3. Vectors - 3D

Lines

Notation

A line in 3D is determined by a point $P = (x_{0}, y_{0}, z_{0})$ and a direction vector $\vec{d} = [\begin{matrix} d_{1} \\ d_{2} \\ d_{3} \end{matrix}]$

\vec{x} = P + t \vec{d}

Note

This is DIFFERENT from [[#Planes]]
Since it only has 1 direction vector

Quick normal vector in 3D

Lines in 3D has infinite number of normal vectors

Steps (Similar to 2D):

Set one of the coords to 0,
Swap the other 2 coords
Negate one of them

Example

$\vec{d} = [\begin{matrix} 2 \\ 3 \\ 4 \end{matrix}] \to \vec{n} = [\begin{matrix} 0 \\ - 4 \\ 3 \end{matrix}]$

Planes

Notation

Vector equation

Using 2 direction vectors,

\vec{x} = P + s \vec{u} + t \vec{v}

\vec{u}

and

\vec{v}

cannot be parallel

3 Points $\to$ Vector equation

Given $P, Q, R$ ,

\vec{x} = P + s \vec{P Q} + t \vec{P R}

Vector $\to$ Linear equation

Find $\vec{n}$ using $\vec{u} \times \vec{v}$ and then form the Linear equation

\begin{aligned} \vec{n} & = \vec{u} \times \vec{v} \\ \vec{x} \cdot \vec{n} & = \vec{O P} \cdot \vec{n} \end{aligned}

Linear equation

To find a point on the plane, sub 2 variables to any value and solve the last value

Find a point on plane

Given $3 x - 2 y - z = - 4$ , get a point on the plane
Sub x=0, $y = 0$ ,$$\begin{align}
0-0-z&=-4\
z=4 \
P=(0,0,4)
\end{align}$$

Cross product

Used to find a vector that's perpendicular to the 2 vectors

\vec{u} \times \vec{v} = [\begin{matrix} u_{x} \\ u_{y} \\ u_{z} \end{matrix}] \times [\begin{matrix} v_{x} \\ v_{y} \\ v_{z} \end{matrix}] = [\begin{matrix} u_{y} v_{z} - u_{z} v_{y} \\ u_{z} v_{x} - u_{x} v_{z} \\ u_{x} v_{y} - u_{y} v_{x} \end{matrix}]

To check:

(\vec{u} \times \vec{v}) \cdot \vec{u} = 0

Cross product Laws

Law	Note
$\vec{u} \times \vec{v} = - (\vec{v} \times \vec{u})$	Swap $\vec{u}$ and $\vec{v}$ when negating
$\vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w}$
$(\vec{u} + \vec{v}) \times \vec{w} = \vec{u} \times \vec{w} + \vec{v} \times \vec{w}$
$\vec{u} \times \vec{v} = \vec{0}$ if $\vec{u}$ and $\vec{v}$ are parallel	Also applies to $\vec{u} \times \vec{u} = 0$ as $\vec{u}$ is parallel to itself
$\vec{u} \times \vec{0}$
$c (\vec{u} \times \vec{v}) = (c \vec{u}) \times \vec{v} = \vec{u} \times (c \vec{v})$	The scalar ( $c$ ) only applies to 1 vector

Finding area of triangle with cross product

Given 3 points - P, Q and R

Area of △ P Q R = \frac{1}{2} | | \vec{P Q} \times \vec{P R} | |

Calculating Distances

Distances for lines

Point to Line

--- Using cross product (recommended)
$d(P,l)=\frac{\text{Area of parallelogram}}{||\vec{d}||}=\frac{||\vec{QP}\times \vec{d}||}{||\vec{d}||}$

--- Using projection
$
\begin{align}
\vec{QH}&=proj_{\vec{d}}{\vec{QP}} \\
\vec{HP}&=\vec{HQ}+\vec{QP} \\
d(P,l)&=||\vec{HP}||
\end{align}
$

Cannot use the same method used in 2D (projection onto normal)

As a line in 3D has infinite normals

Lines

Notation

Quick normal vector in 3D

Planes

Notation

Vector equation

3 Points → Vector equation

Vector → Linear equation

Linear equation

Cross product

Cross product Laws

Finding area of triangle with cross product

Calculating Distances

Distances for lines

Point to Line

Calculating Angles

3 Points $\to$ Vector equation

Vector $\to$ Linear equation