Vectors and Lines

You can put a vector in 3 dimensional space and draw a vector (directed line) to that point P(x & y & z) which turns into the vector v$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ in ${\mathbb{R}}^3$.

v is represented as $\underline{\text{P}}(x,y,z)$

$||v||$ notation says this is the length of vector v. That symbol is called “norm”.

$||v||=\sqrt{x^2+y^2+z^2}$

Ex: v and w are two vectors, so to sum them, we create a parallelogram and calculate the width(?) of the resulting parallelogram.

To subtract, we build a triangle. $v=w$ iff v and w have the same direction and the same length.

The distance between $v$ and $w$ is $||v-w||$ This is the same as:

$$\begin{array}{r}||(v_1-w_1)+(v_2-w_2)+(v_3-w_3)||\end{array}$$

$=\sqrt{(v_1-w_1{)}^2+(v_2-w_2{)}^2+(v_3-w_3{)}^2}$

Example: Find: the distance between ${\underline{\text{P}}}_1(2,-1,3)$ and ${\underline{\text{P}}}_2(1,1,4)$.

Always subtract the second one from the first one. $\overrightarrow{{\underline{\text{P}}}_1{\underline{\text{P}}}_2}={\underline{\text{P}}}_2-{\underline{\text{P}}}_1$

$$\left[\begin{array}{c}1-2\\ \\ 1--1\\ \\ 4-3\end{array}\right]$$

Which becomes

$$\left[\begin{array}{c}-1\\ \\ 2\\ \\ 1\end{array}\right]$$

Scalar multiple

If a is a real number and $v\ne 0$ is a vector, then:

1. $||av||=|a|||v||$ (absolute value of a times the length of v)
2. If $av\ne 0$, the direction of $av$ is: a. the same as v if a > 0 b. opposite to v if a < 0

Unit vector

A vector is a unit vector if it has length [1], which means the norm is equal to [1].

If $v\ne 0$ then $\frac{\vec{v}}{||v||}$ is a unit vector.

Since: $||\frac{v}{||v||}||$

is the same as:

$$\begin{array}{r}||\frac{1}{||v||}\cdot ||v||||\end{array}$$

and we know that we can pull that out from the scalar multiple line.

Then it turns into $\frac{||v||}{||v||}$ which is equal to 1.

Given a non-zero vector v, normalize = take the unit vector $\frac{\vec{v}}{||v||}$.

Normalize $v=\left[\begin{array}{c}0\\ 4\\ -3\end{array}\right]$ Sol:

$$\begin{array}{rl}||v||& =\sqrt{0^2+4^2+-3^2}\\ & \\ & =5\end{array}$$

so

$$\begin{array}{r}\frac{v}{||v||}=\frac{1}{5}\left[\begin{array}{c}0\\ 4\\ -3\end{array}\right]\\ \\ =\left[\begin{array}{c}0\\ \frac{4}{5}\\ -\frac{3}{5}\end{array}\right]\\ \end{array}$$

—

Parallel

Two nonzero vectors re called parallel if they have the same or opposite [direction].

As a consequence, two non-zero vectors are parallel (symbol is: ||) iff one is a scalar multiple of another.

Direction Vector

A vector that is parallel (||) to any two distinct vectors on the line.

I think this is really saying:

Given two distinct vectors, there is a third vector which goes through a different point on each line. The direction vector is parallel to that vector.

Lines

There is one line that passes through a particular point ${\underline{\text{P}}}_0(x_0,y_0,z_0)$ and has a direction vector $d=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$.

This means that $\overrightarrow{P_{0}P}||d$ (vector from $P_0$ to P is parallel to d) iff $\overrightarrow{P_{0}P}=td$ for $t\in \mathbb{R}$.

$$\begin{array}{r}\left[\begin{array}{c}x-x_0\\ y-y_0\\ z-z_0\end{array}\right]=t\left[\begin{array}{c}a\\ b\\ c\end{array}\right]\\ \\ \{\begin{array}{l}x=x_0+ta\\ \\ y=y_0+tb\\ \\ z=z_0+tc\\ \end{array}\end{array}$$

—

$$\begin{array}{r}\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}{x}_0\\ y_0\\ z_0\end{array}\right]+t\left[\begin{array}{c}a\\ b\\ c\end{array}\right]\end{array}$$

is the line through ${\underline{\text{P}}}_0$ in the direction d that is:

$$\begin{array}{r}\left[\begin{array}{c}x-x_0\\ y-y_0\\ z-z_0\end{array}\right]=t\left[\begin{array}{c}a\\ b\\ c\end{array}\right]\end{array}$$

for $t\in \mathbb{R}$

Parametric equations of a line

The line through ${\underline{\text{P}}}_0(x_0,y_0,z_0)$ with direction $d=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]\ne 0$ is given by:

$$\{\begin{array}{l}x=x_0+ta\\ \\ y=y_0+tb\\ \\ z=z_0+tc\\ \end{array}$$

EX: Find the equations of the line which pass through ${\underline{\text{P}}}_0(2,0,1)$ and ${\underline{\text{P}}}_1(4,-1,1)$

Sol: $d=\overrightarrow{P_0{P}_1}$

$$\left[\begin{array}{c}{x}_1-x_0\\ y_1-y_0\\ z_1-z_0\end{array}\right]=\left[\begin{array}{c}2\\ \\ -1\\ \\ 0\end{array}\right]$$

(which is d)

The line is: x = 2+2t y = 0+t z = 1

Works out to have: ${\underline{\text{P}}}_0$ coords - d$ in equation form.

Ex:

Determine whether the following lines intersect and, if so, find the point of intersect.

x=1-3t y=2+5t z=1+t

x=-1+s y=3-4s z=1-s

Sol: Let x,y,z be the point of intersection, ie suppose P(x,y,z) lies on both lines.

The lines intersect iff

1-3t = -1+s 2+5t = 3-4s 1+t = 1-s

$$\{\begin{array}{l}s+3t=2\\ \\ -4s-5t=-1\\ \\ -s-t=0\\ \end{array}$$

bottom one says s= -t

$$\{\begin{array}{l}t=1\\ \\ s=-1\\ \\ s=-t\end{array}$$

So the point of intersection is: t: 1-3 = -2 2+5 = 7 1+1 = 2

s: -1+-1 = -2 3+4 = 7 1—1 = 2