12.5 Equations of Lines and Planes

Lines

On a 2D plane, a line is represented by $y=mx+b$, $m$ representing the slope, or direction on plane and $b$ being the y-intercept, or a point on a line. To make a line in 3D, you also need a point, $P_{0}(x_{0}, y_{0}, z_{0})$ and a direction along the line, $\vec{v}$.

Vector Equation of a Line

The "vector equation of $L$": $r=r_{0}+t\vec{v}$ This is not just a fuction. Lines in 3D must be represented in terms of a parameter. Each value of parameter $t$ gives the position vector $\vec{r}$ a point on $L$. As $t$ varies, the line is traced out by the tip of vector $\vec{r}$. $r_{0}$ can be a point the line passes through.

Parametric Equations of a Line

The three scalar equations, called "parametric equations of $L$" through $P_{0}(x_{0}, y_{0}, z_{0})$ and parallel to $\vec{v}$=<a,b,c> are: $x=x_{0}+at$ $y=y_{0}+bt$ $z=z_{0}+ct$

Planes

Planes in space are defined by a point $P_{0}(x_{0}, y_{0}, z_{0})$ and vector $\vec{n}$, called the "normal vector", that is orthogonal (perpendicular) to the plane.

Vector Equation of a Plane

$\vec{n}$ is orthogonal to any vector in the plane, so $\vec{n} \cdot (\vec{r} - \vec{r_{0}})=0$ or $\vec{n} \cdot \vec{r} = \vec{n_{0}} \cdot \vec{r_{0}}$ This is called the "vector equation of a plane".

Scalar Equation of a Plane

Write $\vec{n}=<a,b,c>$, $\vec{r}=<x,y,z>$ (arbitrary point), and $\vec{r_{0}}=<x_{0},y_{0},z_{0}>$ (The known point on the plane). Distribute $\vec{n} \cdot (\vec{r} - \vec{r_{0}})=0$ to get the "scalar equation of a plane": $a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0$

Linear Equation of Plane

The scalar euqation can be rewritten to form the "linear equation of the plane": $ax+by+cz+d=0$ where $d=(-ax_{0}+by_{0}+cz_{0})$

Ex: Equation Given Point and Vector

Given $\vec{r_{0}}=<2,4,-1>$ and $\vec{n}=<2,3,4>$ $2(x-2)+3(y-4)+4(z+1)=0$ $2x+3y+4z=12$

Angle Between Two Planes

Two planes are parallel if their normal vectors are parallel to each other. If two planes are not parallel, they interset on a line with some angle between them. Find the angle, $\theta$, betwteen two planes with normal vectors $\vec{n_{1}}$ and $\vec{n_{2}}$ with the same formula to find the angle between two vectors: $\cos{\theta}=\frac{ \vec{n_{1}} \cdot \vec{n_{2}}}{|\vec{n_{1}}||\vec{n_{2}}|}$

Line at Intersection of Two Planes

1. Find the point of intersection by setting $z=0$ and solving for $x$ and $y$.
2. Find the vector of the line by finding $\vec{v}=\vec{n_{1}} \times \vec{n_{2}}$ because orthogonal to both normal vectors would be the direction of the line.
3. Write the equation as $\frac{x-x_{0}}{a}=\frac{y-y_{0}}{b}=\frac{z-z_{0}}{c}$

Distance between Point and Plane

To find the distance, $D$, between point $P(x_{1}, y_{1}, z_{1})$ and the plane $ax+by+cz+d=0$, Let $P_{0}(x_{0}, y_{0}, z_{0})$ be any point on the plane and $\vec{b}$ the vector $\vec{P_{0}P}$ $\vec{b} = <x-x_{0}, y-y_{0}, z-z_{0}>$ $D=comp_{n}\vec{b}=\frac{|\vec{n} \cdot \vec{b}|}{|\vec{n}|}=\frac{|ax_{1}+by_{1}+cz_{1}+d|}{\sqrt{a^2+b^2+c^2}}$