# Line intersects plane

## Problem

We have points $$(1,2,3),(2,-3,0)$$ and $$(0,-1,2)$$ that make up a plane in $$\mathbb{R}^3$$ and points $$(1,1,1)$$ and $$(2,3,2)$$ that make up a line. Define intersection of plane and a line.

## Attempt to solve

Line in vector form can be express as:

$$\vec{x}=\vec{p_1}+t(\vec{p_1}-\vec{p_2}) \text{ when } t \in \mathbb{R}$$

where $$\vec{p_1}$$ and $$\vec{p_2}$$ are points that belong to the line.

$$\vec{x}= \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + t\begin{bmatrix} -1 \\ -2 \\ -1 \end{bmatrix}$$

A normal vector for $$\vec{x}$$:

$$\vec{n}= \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$$

Now how do i get to regular form (or equation for line form) from this. Our lecture notes contained something like this:

regular form for line is:

$$\vec{n}\cdot \begin{bmatrix} x \\ y \\z \end{bmatrix} = \vec{n} \cdot \vec{p_1}$$

in this case when $$\vec{n}$$ and $$\vec{p_1}$$ are known.

$$\begin{bmatrix} 1 \\ -1 \\ 1\end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1\end{bmatrix} \cdot\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$

$$x-y+z=1$$

Now this would be equation for a plane and not a line right ? I don't exactly know how this should work or have intuition behind on what is going on here.

If i would have equations for line and a plane i could form system of equations and solve it ? Which would mean i get the intersection point ?

• How did you find $\vec{n}$? Nov 5, 2018 at 20:29
• Was just a guess. Their dot product is $\begin{bmatrix} 1 \\ -1 \\1 \end{bmatrix} \cdot \begin{bmatrix} -1 \\ -2 \\ -1 \end{bmatrix} = 0$. It is simply made up @Nosrati
– Tuki
Nov 5, 2018 at 20:32
• Why "guess"? make two vectors with three points and find cross product of them! Nov 5, 2018 at 20:34

The normal vector of a plane defined by three points $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$ is
$$\vec{n} = (\vec{b}-\vec{a}) \times (\vec{c}-\vec{a}) = \vec{a}\times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$$
In fact the area of the triangle is $$A = \frac{1}{2} \| \vec{n} \|.$$
Find the intersection point of the line $$\vec{x}(t)$$ by solving $$\vec{n} \cdot \vec{x}(t) = \vec{n} \cdot \vec{a}$$ for $$t$$.