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# Lecture Notes

Below are the lecture notes from my course on multivariable calculus, as well as some handouts I give to students for some of the experiments in the introductory physics lab courses. Note that I update my notes each term, but this website may not link to the latest editions.

### Notes on MTH3015: Calculus III

$\newcommand{\vect}{{\mathbf{#1}} }$ $\renewcommand{\vec}{{\overrightarrow{#1}} }$ $\newcommand{\abs}{{\lvert#1\rvert} }$ $\newcommand{\norm}{{\lVert#1\rVert} }$ $\newcommand{\D}{{{\mathcal D}} }$ $\newcommand{\R}{{\mathbb{R}} }$ $\newcommand{\Rtwo}{{\mathbb{R}^2} }$ $\newcommand{\Rthree}{{\mathbb{R}^3} }$ $\newcommand{\V}{{\mathbb{V}} }$ $\newcommand{\nhat}{{\vect{\hat{n}}} }$ $\DeclareMathOperator{\sgn}{sgn}$ $\DeclareMathOperator{\dist}{dist}$

Week 2: Parametrization

Sujeet Akula

Planes

From the study of Euclidean geometry, we know that there are several ways of determining a plane. For example, three points that are not all collinear determine a plane. We also already know that the equation of a plane can always be written in the form \begin{equation} Ax + By + Cz + D = 0 , \end{equation} and that every equation in this form is the equation of some plane. In our vector formalism, we will define a plane by a single point that lies on the plane, and a vector based at that point that is perpendicular to the plane (a normal vector, defined below).

Definition Let $\Pi$ be a plane and let $P$ be a point in the plane. A normal vector to the plane $\Pi$, based at $P$ is a vector with a non-zero norm and is orthogonal to every vector which is based at $P$ and lies in $\Pi$.

Remark If a vector \vect{n} is a normal vector to $\Pi$ at $P$, then so is every non-zero scalar multiple of \vect{n}. Thus, one usually chooses the normal vector to be of unit norm. This is typically written as \nhat. Of course this is still not unique, since both $+\nhat$ and $-\nhat$ are unit normal vectors based at $P$ that are normal to the plane $\Pi$.

The Equation of a Plane

Now, we are ready to parametrize a plane using vectors. Suppose we are given a point $P = (x_0, y_0, z_0)$ that lies on the plane $\Pi$. Further, we are given the vector $\vect{n} = (n_1, n_2, n_3)_P$ is normal to $\Pi$. Then, to write the equation of the plane, we consider an arbitrary point $Q=(x,y,z)$, and constrain it to lie on the plane. First, consider the displacement vector \vect{v} from $P$ to $Q$. According to the definition of the normal vector given above, we must require that $\vect{n}\cdot\vect{v} = 0$. Explicitly, $\vect{v} = (x-x_0, y-y_0, z-z_0)_P$ so, $Q$ will line on $\Pi$, if \begin{align} \vect{n}\cdot(x-x_0, y-y_0, z-z_0)_P &= 0 , \\ n_1(x-x_0) + n_2(y-y_0) + n_3(z-z_0) &= 0 , \\ n_1 x + n_2 y + n_3 z - (n_1x_0 + n_2y_0 + n_3z_0) &= 0 . \\ \end{align}Thus, we have reproduced the equation of a plane by choosing\begin{align} A = n_1 \text{ , } B = n_2 \text{ , } C = n_3 \text{ , and } D=-(n_1&x_0 + n_2y_0 + n_3z_0) . \end{align}

Example Find the equation of the plane through $P=(x_0,y_0,z_0) = (1,-2,5)$ normal vector $\vect{n} = (n_1, n_2, n_3)_P =(-2,2,1)_P$.\\ First, we calculate the coefficients of the equation of the plane \begin{align} A = n_1 &= -2 \\ B = n_2 &= 2 \\ C = n_3 &= 1 \\ D = -(n_1x_0 + n_2y_0 + n_3z_0) &= -(-2)(1) - (2)(-2) - (1)(5) = 1 . \\ \end{align}So, the equation of the plane is\begin{align} -2x + 2y + z +1 &= 0 \end{align}

Example Find the equation of the plane containing the three points $P=(1,0,0)$, $Q=(0,1,0)$, and $R=(0,0,1)$. \\ First, we must convert this information into a single point in the plane and a vector normal to the plane. Since we are given three points in the plane, we can choose any of them to be our base point in the plane. Let's pick $P$. Let \vect{v} be the displacement vector from $P$ to $Q$, and let \vect{w} be the displacement vector from $P$ to $R$. Then, both \vect{v} and \vect{w} will be based at $P$ and will lie on the plane. Now, we can write a vector normal to the plane $\vect{n} = \vect{v}\times\vect{w}$, since the cross product of two vectors will be orthogonal to both vectors. \\ Since, \begin{align} \vect{v} = (-1,1,0)_P \text{ and } \vect{w} &= (-1,0,1)_P ,\\ \vect{n} = \vect{v} \times \vect{w} &= (1,1,1)_P . \end{align}So,\begin{align} A = 1 \text{ , } B = 1 \text{ , } &C = 1 , \end{align}and,\begin{align} D = -(1)(1) - (1)(0) - (1)(0) &= -1 . \end{align}So, the equation of the plane is\begin{align} x + y + z &= 1 . \end{align}

Distance from a Point to a Plane

From the equation of a plane, it is easy to calculate the distance from an arbitrary point in \Rthree to the plane.

Definition The distance from a point $P$ to a plane $\Pi$ is the distance from $P$ to the foot of the perpendicular from $P$ to $\Pi$.

Theorem The distance from a given point $Q=(x_1,y_1,z_1)$ to $\Pi$ if the equation for $\Pi$ is $Ax + By + Cz + D = 0$ is \begin{equation} \dist(Q,\Pi) = \frac{\abs{Ax_1 + By_1 + Cz_1 + D}}{\sqrt{A^2 + B^2 + C^2}} \end{equation}

Lines

We will use two pieces of information to determine a line: a single point on the line, and a vector representing the direction of the line from that point. However, we will not base this vector at that point, but instead, we will base this vector at the origin so that we can be add several of these vectors together. (Recall that two vectors must be based at the same point in order for them to be added together, or dotted together, etc.) To this end, we introduce the notion of a position vector.

Definition The position vector of a point $P$ is the displacement vector from the origin to $P$, and is based at the origin. The direction vector of a line is a vector based at the origin which is parallel to the line. To be clear, the position vector of any given point is based at the origin and the components of the vector are just the (Cartesian) co-ordinates of the point. As an example, the position vector of the point $P=(1,2,5)$ is the vector $\vect{v}=(1,2,5)_{(0,0,0)}$.

Vector Parametric Form

Suppose that the line $l$ has direction vector \begin{equation} \vect{d} = (d_1,d_2,d_3)_O , \end{equation} and that the fixed point $P=(x_0,y_0,z_0)$ is a point on $l$. The position vector of $P$ is \begin{equation} \vect{r_0} = (x_0,y_0,z_0)_O . \end{equation} Then, the line $l$ is described by a vector \vect{r} which serves as the position vector for every point on the line, according to the equation \begin{equation} \vect{r} = \vect{r_0} + t \vect{d} . \label{vecparform} \end{equation} This is the vector parametric form of the line $l$. Here, $t$ is a real number, and serves as a parameter: by varying $t$, \vect{r} becomes the position vector to every point on $l$. This starts with the choice $t=0$, which makes \vect{r} the position vector to $P$. For every other choice of $t$, a scalar multiple of the direction vector \vect{d} is added to the position vector of $P$, giving another point on $l$. ($t$ takes both positive and negative values.)

Scalar Parametric Form

The scalar parametric form of a line $l$ is simply the set of scalar equations for the components of the vector parametric from. That is, the components of Eq. \eqref{vecparform}. Letting $\vect{r}=(x,y,z)_O$, we then have: \begin{equation} \text{Scalar Parametric Form } \begin{cases} x = x_0 + td_1 \\ y = y_0 + td_2 \\ z = z_0 + td_3 \end{cases} \label{scalparform} \end{equation}

Symmetric Form

We can also write a non-parametric form a line $l$. This is by eliminating the parameter $t$ from the scalar parametric form presented in Eq. \eqref{scalparform}. This gives \begin{equation} \frac{x-x_0}{d_1} = \frac{y-y_0}{d_2} = \frac{z-z_0}{d_3} , \end{equation} the symmetric form of $l$. Note that it is not always possible to write the equations this way. If any component of the direction vector, \vect{d}, of $l$ is zero, then the symmetric form will not be a double equality.

Example Write the symmetric form of the line which has the vector parametric form $$\vect{r} = (2,0,-1)_O + t(0,4,7)_O .$$ We begin by writing this in the scalar parametric form, where we simply take each component of the equation and write it as separate equation. This gives \begin{align} x &= 2 \\ y &= 4t \\ z &= -1 + 7t . \end{align} We can eliminate the parameter $t$ from the latter two equations, to obtain the symmetric form: $$x=2, \frac{y}{4} = \frac{z+1}{7} .$$

The Intersection of Two Lines

Since most of your studies have involved lines in a plane, you would expect that two lines must intersect at exactly one point, if they are not parallel. This is not so in \Rthree; two lines which are not parallel may never intersect. Still, it is necessary to consider the strategy to compute the point of intersection of two lines. Suppose that line $l_1$ is parametrized by $\vect{r}(t) = \vect{r_0} + t\vect{d}$, and that $l_2$ is parametrized by $\vect{R}(u) = \vect{R_0} + u\vect{D}$. If there exists values of $t$ and $u$ that satisfy \begin{equation} \vect{v} = \vect{r}(t) = \vect{R}(u) , \label{lineint} \end{equation} then the $l_1$ and $l_2$ intersect, and \vect{v} is the position vector of the point at which they intersect. Note that Eq. \eqref{lineint} is a set of three scalar equations, with only two unknowns ($t$, and $u$), thus it is not necessary for solutions to exist.

Example Let $l_1$ and $l_2$ be lines in \Rthree with vector parametrizations as given below. Find the point at which they intersect. \begin{align} l_1: \vect{r}(t) &= (1,2,3)_O + t(-1,7,4)_O \\ l_2: \vect{R}(u) &= (-6,3,-5)_O + u(2,10,10)_O \end{align} We begin by substituting the parametrization of the two lines into Eq. \eqref{lineint}. \begin{align} (1,2,3)_O + t(-1,7,4)_O &= (-6,3,-5)_O + u(2,10,10)_O \\ (7-t-2u, -1+7t&-10u,8+4t-10u)_O = 0 \\ \end{align}We can separate the components into three equations:\begin{align} 7 -t -2u &= 0 \\ -1 + 7t - 10u &= 0 \\ 8 + 4t - 10u &= 0 . \end{align}Let's temporarily ignore any one of the equations--say, the third equation. This will allow us to determine values for $t$ and $u$, which may or may not be consistent with the third equation. Only if the values are consistent with the third equation may we say that the lines intersect. So, for the moment we wish to solve the simple simultaneous linear system \begin{align} t + 2u = 7 \text{ and } & 7t-10u=1 . \end{align}Solving these equations yields that \begin{align} t=3 \text{ and } & u=2 . \end{align}We not test if these choices of $t$ and $u$ also satisfy the third equation, $8 + 4t - 10u = 0$, and one can easily see that they do! Thus, indeed $l_1$ and $l_2$ intersect, and the position vector of the point at which they intersect is\begin{align} \vect{v} = \vect{r}(3) &= \vect{R}(2) = (-2,23,15)_O \end{align}So, the lines intersect at the point $(-2,23,15)$.\begin{align} \end{align}

Distance Between a Point and a Line

We present here the precise definition of what is meant by the distance from a given point to a line, as well as a formula for its computation.

Definition The distance from a given point $P$ to a line $l$ is the shortest distance from $P$ to any point on $l$. Formally, this is expressed as \begin{equation} \dist(P,l) := \min\limits_{Q}\{\dist(P,Q) : Q \text{ is on } l\} . \end{equation}

Remark The distance from a point to a line geometrically is the length of the segment that extends from the point and is perpendicularly incident on the line.

Theorem Let $P$ be a point in \Rthree with position vector $\vect{r}=(x,y,z)_O$, and let $l$ be a line in \Rthree that is written in vector parametric form as \begin{equation} \vect{R}(t) = \vect{R_0} + t\vect{D} . \end{equation} Then, the distance from $P$ to $l$ is given by \begin{equation} \dist(P,l) = \frac{\norm{(\vect{r} - \vect{R_0})\times\vect{D}}}{\norm{\vect{D}}} . \end{equation}

Distance Between Two Lines

Here we define carefully the distance between two lines, and use the result of the previous section to calculate the distance between two lines.

Definition The distance between two lines $l_1$ and $l_2$ is the shortest distance between any two points from each line. This is written formally as \begin{equation} \dist(l_1,l_2) := \min\limits_{P,Q}\{\dist(P,Q) : P \text{ is on } l_1 \text{ and } Q \text{ is on } l_2\} . \end{equation}

Remark The distance between two lines is geometrically the length of the segment that has an endpoint on each line, and is perpendicular to both lines.

Theorem Let $l_1$ and $l_2$ be two non-parallel lines, with equations \begin{align} l_1: \vect{r}(t) &= \vect{r_0} + t\vect{d} \\ l_2: \vect{R}(t) &= \vect{R_0} + t\vect{D} . \end{align}Then, the distance from $l_1$ to $l_2$ is\begin{align} \dist(l_1,l_2) = \frac{\abs{(\vect{r_0}-\vect{R_0})\cdot(\vect{d}\times\vect{D})}}{\norm{\vect{d}\times\vect{D}}} . \end{align} Note that here, we require that the lines not be parallel, since we divide by $\norm{\vect{d}\times\vect{D}}$, the norm of the cross product of the two direction vectors, which would be zero if the two lines were parallel (since that means that \vect{d} and \vect{D} would also be parallel. In the case that they are parallel, the problem is actually much easier: simply choose any point on $l_1$, then calculate the distance from that point to $l_2$, as in the last section. You will get the same answer regardless of which point you originally chose on $l_1$. (Can you explain why?)