(\frac{\mathbf{u}\cdot \mathbf{v}}{\mathbf{v}\cdot \mathbf{v}}\right)\mathbf{v}$$

The scalar coefficient $\mathbf{u}\cdot {\mathbf{e}}_{\mathbf{u}}$ is called the parallel component of $\mathbf{u}$ along $\mathbf{v}$. (Or, sometimes, ${\mathbf{u}}_{||}$ is called the component.)

The projection ${\mathbf{u}}_{||}$ points in the direction of $\mathbf{v}$, and it says how much of $\mathbf{u}$ lies in that direction.

Another vector called the perpendicular component is given by subtracting the parallel part:

$${\mathbf{u}}_{\perp }=\mathbf{u}-{\mathbf{u}}_{||}$$

It is called ‘perpendicular’ because:

$${\mathbf{u}}_{\perp }\cdot \mathbf{v}=\mathbf{u}\cdot \mathbf{v}-\left(\frac{\mathbf{u}\cdot \mathbf{v}}{\mathbf{v}\cdot \mathbf{v}}\right)\mathbf{v}\cdot \mathbf{v}=0$$ Link to original

Exercise

06

Projection by hand

Let $\mathbf{u}=(5,1,-3)$ and $\mathbf{v}=(4,4,2)$.

• (a) Find the projection of $\mathbf{u}$ to the direction of $\mathbf{v}$.
• (b) Write the vector $\mathbf{u}$ in terms of parts parallel and perpendicular to $\mathbf{v}$.

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Dot product (part 4)

Cauchy-Schwarz inequality
The Cauchy-Schwarz inequality says that for any vectors $\mathbf{u}$, $\mathbf{v}$, we know:

$$|\mathbf{u}\cdot \mathbf{v}|\le |\mathbf{u}||\mathbf{v}|.$$ Link to original

Exercise

07

Cauchy-Schwarz

Prove the Cauchy-Schwarz inequality using the angle formula.

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Dot product (part 5)

The Cauchy-Schwarz inequality is very important in math. There is a version of this inequality in many other settings where it is harder to prove.

Triangle inequality
The Triangle inequality says that for any vectors $\mathbf{u}$, $\mathbf{v}$, we know:

$$|\mathbf{u}+\mathbf{v}|\le |\mathbf{u}|+|\mathbf{v}|.$$ Link to original

Exercise

08

Cauchy-Schwarz $\Rightarrow$ Triangle

Prove the Triangle inequality using the Cauchy-Schwarz inequality and an expansion of $|\mathbf{u}+\mathbf{v}{|}^2$ with the dot product.

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Exercise

09

Orthogonality identity

Suppose that vectors $\mathbf{u}$ and $\mathbf{v}$ satisfy $|\mathbf{u}+\mathbf{v}{|}^2=|\mathbf{u}{|}^2+|\mathbf{v}{|}^2$. Show that they must be orthogonal.

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Cross product

Cross product

Another product is defined for vectors in ${\mathbb{R}}^3$ called the cross product. It is also called the vector product. The simplest definition is in terms of components:

$$\mathbf{u}\times \mathbf{v}=(a_2{b}_3-a_3{b}_2,-a_1{b}_3+a_3{b}_1,a_1{b}_2-a_2{b}_1)$$

assuming that $\mathbf{u}=(a_1,a_2,a_3)$ and $\mathbf{v}=(b_1,b_2,b_3)$.

There are some mnemonic formulations that can be helpful:

Or:
“Twist the others, minus in the middle.”

Or:

$$\mathbf{A}\times \mathbf{B}=+(A_y{B}_z-A_z{B}_y){\mathbf{e}}_{\mathbf{i}}-(A_x{B}_z-A_z{B}_x){\mathbf{e}}_{\mathbf{j}}+(A_x{B}_y-A_y{B}_x){\mathbf{e}}_{\mathbf{k}}$$

Or:

$$(a_1,a_2,a_3)\times (b_1,b_2,b_3)=\left|\begin{array}{ccc}{\mathbf{e}}_{\mathbf{i}}& {\mathbf{e}}_{\mathbf{j}}& {\mathbf{e}}_{\mathbf{k}}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$$

This is a $3\times 3$ determinant. The first row of vector entries should be treated as if scalars when taking this determinant.

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Exercise

10

Cross products by hand

Let $\mathbf{u}=(1,-2,1)$ and $\mathbf{v}=(3,1,-2)$. Compute in coordinates:

• (a) $\mathbf{u}\times \mathbf{v}$
• (b) $\mathbf{v}\times \mathbf{u}$
• (c) $\mathbf{u}\times \mathbf{u}$

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Cross product (part 2)

Algebraic properties
The cross product is not commutative and not associative. But it does satisfy some relations:

• $\mathbf{u}\times \mathbf{v}=-\mathbf{v}\times \mathbf{u}$ ( thus $\mathbf{v}\times \mathbf{v}=0$ )
• $(\lambda \mathbf{u})\times \mathbf{v}=\lambda (\mathbf{u}\times \mathbf{v})$
• $(\mathbf{u}+\mathbf{v})\times \mathbf{w}=\mathbf{u}\times \mathbf{w}+\mathbf{u}\times \mathbf{w}$
• $\mathbf{u}\cdot (\mathbf{u}\times \mathbf{v})=0$ ( thus $\mathbf{v}\cdot (\mathbf{u}\times \mathbf{v})=0$ )
• $\mathbf{u}\times \mathbf{v}=0$ implies $\mathbf{u}=\lambda \mathbf{v}$ for some $\lambda$

Geometry
The norm of the cross product satisfies an angle formula analogous to that of the dot product:

$$|\mathbf{u}\times \mathbf{v}|=|\mathbf{u}||\mathbf{v}|\sin (\theta )$$

The direction of the cross product is perpendicular to the plane spanned by $\mathbf{u},\mathbf{v}$ and it creates a right-handed system when combined with $\mathbf{u},\mathbf{v}$.

I.e. $(\mathbf{u},\mathbf{v},\mathbf{u}\times \mathbf{v})$ is right-handed.

The norm of the cross product $\mathbf{u}\times \mathbf{v}$ is also the area of the parallelogram formed by $\mathbf{u},\mathbf{v}$.

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Exercise

11

Cross product $⇝$ left-handed system

Use the cross product to find a unit vector $\mathbf{v}$ that is perpendicular to both $\mathbf{a}=(1,-4,1)$ and $\mathbf{b}=(2,3,0)$ such that the triple $(\mathbf{a},\mathbf{v},\mathbf{b})$ forms a left-handed system.

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Cross product (part 3)

Unit vectors

$$\begin{array}{r}\begin{array}{cc}{\mathbf{e}}_{\mathbf{i}}\times {\mathbf{e}}_{\mathbf{j}}& ={\mathbf{e}}_{\mathbf{k}}\\ {\mathbf{e}}_{\mathbf{j}}\times {\mathbf{e}}_{\mathbf{k}}& ={\mathbf{e}}_{\mathbf{i}}\\ {\mathbf{e}}_{\mathbf{k}}\times {\mathbf{e}}_{\mathbf{i}}& ={\mathbf{e}}_{\mathbf{j}}\end{array}\end{array}$$

These products are right-handed. Left-handed products include a minus sign, such as ${\mathbf{e}}_{\mathbf{i}}\times {\mathbf{e}}_{\mathbf{k}}=-{\mathbf{e}}_{\mathbf{j}}$.

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Exercise

12

Cross product via unit vectors

Compute the cross product of $4{\mathbf{e}}_{\mathbf{i}}-3{\mathbf{e}}_{\mathbf{j}}$ and ${\mathbf{e}}_{\mathbf{i}}+2{\mathbf{e}}_{\mathbf{j}}+{\mathbf{e}}_{\mathbf{k}}$ using the unit vector cross products.

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Cross product (part 4)

Dot-cross norm identity
The dot product and norm of cross product satisfy a simple relationship:

$$|\mathbf{u}\times \mathbf{v}{|}^2+(\mathbf{u}\cdot \mathbf{v}{)}^2=|\mathbf{u}{|}^2|\mathbf{v}{|}^2$$

This identity follows quickly from the angle formulas for dot and cross products.

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Triple products

Triple products

There are some general facts about products taken with three or more vectors.

Dot-cross triple product
Given vectors $\mathbf{u}=(a_1,a_2,a_3)$, $\mathbf{v}=(b_1,b_2,b_3)$, $\mathbf{w}=(c_1,c_2,c_3)$, consider the scalar quantity $\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})$. By combining the effect of the dot product with the determinant formula, we have a determinant formula for this scalar:

$$\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$$

and this formula immediately implies equalities among possible triple products:

$$\begin{array}{rl}\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})& =\mathbf{v}\cdot (\mathbf{w}\times \mathbf{u})=\mathbf{w}\cdot (\mathbf{u}\times \mathbf{v})\\ =-\mathbf{u}\cdot (\mathbf{w}\times \mathbf{v})& =-\mathbf{w}\cdot (\mathbf{v}\times \mathbf{u})=-\mathbf{v}\cdot (\mathbf{u}\times \mathbf{w})\end{array}$$

(To figure the signs: right-handed triples are equal, and left-handed triples are equal, and right and left differ by a sign.)

The determinant formula also implies a geometric interpretation of the dot-cross product: $\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})$ is the volume of the parallelepiped spanned by $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$.

Cross-cross triple product
Given vectors $\mathbf{a}=(a_1,a_2,a_3)$, $\mathbf{b}=(b_1,b_2,b_3)$, $\mathbf{c}=(c_1,c_2,c_3)$, we can form the triple product $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})$. (One should maybe not try to think about this product geometrically. It does arise in electromagnetism.) There is a useful shortcut called the Lagrange triple product identity for computing this product:

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

The abnormal order on the right (vector then scalar coefficient) facilitates a famous mnemonic: “Bac–Cab”.

$$\begin{array}{c}\mathbf{a}\times (\mathbf{b}\times \mathbf{c})=\\ \mathbf{b}(\mathbf{a}\cdot \mathbf{c})-\mathbf{c}(\mathbf{a}\cdot \mathbf{b})\end{array}$$ Link to original

Exercise

13

Jacobi identity

Use the Lagrange triple product identity to prove the Jacobi identity:

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

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Triple products (part 2)

Non-associativity

While the cross product is not associative, the difference has a simple expression using the dot product:

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

This identity can be proved using the Lagrange triple product identity.

$$\begin{array}{c}(\mathbf{a}\times \mathbf{b})\times \mathbf{c}\\ \ne \mathbf{a}\times (\mathbf{b}\times \mathbf{c})\end{array}$$Link to original

Geometric applications of vectors

Lines and planes

Lines and planes

Lines
Given a point identified with a vector ${\mathbf{r}}_0$ and a direction vector $\mathbf{v}$, the line through ${\mathbf{r}}_0$ in direction $\mathbf{v}$ is given parametrically by the formula:

$$\mathbf{r}(t)={\mathbf{r}}_0+t\mathbf{v}$$

Given two points ${\mathbf{r}}_0$ and ${\mathbf{r}}_1$, the line passing through these two points is given parametrically by:

$$\mathbf{r}(t)=(1-t){\mathbf{r}}_0+t{\mathbf{r}}_1$$

The line segment from ${\mathbf{r}}_0$ to ${\mathbf{r}}_1$ corresponds to $t\in [0,1]$. The point $\mathbf{r}(t)$ may be considered to be a weighted average of ${\mathbf{r}}_0$ and ${\mathbf{r}}_1$ with weights $1-t$ and $t$.

Planes
A plane in ${\mathbb{R}}^3$ can be determined as the points $(x,y,z)$ satisfying a single linear equation, such as:

$$ax+by+cz+d=0$$

Other formulations of this equation can display more meaning.

A plane is determined in vector form by the data of a single point ${\mathbf{r}}_0$ through which it passes together with a normal vector $\mathbf{n}$ that is perpendicular to the plane. The equation of the plane can be formed by simply stating that vectors in the plane from ${\mathbf{r}}_0$ are perpendicular to $\mathbf{n}$:

$$(\mathbf{r}-{\mathbf{r}}_0)\cdot \mathbf{n}=0$$

If we have components $\mathbf{r}=(x,y,z)$ and ${\mathbf{r}}_0=(x_0,y_0,z_0)$ and $\mathbf{n}=(a,b,c)$, then this formula can be rewritten as:

$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$$

This equation implies that the $a$, $b$, $c$ in the linear equation of the plane are just the components of its normal vector $\mathbf{n}$.

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Exercise

01

Vector plane from scalar plane

Find the vector equation of a plane with scalar equation $7x-4y+2z+10=0$.

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Exercise

02

Scalar plane from vector plane

Find the scalar equation of a plane passing through ${\mathbf{r}}_0=(3,1,0)$ and with normal vector $\mathbf{n}=(3,2,-5)$. (Start with the vector form.)

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Exercise

03

Line and plane intersection

Find the point at which the plane $3x-9y+2z-7=0$ and the line $\mathbf{r}(t)=(1,2,1)+t(-2,0,1)$ intersect.

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Exercise

04

Angle between planes

Find a formula for the angle formed between two planes. Use this formula to find the angle between the planes passing through $(0,0,0)$ with normal vectors ${\mathbf{n}}_1=(1,2,1)$ and ${\mathbf{n}}_2=(4,1,3)$, respectively.

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Distances

Distances

Point to point
The distance between $\mathbf{p}$ and $\mathbf{q}$ (more accurately, between the point identified by $\mathbf{p}$ and that identified by $\mathbf{q}$) is $|\mathbf{p}-\mathbf{q}|$.

Point to line
The vector $\mathbf{u}$ from a point $\mathbf{p}$ to a line $\mathbf{r}(t)={\mathbf{r}}_0-t\mathbf{v}$, namely to the point on the line closest to $\mathbf{p}$, is given by taking the vector from $\mathbf{p}$ to ${\mathbf{r}}_0$ and removing the component parallel to $\mathbf{v}$:

$$\mathbf{u}=({\mathbf{r}}_0-\mathbf{p})-(({\mathbf{r}}_0-\mathbf{p})\cdot {\mathbf{e}}_{\mathbf{v}}){\mathbf{e}}_{\mathbf{v}}$$

The distance from $\mathbf{p}$ to $\mathbf{r}$ is then $|\mathbf{u}|$.

Projection gives minimal distance

One can verify the formula for $\mathbf{u}$ by minimizing the squared distance from $\mathbf{p}$ to $\mathbf{r}(t)$ using the standard calculus technique.

Point to plane
The vector $\mathbf{u}$ from a point $\mathbf{p}$ to the plane through ${\mathbf{r}}_0$ with normal vector $\mathbf{n}$ (that is, to the nearest point on the plane) is given by projecting $\mathbf{p}-{\mathbf{r}}_0$ onto $\mathbf{n}$, that is:

$$\mathbf{u}=((\mathbf{p}-{\mathbf{r}}_0)\cdot {\mathbf{e}}_{\mathbf{n}}){\mathbf{e}}_{\mathbf{n}}$$

So the distance from $\mathbf{p}$ to this plane is:

$$|\mathbf{u}|=|(\mathbf{p}-{\mathbf{r}}_0)\cdot {\mathbf{e}}_{\mathbf{n}}|=(\mathbf{p}-{\mathbf{r}}_0)\cdot \frac{\mathbf{n}}{|\mathbf{n}|}$$

Plane to plane

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Exercise

05

Distance between parallel planes

Find the distance from the plane given by $15x+3y-3z-10=0$ to the parallel plane given by $5x+y-z-3=0$.

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Quaternions

Introduction

[Optional material.]
One can combine 3d vectors and 1d scalers into 4d quantities called quaternions, and written $\mathbb{H}$. Quaternions can be added and multiplied and divided.

Quaternions have real number coefficients for each of four basic units:

$$a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$$

The coefficient $a$ is called the scalar part and the coefficients $(b,c,d)$ are together called the vector part of the quaternion.

Multiplication rules
Multiplication for $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ follows the cross product rules (written formerly using ${\mathbf{e}}_{\mathbf{i}}$ etc. instead), with an additional rule for squares:

$${\mathbf{i}}^2={\mathbf{j}}^2={\mathbf{k}}^2=-1$$

Using these rules, one can multiply any quaternion sums.

Multiplication charts

$$\begin{array}{rl}& \\ \mathbf{i}\cdot \mathbf{j}\cdot \mathbf{k}& =-1\\ {\mathbf{i}}^2={\mathbf{j}}^2={\mathbf{k}}^2& =-1\end{array}$$

If two vector quaternions (with zero scalar part) are multiplied, then the resulting scalar part is the negative dot product, and the resulting vector part is the cross product, of the vector parts considered as vectors using their components.

Quaternions can also be divided:

$$\frac{1}{a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}}=\frac{a-b\mathbf{i}-c\mathbf{j}-d\mathbf{k}}{a^2+b^2+c^2+d^2}$$

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Surfaces: basic examples

Introduction

Quadric surfaces are given in ${\mathbb{R}}^3$ as the set of points satisfying a quadratic equation in three variables. The most general form of equation is:

$$A{x}^2+B{y}^2+C{z}^2+Dxy+Eyz+Fxz+Gx+Hy+Iz+J=0$$

After applying a rigid motion to ${\mathbb{R}}^3$, any such equation can be converted to one of these simpler equations:

$$\begin{array}{rlr}A{x}^2+B{y}^2+C{z}^2+J=0& & (1)\\ A{x}^2+B{y}^2+Iz=0& & (2)\end{array}$$

The surfaces satisfying Equation (1) include spheres, cones, cylinders, ellipsoids, and hyperboloids. The surfaces satisfying Equation (2) include elliptic paraboloids and hyperbolic paraboloids.

The surface type can be deduced from the coefficients $A,B,C,J$ or $A,B,I$.

One way to study equations of quadrics is to consider their traces, which are conic sections in two variables given by setting the third variable constant. A trace is the intersection of the quadric surface with the plane determined by the constant variable.

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Exercise

01

Translating away linear terms of a quadric

Show how you can eliminate the linear part $Gx+Hy+Iz$ from the general equation of a quadric by completing squares.

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Introduction (part 2)

Rotating away cross-variable terms of a quadric

Libretext link for an explanation of how to eliminate cross-variable terms by applying rotations.

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Spheres

Spheres

The general equation of a sphere in ${\mathbb{R}}^3$ with center at $(x_0,y_0,z_0)$ and radius $r$ is given by:

$$(x-x_0{)}^2+(y-y_0{)}^2+(z-z_0{)}^2=r^2$$

This equation is justified by the generalized Pythagorean theorem for three dimensions.

Spherical coordinates $(r,\theta ,\phi )$ parametrize a sphere at radius $r$. The angle $\theta$ is the standard angle in the $xy$-plane, and the angle $\phi$ measures declination from the north pole. These coordinates may be converted to standard coordinates:

$$(x,y,z)=(r\cos (\theta )\sin (\phi ),r\sin (\theta )\sin (\phi ),r\cos (\phi ))$$

Warning

Note that $(r,\theta ,\phi )$ give coordinates for points in ${\mathbb{R}}^3$, but the numbers $r,\theta ,\phi$ are not vector components. Do not perform vector operations componentwise on $(r,\theta ,\phi )$!

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Cylinders generalized

Cylinders generalized

A simple cylinder is a tube extended in the $z$-coordinate with a circle for each $xy$-cross-section. No constraint on the $z$-coordinate is needed, and the $xy$-coordinates satisfy the equation of a circle: $x^2+y^2=r^2$.

More generally, the set of points $(x,y,z)\in {\mathbb{R}}^3$ satisfying an equation in terms of only two variables gives a generalized cylinder, extended in the coordinate of the missing variable. For example, $x^2-z=0$ gives a parabolic cylinder extended in the $y$-coordinate:

For any cylindrical surface, it is possible to find new coordinates in which the surface is defined using only two of them.

Note that quadric surfaces written with two variables are generalized cylinders, but most generalized cylinders are not quadratic!

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Exercise

02

Describe surfaces

Describe the surface accurately using as few words as possible:

• (a) $y=5$
• (b) $y^2+z^2=9$
• (c) $x^2-y=0$
• (d) $4x^2+y^2=4$
• (e) $z-\sin (y)=0$

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Exercise

03

Describe surfaces

Describe the surface accurately using as few words as possible:

• (a) $z=2$
• (b) $x^2+y^2=25$
• (c) $y^2+z=6$
• (d) $y^2-z^2=16$
• (e) $z-e^y=0$

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Exercise

04

Surfaces: rectangular to spherical coordinates

Find an equation in spherical coordinates of the form $r=f(\theta ,\phi )$ for the following surfaces given in rectangular coordinates:

• (a) $x^2+y^2-3z^2=0$
• (b) $y=2$
• (c) $x^2+y^2=4y$
• (d) $x^2+y^2=36$

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Ruled surfaces

Ruled surfaces

Generalized cylinders are examples of ruled surfaces, which are surfaces created by the smooth motion through space of a straight line. For generalized cylinders, the lines must remain parallel, i.e. the motion must be perpendicular to the lines.

Other ruled surfaces that are not generalized cylinders:
Helicoid:

Mobius strip:

Hyperboloid:

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Quadrics: ellipsoids

Quadrics - ellipsoids

Ellipsoids are quadrics surfaces of type $(1)$ with $A,B,C>0$ and $J<0$.

Under these conditions, the quadric equation can be rewritten into a more expressive form:

$${\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2+{\left(\frac{z}{c}\right)}^2=1$$

As with an ellipse, the constants $a,b,c$ determine the three axis lengths of the ellipsoid. When $a=b=c$, the ellipsoid is actually a sphere.

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Quadrics: hyperboloids

Quadrics - hyperboloids

Hyperboloids are quadric surfaces of type $(1)$ with one of $A,B,C$ less than $0$, and the other two greater.

By exchanging the roles of $x,y,z$ and rewriting, we can assume the equation takes the form:

$${\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2-{\left(\frac{z}{c}\right)}^2=J$$

There are three kinds of hyperboloids: cones, hyperboloids of one sheet, and hyperboloids of two sheets. They are given by equations with $J=0$, $J>0$, and $J<0$, respectively.

One-sheeted paraboloids include a circle at $z=0$, which is possible because $J>0$. Shrinking $J$ shrinks the circle at $z=0$ until it is a point at $J=0$ (creating a cone, or more descriptively, a double-cone), and when $J$ goes negative, the surface does not exist at $z=0$.

${\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2-{\left(\frac{z}{c}\right)}^2=1$:

${\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2-{\left(\frac{z}{c}\right)}^2=0$:

${\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2-{\left(\frac{z}{c}\right)}^2=-1$:

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Quadrics: paraboloids

Quadrics - paraboloids

Paraboloids are quadric surfaces of type $(2)$. For this type, $I\ne 0$, so we may divide by $-I$ and thereby assume $I=-1$. We can rewrite the general form:

$$\pm {\left(\frac{x}{a}\right)}^2\pm {\left(\frac{y}{b}\right)}^2=z$$

The paraboloid is elliptic when both signs are plus (opening up) or minus (opening down). The paraboloid is hyperbolic when the signs are opposite.

Elliptic and hyperbolic paraboloids are often useful to illustrate optimization methods. A paraboloid has a horizontal tangent plane at the bottom (both signs positive), or the top (both negative), or the saddle point (opposite signs).

${\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2=z$:

${\left(\frac{x}{a}\right)}^2-{\left(\frac{y}{b}\right)}^2=z$:

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Exercise

05

Hyperboloid with circular traces

Suppose a hyperboloid is such that every intersection with a horizontal plane is a circle. Give a simplified form of the general equation for such a surface.

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