All my math notes, now in Markdown.

You should be able to

- Plot points and graph planes
- Evaluate midpoint and distance of points
- Write equations of spheres
- Write vectors (component form / linear combination form)
- Find the unit vector
- Evaluate the dot product
- Write parametric / vector equations of lines

- Evaluate the angle between vectors
- Evaluate the cross product
- Write the equation of a plane
- Find the angle between planes
- Find the equation of the line of intersection of two planes
- Find the distance between a point and a plane

Sketch gridlines in the $xy$ plane and vertically to a point to show perspective. *Make sure they are parallel.* Right-hand rule axis.

Plot intercepts, sketch the “triangle” which represents a plane in one octant.

The **distance** $d$ between the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\)

The **midpoint** $M$ of line segment $PQ$ with endpoints $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is
\((M_x M_y M_z) = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2})\)

The **equation of a sphere** with point $P(x, y, z)$ on the sphere and center $(h, k, l)$ and radius $r$ is
\((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\)

An **equation of a plane** can be written as
\(Ax + By + Cz + D = 0\)
provided $A$, $B$, and $C$ are not all zero.

The **equation of a line** through the point $P_0 (x_0, y_0, z_0)$ in the direction of $\vec{v} = \langle a, b, c \rangle$

**Parametric form:**
\(x = x_0 + at, y = y_0 + bt, z = z_0 + ct\)

**Vector form:**
\(\langle x, y, z \rangle = \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle\)

If $\vec{u} = u_1 \vec{\imath} + u_2 \vec{\jmath} + u_3 \vec{k}$ and $v_1 \vec{\imath} + v_2 \vec{\jmath} + v_3 \vec{k}$ are two vectors in space, then the *cross product* is
\(\vec{u} \times \vec{v} = (u_2 v_3 - u_3 v_2) \vec{\imath} + (u_3 v_1 - u_1 v_3) \vec{\jmath} + (u_1 v_2 - u_2 v_1) \vec{k}\)

$\vec{u} \times \vec{v}$ is a vector that is orthogonal to $\vec{u}$ and $\vec{v}$ $\lvert \vec{u} \times \vec{v} = \lvert\vec{u}\rvert \lvert\vec{v}\rvert \sin{\theta}$, where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$.

$\lvert\vec{u} \times \vec{v} \rvert$ is the area of the parallelogram having non-zero vectors $\vec{u}$ and $\vec{v}$ as adjacent sides.

NOTE: Direction of vector is determined by the **right-hand rule** (order matters).

**Torque** (in physics) is a force, and is how much turning power you have. It is the cross product of the radius vector and force vector. Units are Newton-meter or lbs-ft, not same as work, which is energy, and Newton-meter is joule.

(Distance not in same direction, it’s perpendicular; multiply by unitless rotation (radians) to get work.)

The **standard equation of a plane** containing the point $P (x_1, y_1, z_1)$ and having nonzero normal (perpendicular) vector $\vec{n} = \langle a, b, c \rangle$ is
\(a(x - x_1) + b(y - y_1) + c(z - z_1) = 0\)

The general form after being simplified is \(ax + by + cz + d = 0\)

The **angle between two planes** can be found by
\(\cos{\theta} = \frac{n_1 \bullet n_2}{\lvert n_1\rvert \lvert n_2\rvert}\)
where $n_1$ and $n_2$ are the normal vectors of the two planes.

The **equation of the line of intersection of the two planes**, in parametric form, can be found by using the cross product of the two normal vectors of the planes and a point on the line. The cross product vector will be parallel to both planes, and will be in the direction of the line.

The **distance between a plane and a point $Q$** (that is not in the plane) is
\(d = \lvert proj_n\overrightarrow{PQ}\rvert\)
where $P$ is a point in the plane and $n$ is a vector normal to the plane.