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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 ?

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    $\begingroup$ How did you find $\vec{n}$? $\endgroup$
    – Nosrati
    Nov 5, 2018 at 20:29
  • $\begingroup$ 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 $\endgroup$
    – Tuki
    Nov 5, 2018 at 20:32
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    $\begingroup$ Why "guess"? make two vectors with three points and find cross product of them! $\endgroup$
    – Nosrati
    Nov 5, 2018 at 20:34

1 Answer 1

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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 $t$.

The point of the line must project along the plane normal as any point on the plane.

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