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
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)
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