This unit covers the following ideas. In preparation for the quiz and exam, make sure you have a lesson plan containing examples that explain and illustrate the following concepts.

- Define, draw, and explain what a vector is in 2 and 3 dimensions.
- Add, subtract, multiply (scalar, dot product, cross product) vectors. Be able to illustrate each operation geometrically.
- Use vector products to find angles, length, area, projections, and work.
- Use vectors to give equations of lines and planes, and be able to draw lines and planes in 3D.

See Larson Sections 11.1 and 11.2
for more information about vector operations

Learning to work with vectors will be key tool we need for our work in high dimensions. Let's start with some problems related to finding distance in 3D, drawing in 3D, and then we'll be ready to work with vectors.

To find the distance between two points $(x_1,y_1)$ and
$(x_2,y_2)$ in the plane, we create a triangle connecting the
two points. The base of the triangle has length $\Delta
x=(x_2-x_1)$ and the vertical side has length $\Delta
y=(y_2-y_1)$. The Pythagorean theorem gives us the distance
between the two points as $\sqrt{\Delta x^2+\Delta
y^2}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Show that the distance
between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in
3-dimensions is $\sqrt{\Delta x^2+\Delta y^2+\Delta
z^2}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$.

See 12.1:41-58.
Find the distance between the two points $P=(2,3,-4)$ and
$Q=(0,-1,1)$. Then find an equation of the sphere passing
though point $Q$ whose center is at $P$. [A sphere is the set of
all points equidistant from a center point; in this problem, the
sphere is the set of all points that are the same distance from
$P$ as point $Q$ is.]

See 12.1:1-40.
Be careful to label your axes $x$, $y$, or
$z$.
For each of the following, construct a rough sketch of
the set of points in space (3D) satisfying:

- $2\leq z\leq 5$
- $x=2$ and $y=3$
- $x^2+y^2+z^2=25$

In linear algebra, you'll study more
general vectors.
In our class, a vector is a magnitude in a certain direction. If $P$ and $Q$
are points, then the vector $\vec{PQ}$ is the directed line
segment from $P$ to $Q$. This definition holds in 1D, 2D, 3D,
and beyond. If $V=(v_1,v_2,v_3)$ is a point in space, then to
talk about the vector $\vec v$ from the origin $O$ to $V$ we'll
use any of the following notations: $$\vec v =
\vec{OV}=\left<v_1,v_2,v_3\right> = v_1{\bf i}+v_2{\bf
j}+v_3{\bf k} = (v_1,v_2,v_3) = \begin{pmatrix}
v_1\\ v_2\\ v_3\end{pmatrix}.$$
The entries of the vector are called the $x$, $y$,
and $z$ components of the vector.

Note that $(v_1,v_2,v_3)$ could refer to either the point $V$ or the vector $\vec v$. The context of the problem we are working on will help us know if we are dealing with a point or a vector.

Let $\mathbb{R}$ represent the set real numbers. Real numbers
are actually 1D vectors. Let $\mathbb{R}^2$ represent the set
of vectors $(x_1,x_2)$ in the plane. Let $\mathbb{R}^3$
represent the set of vectors $(x_1,x_2,x_3)$ in space. There's
no reason to stop at 3, so let $\mathbb{R}^n$ represent the set
of vectors $(x_1,x_2,\ldots,x_n)$ in $n$ dimensions.

In first semester calculus and before, most of our work dealt with problem in $\mathbb{R}$ and $\mathbb{R}^2$. Most of our work now will involve problems in $\mathbb{R}^2$ and $\mathbb{R}^3$. We've got to learn to visualize in $\mathbb{R}^3$.

The magnitude, or length, or norm of a vector $\vec v =
\left<v_1,v_2,v_3\right>$ is $|\vec v| =
\sqrt{v_1^2+v_2^2+v_3^2}$. It is just the distance from the
origin to the point $(v_1,v_2,v_3)$. A unit vector is a vector
whose length is one unit. The standard unit vectors are ${\bf
i}=\left<1,0,0\right>$, ${\bf
j}=\left<0,1,0\right>$, ${\bf
k}=\left<0,0,1\right>$.

Note that in one dimension, the length of the vector $\left<-2\right>$ is simply $|-2|=\sqrt{(-2)^2}=2$, the distance to 0. Our use of the absolute value symbols is appropriate, as it generalizes the concept of absolute value (distance to zero) to all dimensions.

Suppose $\vec x=\left<x_1,x_2,x_3\right>$ and $\vec
y=\left<y_1,y_2,y_3\right>$ are two vectors, and
$c$ is a real number. We define vector addition and scalar
multiplication as follows:

- Vector addition: $\vec x+\vec y = (x_1+y_1,x_2+y_2,x_3+y_3)$ (add component-wise).
- Scalar multiplication: $c\vec x = (cx_1,cx_2,cx_3)$ (multiply each component by the scalar).

See 12.2:23-24.
Consider the vectors $\vec u=(1,2)$ and $\vec
v=\left<3,1\right>$. Draw $\vec u$, $\vec v$, $\vec
u+\vec v$, and $\vec u-\vec v$ with their tail placed at the
origin. Then draw $\vec v$ with its tail at the head of $\vec
u$.

See 11.1: 3,4.

Consider the vector $\vec v=(3,-1)$. Draw $\vec v$,
$-\vec v$, and $3\vec v$. Suppose a donkey travels along the
path given by $(x,y)=\vec v t = (3t,-t)$, where $t$ represents
time. Draw the path followed by the donkey. Where is the donkey
at time $t=0,1,2$? Put markers on your graph to show the
donkey's location. Then determine how fast the donkey is
traveling.
In the previous problem you encountered $(x,y)=(3t,-t)$. This is an example of a function where the input is $t$ and the output is a vector $(x,y)$. For each input $t$, you get a single vector output $(x,y)$. Such a function is called a parametrization of the donkey's path. Because the output is a vector, we call the function a vector-valued function. Often, we'll use the variable $\vec r$ to represent the radial vector $(x,y)$, or $(x,y,z)$ in 3D. So we could rewrite the position of the donkey as $\vec r(t)=(3,-1)t$. We use $\vec r$ instead of $r$ to remind us that the output is a vector.

See 12.2: 1.

Suppose a horse races down a path given by the
vector-valued function $\vec r(t) = (1,2)t+(3,4)$. (Remember
this is the same as writing $(x,y) = (1,2)t+(3,4)$ or similarly
$(x,y)=(1t+3,2t+4)$.)
- Where is the horse at times $t=0,1,2$? Put markers on your graph to show the horse's location.
- Draw the path followed by the horse.
- Give a unit vector that tells the horse's direction.
- Then determine how fast the horse is traveling.

See 12.2: 9,17,25,33 and surrounding.

Consider the two points $P=(1,2,3)$ and $Q=(2,-1,0)$.
- Write the vector $\vec {PQ}$ in component form $(a,b,c)$.
- Find the length of vector $\vec {PQ}$.
- Then find a unit vector in the same direction as $\vec{PQ}$.
- Finally, find a vector of length 7 units that points in the same direction as $\vec{PQ}$.

See 12.5: 1-12.

A raccoon is sitting at point $P=(0,2,3)$. It starts to
climb in the direction $\vec v=\left<1,-1,2\right>$.
Write a vector equation $\vec r(t)=(?,?,?)t+(?,?,?)$ for the line that
passes through the point $P$ and is parallel to $\vec v$.
[Hint, study the horse race problem, and base your work off
of what you saw there. It's almost identical.] Then generalize
your work to give an equation of the line that passes through
the point $P=(x_1,y_1,z_1)$ and is parallel to the vector $\vec
v=(v_1,v_2,v_3)$.
Make sure you ask me in class to show you how to connect the equation developed above to what you have been doing since middle school. If you can remember $y=mx+b$, then you can quickly remember the equation of a line. If I don't show you in class, make sure you ask me (or feel free to come by early and ask before class).

See 12.5: 13-20.

Let $P=(3,1)$ and
$Q=(-1,4)$.
- Write a vector equation $\vec r(t)=(x,y)$ for (i.e, give a parametrization of) the line that passes through $P$ and $Q$, with $\vec r(0)=P$ and $\vec r(1)=Q$.
- Write a vector equation for the line that passes through $P$ and $Q$, with $\vec r(0)=P$ but whose speed is twice the speed of the first line.
- Write a vector equation for the line that passes through $P$ and $Q$, with $\vec r(0)=P$ but whose speed is one unit per second.

See Larson section 11.3 for more
information about the dot product

Now that we've learned how to add and subtract vectors, stretch them by scalars, and use them to find lines, it's time to introduce a way of multiplying vectors called the dot product. We'll use the dot product to help us find find angles. First, we need to recall the law of cosines.

The Law of
Cosines

Consider a triangle with side lengths $a$, $b$, and $c$. Let $\theta$ be the angle between the sides of length $a$ and $b$. Then the law of cosines states that $$c^2=a^2+b^2-2ab\cos\theta.$$ If $\theta=90^\circ$, then $\cos\theta=0$ and this reduces to the Pythagorean theorem.

The Dot Product
If $\vec u = (u_1,u_2,u_3)$ and $\vec v= (v_1,v_2,v_3)$ are
vectors in $\mathbb{R}^3$, then we define the dot product of
these two vectors to be $$\vec u\cdot \vec v = u_1 v_1+ u_2
v_2+ u_3 v_3.$$ A similar definition holds for vectors in
$\mathbb{R}^n$, where $\vec u\cdot \vec v = u_1 v_1+ u_2
v_2+\cdots+ u_n v_n.$ You just multiply corresponding
components together and then add. It is the same process used
in matrix multiplication.

Page 693 has the solution if you are struggling.

If $\vec u = (u_1,u_2,u_3)$ and $\vec v= (v_1,v_2,v_3)$
are vectors in $\mathbb{R}^3$ (which is often written $\vec
u,\vec v\in\mathbb{R}^3$), then show that $$|\vec u-\vec v|^2 =
|\vec u|^2-2\vec u\cdot \vec v +|\vec v|^2.$$ [Hint: calculate
both sides using the equation for magnitude and the definition of
the dot product and show that they are equal.]
See 12.3: 9-12.

Sketch in $\mathbb{R}^2$ the vectors
$\left<1,2\right>$ and $\left<3,5\right>$. Use the
law of cosines to find the angle between the vectors.
See page 693.

Let $\vec u,\vec v\in\mathbb{R}^3$. Let $\theta$ be the
angle between $\vec u$ and $\vec v$.
- Use the law of cosines to explain why $|\vec u-\vec v|^2=|\vec u|^2+|\vec v|^2-2|\vec u||\vec v|\cos\theta$.
- Use the previous part together with the problem above to explain why $$\vec u\cdot \vec v=|\vec u||\vec v|\cos\theta.$$

See 12.3: 9-12.

Sketch in $\mathbb{R}^3$ the vectors
$\left<1,2,3\right>$ and $\left<-2,1,0\right>$. Use
the law of cosines to find the angle between the vectors. Then
use the formula $\vec u\cdot \vec v=|\vec u||\vec v|\cos\theta$
to find the angle between them. Which was easier?
We say that the vectors $\vec u$ and $\vec v$ are orthogonal if
$\vec u\cdot \vec v=0$.

Find two vectors orthogonal to $(1,2)$. Then find 4 vectors orthogonal to $(3,2,1)$.

Press "Evaluate" to check your answers for the second part.
Mark each statement true or false.
Explain. You can assume that $\vec u,\vec v,\vec
w\in\mathbb{R}^3$ and that $c\in\mathbb{R}$.

- $\vec u\cdot \vec v=\vec v\cdot \vec u$.
- $\vec u\cdot (\vec v\cdot \vec w)=(\vec u\cdot\vec v)\cdot\vec w$.
- $c(\vec u\cdot \vec v)=(c\vec u)\cdot \vec v=\vec u\cdot (c\vec v)$.
- $\vec u+(\vec v\cdot \vec w)=(\vec u+\vec v)\cdot(\vec u+\vec w)$.
- $\vec u\cdot (\vec v+ \vec w)=(\vec u\cdot \vec v)+(\vec u\cdot\vec w)$.
- $\vec u\cdot \vec u= |\vec u|^2$.

See page 694.

Show that if two nonzero vectors $\vec u$ and $\vec v$
are orthogonal, then the angle between them is 90$^\circ$. Then
show that if the angle between them is 90$^\circ$, then the
vectors are orthogonal.
The dot product provides a really easy way to find when two vectors meet at a right angle. The dot product is precisely zero when this happens.

See the last part of Larson
section 11.3 for more examples

Suppose a heavy box needs to be lowered down a ramp. The box exerts a downward force of 200 Newtons, which we will write in vector notation as $\vec F=\left<0,-200\right>$. The ramp was placed so that the box needs to be moved right 6 m, and down 3 m, so we need to get from the origin $(0,0)$ to the point $(6,-3)$. This displacement can be written as $\vec d=\left<6,-3\right>$. The force $F$ acts straight down, which means the ramp takes some of the force. Our goal is to find out how much of the 200N the ramp takes, and how much force must be applied to prevent the box from sliding down the ramp (neglecting friction). We are going to break the force $\vec F$ into two components, one component in the direction of $\vec d$, and another component orthogonal to $\vec d$.

Read the preceding paragraph. We want to
write $\vec F$ as the sum of two vectors $\vec F = \vec w+\vec
n$, where $\vec w$ is parallel to $\vec d$ and $\vec n$ is
orthogonal to $\vec d$. Since $\vec w$ is parallel to $\vec d$,
we can write $\vec w=c\vec d$ for some unknown scalar $c$. This
means that $\vec F=c\vec d+\vec n$. Use the fact that $\vec n$
is orthogonal to $\vec d$ to solve for the unknown scalar $c$.
[Hint: dot each side of $\vec F=c\vec d+\vec n$ with $\vec d$.
This should turn the vectors into numbers, so you can use
division.]

The solution to the previous problem gives us the definition of a projection.

The projection of $\vec F$ onto $\vec d$, written $\proj_{\vec
d}\vec F$, is defined as $$\proj_{\vec d}\vec F =
\left(\frac{\vec F\cdot \vec d}{\vec d\cdot \vec d}\right)\vec
d.$$

See 12.3:1-8 (part d).

Let $\vec u=(-1,2)$ and $\vec v=(3,4)$. Draw $\vec u,$
$\vec v$, and $\proj_{\vec v}\vec u$. Then draw a line segment
from the head of $\vec u$ to the head of the projection. Now
let $\vec u=(-2,0)$ and keep $\vec v=(3,4)$. Draw $\vec u,$
$\vec v$, and $\proj_{\vec v}\vec u$. Then draw a line segment
from the head of $\vec u$ to the head of the projection.
One final application of projections pertains to the concept of work. Work is the transfer of energy. If a force $F$ acts through a displacement $d$, then the most basic definition of work is $W=Fd$, the product of the force and the displacement. This basic definition has a few assumptions.

- The force $F$ must act in the same direction as the displacement.
- The force $F$ must be constant throughout the entire displacement.
- The displacement must be in a straight line.

Before the semester ends, we will be able to remove all 3 of these assumptions. The next problem will show you how dot products help us remove the first assumption. Recall the set up to this problem. We want to lower a box down a ramp (which we will assume is frictionless). Gravity exerts a force of $\vec F=\left<0,-200\right>$ N. If we apply no other forces to this system, then gravity will do work on the box through a displacement of $\left<6,-3\right>$ m. The work done by gravity will transfer the potential energy of the box into kinetic energy (remember that work is a transfer of energy). How much energy is transferred?

See 12.3: 24, 41-44.

Find the amount of work done
by the force $\vec F=\left<0,-200\right>$ through the
displacement $\vec d=\left<6,-3\right>$. Find this by
doing the following:
- Find the projection of $\vec F$ onto $\vec d$. This tells you how much force acts in the direction of the displacement. Find the magnitude of this projection.
- Since work equals $W=Fd$, multiply your answer above by $|\vec {d}|$.
- Now compute $\vec F\cdot \vec d$. You have just shown that $W=\vec F\cdot \vec d$ when $\vec F$ and $\vec d$ are not in the same direction.

See Larson section 11.4 for more
information about the cross product

The dot product gave us a way of multiplying two vectors together, but the result was a number, not a vectors. We now define the cross product, which will allow us to multiply two vectors together to give us another vector. We were able to define the dot product in all dimensions. The cross product is only defined in $\mathbb{R}^3$.

The Cross Product
This definition is not really a definition. It is actually a
theorem. If you use the formula given as the definition, then
you would need to prove the three facts. We have the tools to
give a complete proof of (1) and (3), but we would need a
course in linear algebra to prove (2). It shouldn't be too
much of a surprise that the cross product is related to area,
since it is defined in terms of determinants
The cross product of two vectors $\vec u =
\left<u_1,u_2,u_3\right>$ and $\vec v =
\left<v_1,v_2,v_3\right>$ is a new vector $\vec u\times
\vec v$. This new vector is (1) orthogonal to both $\vec u$ and
$\vec v$, (2) has a length equal to the area of the
parallelogram whose sides are these two vectors, and (3) points
in the direction your thumb points as you curl the base of your
right hand from $\vec u$ to $\vec v$. The formula for the cross
product is $$\vec u\times \vec v =
\left<u_2v_3-u_3v_2,-(u_1v_3-u_3v_1),u_1v_2-u_2v_1\right>
= \det\begin{bmatrix}
\vec i & \vec j&\vec k\\ u_1&u_2&u_3\\
v_1&v_2&v_3\\
\end{bmatrix}.$$

Practice computing the cross product by clicking "Evaluate".
See 12.4: 1-8.
Let $\vec u=(1,-2,3)$ and $\vec v=(2,0,-1)$.

- Compute $\vec u\times \vec v$ and $\vec v\times \vec u$. How are they related?
- Compute $\vec u \cdot (\vec u\times \vec v)$ and $\vec v \cdot (\vec u\times \vec v)$. Why did you get the answer you got?
- Compute $\vec u \times (2\vec u)$. Why did you get the answer you got?
- Compute $|\vec u \times \vec v|$. Compute the area of the parallelogram formed by $\vec u$ and $\vec v$ using trigonometry and $|\vec u|$, $|\vec v|$, and the angle $\theta$ between the two vectors. Compare your answer with $|\vec u \times \vec v|$.

See 12.4: 15-18.

Let $P=(2,0,0)$, $Q=(0,3,0)$, and $R=(0,0,4)$. Find a
vector that is orthogonal to both $\vec {PQ}$ and $\vec {PR}$.
Then find the area of the triangle $PQR$. Construct a 3D graph
of this triangle.
See 12.3: 9-14.

Consider the vectors ${\ii}=(1,0,0)$, ${2\jj}=(0,2,0)$,
and ${3\kk}=(0,0,3)$.
- Compute $\ii\times {2\jj}$ and ${2\jj}\times {\ii}$.
- Compute ${\ii}\times {3\kk}$ and ${3\kk}\times {\ii}$.
- Compute ${2\jj}\times {3\kk}$ and ${3\kk}\times {2\jj}$.

We will now combine the dot product with the cross product to develop an equation of a plane in 3D. Before doing so, let's look at what information we need to obtain a line in 2D, and a plane in 3D. To obtain a line in 2D, one way is to have 2 points. The next problem introduces the new idea by showing you how to find an equation of a line in 2D.

Suppose the point $P=(1,2)$ lies on line $L$. Suppose that the
angle between the line and the vector $\vec
n=\left<3,4\right>$ is 90$^\circ$ (whenever this happens we
say the vector $\vec n$ is normal to the line). Let $Q=(x,y)$ be
another point on the line $L$. Use the fact that $\vec n$ is
orthogonal to $\vec {PQ}$ to obtain an equation of the line $L$.

See page 709.

See Larson 11.5.

Let $P=(a,b,c)$ be a point on a plane in 3D. Let $\vec
n=(A,B,C)$ be a normal vector to the plane (so the angle
between the plane and and $\vec n$ is 90$^\circ$). Let
$Q=(x,y,z)$ be another point on the plane. Show that an
equation of the plane through point $P$ with normal vector
$\vec n$ is $$A(x-a)+B(y-b)+C(z-c)=0.$$
See 12.5: 21-28.

See Larson 11.5:47--58 for
more practice.

Consider the three points $P=(1,0,0), Q=(2,0,-1),
R=(0,1,3)$. Find an equation of the plane which passes through
these three points. [Hint: first find a normal vector to the
plane.]
See 12.5: 57-60.

See Larson 11.5:91--92 for more practice.

Consider the two planes $x+2y+3z=4$ and $2x-y+z=0$.
These planes meet in a line. Find a vector that is parallel to
this line. Then find a vector equation of the line.
See Larson 11.5:47--58 for more
practice.

Find an equation of the plane containing the lines
$\vec r_1(t)=(1,3,0)t+(1,0,2)$ and $\vec
r_2(t)=(2,0,-1)t+(2,3,2)$.
Consider the points $P=(2,-1,0)$, $Q=(0,2,3)$, and
$R=(-1,2,-4)$.

- Give an equation $\vec r(t)=(?,?,?)t+(?,?,?)$ of the line through $P$ and $Q$.
- Give an equation of the line through $P$ and $R$.
- Give an equation of the plane through $P$, $Q$, and $R$.

Consider the points $P=(2,4,5)$, $Q=(1,5,7)$, and $R=(-1,6,8)$.

- What is the area of the triangle $PQR$.
- Give a normal vector to the plane through these three points.
- What is the distance from the point $A=(1,2,3)$ to the plane $PQR$. [Hint: Compute the projection of $\vec {PA}$ onto $\vec n$. How long is it?]

Show that the distance from a point $Q$ to a line (with
direction vector $\vec v$ passing through $P$) is
$|\overrightarrow{PQ}-\proj_{\vec v}\overrightarrow {PQ}|$.
Draw a diagram illustrating your reasoning.

Show that the distance from a point $Q$ to a plane (with normal
vector $\vec n$ and a point $P$) is given by $|\proj_{\vec
n}\overrightarrow {PQ}|$. Draw a diagram illustrating your
reasoning.

Show that the distance from a line (with direction vector $\vec
v_1$ passing through $P_1$) to a line (with direction vector
$\vec v_2$ passing through $P_2$) is $|\proj_{\vec
v_1\times\vec v_2}\overrightarrow {P_1P_2}|$. Draw a diagram
illustrating your reasoning.

If the statement is true, explain why. If it is false, give a
counterexample.

- If $\vec u \neq 0$ and $\vec u \times \vec v = \vec u \times \vec w$, then is $\vec v = \vec w$ always?
- If $\vec u \neq 0$, $\vec u\cdot \vec v = \vec u \cdot \vec w$, and $\vec u \times \vec v = \vec u \times \vec w$, then is $\vec v = \vec w$ always?