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?
Hello, theowne!
What point ??If you are given a line: .$\displaystyle r\:=\:[4,-5,6] + t[2,0,-1]$
how can you, from that info, figure out 3 planes which intersect at that point?
If you're referring to the point $\displaystyle (4,-5,6),$
. . write three equations of the form: $\displaystyle ax + by + cz \:=\:d$
. . which have $\displaystyle (4,-5,6)$ as their solution.
Here's one system: .$\displaystyle \begin{array}{ccc}x + y + z &=& 5 \\ x - y + z &=& 15 \\ x + y - z &=& \text{-}7 \end{array}$
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:
$\displaystyle p: \vec r = \langle 4,-5,6\rangle + t \langle 2,0,-1\rangle + s \langle a,b,c \rangle$
a = 1, b = 2, c = 3 will yield:
$\displaystyle p: \vec r = \langle 4,-5,6\rangle + t \langle 2,0,-1\rangle + s \langle 1,2,3 \rangle$
General solution:
The normal vector of the new plane is $\displaystyle \vec n = \langle 2,0,-1\rangle \times \langle a,b,c \rangle = \langle b~,~ -2c-a~,~ 2b \rangle$
Considering that the point A(4, -5, 6) must be in the new plane you'll get the general equation:
$\displaystyle p: bx-(2c+a)y + 2bz=5a+16b+10c$