Lecture 005

Lines in \mathbb{R}^3

Conditions:

• intersect

• parallel

• skew

Line: a point $p = (x_0, y_0, z_0)$ and a direction vector $\langle{a, b, c}\rangle$, then a line is $\begin{cases} x = x_0 + at\\ y = y_0 + bt\\ z = z_0 + ct\\ \end{cases}$ or $\langle{x, y, z}\rangle = \langle{x_0, y_0, z_0}\rangle + t\langle{a, b, c}\rangle$

• or you can solve for $t$, then $\begin{cases} t = \frac{x - x_0}{a}\\ t = \frac{y - y_0}{b}\\ t = \frac{z - z_0}{c}\end{cases}$

• symmetric equation: $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$

Parallel Vector: for vector $\overrightarrow{u}, \overrightarrow{v}$, they are parallel if $(\exists k \neq 0) \overrightarrow{u} = k \overrightarrow{v}$. (or $\overrightarrow{u} \times \overrightarrow{v} = \overrightarrow{0} \land \overrightarrow{u} \neq \overrightarrow{0} \land \overrightarrow{v} \neq \overrightarrow{0}$)

Parallel Line: $L_1: \overrightarrow{r} = \langle{r_0}\rangle + t \overrightarrow{u}$ and $L_2: \overrightarrow{e} = \langle{e_0}\rangle + t \overrightarrow{v}$ are parallel if $\overrightarrow{u}$ and $\overrightarrow{v}$ are parallel vectors

Intersect: $L_1: \overrightarrow{r} = \langle{r_0}\rangle + t_1 \overrightarrow{u}$ and $L_2: \overrightarrow{e} = \langle{e_0}\rangle + t_2 \overrightarrow{v}$ intersect if there exists $t_0, t_1$ such that $\langle{r_0}\rangle + t_1 \overrightarrow{u} = \langle{e_0}\rangle + t_2 \overrightarrow{v}$ by solving a system of equation

Colinear: in the same line

Plane:

• 3 points not colinear (in the plane)

• 2 crossing lines (in the plane)

• a line and a point (in the plane)

• a normal vector $\overrightarrow{n}$ and a (starting) point $P$

  • the set of points $Q$ so that $\overrightarrow{n} \cdot \overrightarrow{PQ} = 0$ is a point (shift point $P$ to origin, where $n$ starts in origin)
  • if a normal vector $\overrightarrow{n}$ describe a plane $N$, then $-\overrightarrow{n}$ describe the same plane $N$

• scalar equation of plane: $a(x − x_0) + b(y - y_0) + c(z − z_0) = 0$

• general form: $ax + by + cz + d = 0$, where $d = −ax_0 − by_0 − cz_0$

Angle between Two Plane: Angle $(0 < \theta < \frac{\pi}{2})$ between $P_1$ and $P_2$ with normal vectors $n_1$ and $n_2$ is $\cos{\theta} = \frac{|\overrightarrow{n_1} \cdot \overrightarrow{n_2}|}{\|n_1\|\|n_2\|}$

The Distance between a Plane and a Point: $d = |\text{comp}_n \overrightarrow{QP}| = \frac{|\overrightarrow{QP}| \cdot n}{\|n\|}$ where $n$ is the normal vector pass through $Q$ and $P$ is point outside plane. (think about projection onto the normal line)