Finding planes which intersect at a given line

**theowne** · June 7th 2008, 02:59 PM

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?

**Soroban** · June 7th 2008, 04:09 PM

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

**theowne** · June 7th 2008, 04:28 PM

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

**earboth** · June 7th 2008, 08:27 PM

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

**theowne** · June 8th 2008, 11:20 AM

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

**earboth** · June 8th 2008, 10:39 PM

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:

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

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