## Introducing Vectors for Geometry

We will discuss vectors, which for our purposes right now are thought of as a directed line segment. Its length corresponds to its magnitude, and it encodes information regarding orientation. So we can represent, e.g., “The wind blowing 30 miles per hour due East” as a vector pointing East whose magnitude is 30.

We use vectors to discuss angles, and projections.
1. The 3 Dimensional Coordinate System. The ${x}$, ${y}$, ${z}$ axes are perpendicular to each other. We will doodle three dimensions as follows:

The formula for distance between two points ${(x_{0}, y_{0}, z_{0})}$ and ${(x_{1},y_{1},z_{1})}$ would be
(1)$\displaystyle s = \sqrt{(x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}+(z_{1}-z_{0})^{2}}.$
A sphere with its center at ${(a,b,c)}$ would be the points which are a distance ${r}$ away from the center. So we would have
(2)$\displaystyle S^{2} = \{\,(x,y,z) : \sqrt{(x-a)^{2}+(y-b)^{2}+(z-c)^{2}}=r\,\}$
Usually the formula is given as
(3)$\displaystyle (x-a)^{2}+(y-b)^{2}+(z-c)^{2}=r^{2}$
describes a sphere.

2. Vectors. A “Vector” is a directed line segment.

We know a directed line segment has both length (magnitude) and direction, so any two directed line segments with the same length and direction represent the same vector.

Vectors are “transportable” in the sense that we may translate their base point. We will represent the length of a vector ${\vec{u}}$ as ${\|\vec{u}\|}$ or ${|\vec{u}|}$.

The notation for a vector would be ${\langle x,y\rangle}$ (in two dimensions) or ${\langle x,y,z\rangle}$ (in three dimensions). The vector from ${(0,0,0)}$ to ${(x,y,z)}$ is given as ${\langle x,y,z\rangle}$. For ${P=(-1,4,7)}$ and ${Q=(2,5,3)}$, then the vector from ${P}$ to ${Q}$ is denoted ${\overrightarrow{PQ}}$.

Subtraction would amount to ${\vec{A}-\vec{B}=\vec{A}+(-\vec{B})}$, and graphically this is:

Algebraically, if
(4)$\displaystyle \vec{A}=\langle x_{a},y_{a},z_{a}\rangle,\quad \vec{B}=\langle x_{b},y_{b},z_{b}\rangle$
then
(5)\displaystyle \begin{aligned} \vec{A}+\vec{B} &= \langle x_{a}+x_{b}, y_{a}+y_{b}, z_{a}+z_{b}\rangle\\ \vec{A}-\vec{B} &= \langle x_{a}-x_{b}, y_{a}-y_{b}, z_{a}-z_{b}\rangle \end{aligned}
This describes vector addition and subtraction on the components.

Note in two-dimensional space, the vectors
(6)$\displaystyle \widehat{\textbf{\i}}=\langle 1,0\rangle \quad\mbox{and}\quad \widehat{\textbf{\j}}=\langle 0,1\rangle$
are unit vectors (i.e., vectors whose length is ${1}$). They are also called basis vectors since any other vector ${\vec{v}}$ in two-dimensions can be written as
(7)\displaystyle \begin{aligned} \vec{v}&=\langle v_{x}, v_{y}\rangle\\ &=\langle v_{x},0\rangle + \langle0,v_{y}\rangle\\ &=v_{x}\langle1,0\rangle + v_{y}\langle0,1\rangle\\ &=v_{x}\widehat{\textbf{\i}}+v_{y}\widehat{\textbf{\j}} \end{aligned}
where ${v_{x}}$ and ${v_{y}}$ are called the vector’s components. Note that the components of the vector depends on a choice of coordinates (i.e., a choice of basis vectors).

The last notion we will discuss: given any vector ${\vec{v}}$ which is nonzero, then we can construct the unit vector
(8)$\displaystyle \widehat{v} = \frac{\vec{v}}{\|\vec{v}\|}$
which has magnitude 1. We use hats to indicate unit vectors, and arrows for arbitrary vectors.

3. Caution: Everything stated about vectors is a half-truth. Really, these are “tangent vectors” which has a base point and a vector part (i.e., where we stick the line segment, and the directed line segment itself). We can only add/subtract two tangent vectors if they have the same base point. But since we work in Euclidean space (which is flat), we can transport vectors without a problem. This is a very special situation!

Since this is never mentioned, often students become confused when they finish vector calculus and begin studying linear algebra. Linear algebra fixes a base point, and considers the collection of all vectors sharing the same base point. This is the honest definition of a vector.

Remark 1. We will also use the phrase “three-space” instead of “three-dimensional space”, and “two-space” replacing “two-dimensional space”. In general ${n}$-space is ${n}$-dimensional Euclidean space.

### Dot Products

4. Definition. Given two vectors
(9)$\displaystyle \vec{u}=u_{1}\widehat{\textbf{\i}}+u_{2}\widehat{\textbf{\j}}+u_{3}\widehat{\textbf{k}},\quad\mbox{and}\quad \vec{v}=v_{1}\widehat{\textbf{\i}}+v_{2}\widehat{\textbf{\j}}+v_{3}\widehat{\textbf{k}}$
their “Dot Product” is the number
(10)\displaystyle \begin{aligned} \vec{u}\cdot\vec{v} &= \sum_{i} u_{i}v_{i}\\ &= u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3} \end{aligned}
Let ${\vec{u}}$ and ${\vec{v}}$ be given vectors in three-space.

5. Angles. How do we find the angle ${\theta}$ between the two vectors?

We find that
(11)$\displaystyle \theta = \arccos\left(\frac{\|\vec{v}\|}{\|\vec{u}\|}\right)$
How is this? Well, we should recall the law of cosines
(12)$\displaystyle \|\vec{w}\|^{2}=\|\vec{u}\|^{2}+\|\vec{v}\|^{2}-2\|\vec{u}\|\cdot\|\vec{v}\|\cos(\theta)$
which can be written as
(13)\displaystyle \begin{aligned} \vec{w}\cdot\vec{w} &= (u_{1}-v_{1})^{2}+(u_{2}-v_{2})^{2}+(u_{3}-v_{3})^{2}\\ &= \|\vec{u}\|^{2}-2(\vec{u}\cdot\vec{v})+\|\vec{v}\|^{2} \end{aligned}
Setting equals to equals gives us
(14)$\displaystyle -2(\vec{u}\cdot\vec{v})=-2\|\vec{u}\|\cdot\|\vec{v}\|\cos(\theta)$
and thus
(15)$\displaystyle \frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\cdot\|\vec{v}\|}=\cos(\theta)$
Taking the arc cosine of both sides yields
(16)$\displaystyle \arccos\left(\frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\cdot\|\vec{v}\|}\right)=\theta.$
A useful formula worth remembering
(17)$\displaystyle (\vec{u}\cdot\vec{v})=\|\vec{u}\|\cdot\|\vec{v}\|\cos(\theta)$

Example 1. Let ${\vec{u}=\langle3,-1,4\rangle}$ and ${\vec{v}=\langle1,5,-2\rangle}$. What’s the angle between them?

Solution: We first find
(18)\displaystyle \begin{aligned} \vec{u}\cdot\vec{v} &= (3\cdot1)+(-1\cdot5)+(4\cdot-2)\\ &=3-5-8=-10. \end{aligned}
We then compute
(19)$\displaystyle \|\vec{u}\|=\sqrt{9+1+16}=\sqrt{26}$
and
(20)$\displaystyle \|\vec{v}\|=\sqrt{1+25+4}=\sqrt{30}.$
Thus the angle between ${\vec{u}}$ and ${\vec{v}}$ is
(21)$\displaystyle \theta=\arccos\left(\frac{-10}{\sqrt{26}\sqrt{30}}\right)\approx1.937$

### Orthogonality

6. Vectors are perpendicular or “Orthogonal” if ${\vec{u}\cdot\vec{v}=0}$. Sometimes this is denoted ${\vec{u}\bot\vec{v}}$.

Example 2 Consider
(22)$\displaystyle \vec{u}=\langle6,-3,8\rangle,\quad\mbox{and}\quad \vec{v}=\langle-2,4,3\rangle.$
We see
(23)$\displaystyle \vec{u}\cdot\vec{v}=-12-12+24=0$
which implies ${\vec{u}}$ and ${\vec{v}}$ are orthogonal.

7. Consider the following diagram

We’re given vectors ${\vec{u}=\overrightarrow{PQ}}$ and ${\vec{v}=\overrightarrow{PS}}$ in 3-space. Notice that if the angle between the vectors ${\theta}$ is acute, as doodled, then ${\overrightarrow{PR}}$ is the projection of ${\vec{u}}$ onto ${\vec{v}}$.

However, if ${\theta}$ is obtuse, we doodle the situation thus:

Observe the projection of ${\vec{u}}$ onto ${\vec{v}}$ will not fall on ${\vec{v}}$. The projection of ${\vec{u}}$ onto ${\vec{v}}$ is syntactically
(24)$\displaystyle \mathop{\mathrm{proj}}\nolimits_{\vec{v}}\vec{u}$
The natural question: what is the formula for projecting ${\vec{u}}$ onto ${\vec{v}}$? We have
(25)$\displaystyle \|\overrightarrow{PR}\|=\begin{cases}\|\vec{u}\|\cos(\theta) &\mbox{for }\theta\mbox{ acute}\\ -\|\vec{u}\|\cos(\theta) &\mbox{for }\theta\mbox{ obtuse} \end{cases}$
The direction of ${\overrightarrow{PR}}$ depends on whether ${\theta}$ is acute or obtuse; we have its unit vector be
(26)$\displaystyle \widehat{PR}=\begin{cases}\widehat{v}&\mbox{for }\theta\mbox{ acute}\\ -\widehat{v}&\mbox{for }\theta\mbox{ obtuse}\end{cases}$
But now look, for both obtuse and acute ${\theta}$ we have
(27)\displaystyle \begin{aligned} \overrightarrow{PR} &=\|\vec{u}\|\cos(\theta)\frac{\vec{v}}{\|\vec{v}\|}\\ &=\frac{\|\vec{u}\|\|\vec{v}\|\cos(\theta)}{\|\vec{v}\|}\frac{\vec{v}}{\|\vec{v}\|}\\ &=\frac{\vec{u}\cdot\vec{v}}{\|\vec{v}\|}\frac{\vec{v}}{\|\vec{v}\|} \end{aligned}
This describes the projection of ${\vec{u}}$ onto ${\vec{v}}$ for any ${\theta}$:
(28)$\displaystyle \mathop{\mathrm{proj}}\nolimits_{\vec{v}}\vec{u}=\frac{\vec{u}\cdot\vec{v}}{\|\vec{v}\|}\frac{\vec{v}}{\|\vec{v}\|}$
Notice that its magnitude is ${\vec{u}\cdot\vec{v}/\|\vec{v}\|}$.

8. Consider the same situation again. We have a vector ${\vec{w}=\overrightarrow{RQ}}$ as doodled

This vector ${\vec{w}}$ is orthogonal to the projection of ${\vec{u}}$ onto ${\vec{v}}$. For this reason, we write
(29)$\displaystyle \vec{w} = \mathop{\mathrm{orth}}\nolimits_{\vec{v}}\vec{u}$
What is it? Well, using basic vector arithmetic, we find
(30)$\displaystyle \vec{w} = \vec{u} - \mathop{\mathrm{proj}}\nolimits_{\vec{v}}\vec{u}.$
We will conclude our discussion of vectors here, but continue next time.