# Thread: Finding planes which intersect at a given line

1. ## Finding planes which intersect at a given line

If you are given a line (r=[4,-5,6] + t[2,0,-1], how can you, from that info, figure out 3 planes which intersect at that point?

2. Hello, theowne!

If you are given a line: . $r\:=\:[4,-5,6] + t[2,0,-1]$
how can you, from that info, figure out 3 planes which intersect at that point?
What point ??

If you're referring to the point $(4,-5,6),$
. . write three equations of the form: $ax + by + cz \:=\:d$
. . which have $(4,-5,6)$ as their solution.

Here's one system: . $\begin{array}{ccc}x + y + z &=& 5 \\ x - y + z &=& 15 \\ x + y - z &=& \text{-}7 \end{array}$

3. Sorry, I meant to say, "intersect at that line".

4. Originally Posted by theowne
If you are given a line (r=[4,-5,6] + t[2,0,-1], how can you, from that info, figure out 3 planes which intersect at that point?
You only have to add a second direction to this line:

$p: \vec r = \langle 4,-5,6\rangle + t \langle 2,0,-1\rangle + s \langle a,b,c \rangle$

with $\langle a,b,c \rangle \ne k \cdot \langle 2,0,-1\rangle ~,~ k \in \mathbb{R} \setminus \{0 \}$

5. Sorry, I don't understand..can you please clarify? How does this give us 3 planes?

6. Originally Posted by theowne
Sorry, I don't understand..can you please clarify? How does this give us 3 planes?
I've attached a sketch of three planes with a common line (in red). As you can see you only need a second direction vector (colour of the 2nd vector correspond to the colour of the plane's name) to span a plane.

So choose any numbers for a, b and c to get a new plane which contains the given line:

$p: \vec r = \langle 4,-5,6\rangle + t \langle 2,0,-1\rangle + s \langle a,b,c \rangle$

a = 1, b = 2, c = 3 will yield:

$p: \vec r = \langle 4,-5,6\rangle + t \langle 2,0,-1\rangle + s \langle 1,2,3 \rangle$

General solution:
The normal vector of the new plane is $\vec n = \langle 2,0,-1\rangle \times \langle a,b,c \rangle = \langle b~,~ -2c-a~,~ 2b \rangle$

Considering that the point A(4, -5, 6) must be in the new plane you'll get the general equation:

$p: bx-(2c+a)y + 2bz=5a+16b+10c$