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Planes in the Space
Line-Plane Intersection

In analytic geometry, the intersection of a line and a plane can be the empty set, a point or a line:

 No Intersection Point Intersection Line Intersection

How to find the relationship between a line and a plane.

If the line is

 A1x+B1y+C1z+D1=0 A2x+B2y+C2z+D2=0

and the plane is $\pi\;\equiv\;A_3x+B_3y+C_3z+D_3=0\;$

Form a system with the equations and calculate the ranks.

$\left{A_1x+B_1y+C_1z+D_1=0\\A_2x+B_2y+C_2z+D_2=0\\A_3x+B_3y+C_3z+D_3=0$

r = rank of the coefficient matrix
r'= rank of the augmented matrix

The relationship between the line and the plane can be described as follow:

 Case 1. Point intersection r=3 and r'=3 Case 2. No Intersection r=2 and r'=3 Case 3. Line Intersection r=2 and r'=2

State the relationship between the line:

$r\;\equiv\;\frac{x+1}{2}=\frac{y}{1}=\frac{z}{-1}\;$

and the plane $\pi\;\equiv\;x-2y+3z+1=0\;$

Solution:

Form the system of equations and calculate the ranks.

$\left{x-2y=-1\\y+z=0\\x-2y+3z=-1$

$M_1=\left(\begin{matrix}1&-2&0\\0&1&1\\1&-2&3\\\end{matrix}\right)$
 1 -2 0 0 1 1 1 -2 3
$\neq\;0$ r=3

$M_2=\left(\begin{matrix}1&-2&0&-1\\0&1&1&0\\1&-2&3&-1\\\end{matrix}\right)$
 1 -2 0 0 1 1 1 -2 3
$\neq\;0$ r'=3

Point Intersection.

State the relationship between the line:

$r\;\equiv\;\frac{x-1}{5}=\frac{y}{1}=\frac{z+2}{1}\;$

and the plane $\pi\;\equiv\;-x+3y+2z+5=0\;$

Solution:

Form the system of equations and calculate the ranks.

$\left{x-5y=1\\y-z=2\\-x+3y+2z=-5$

$M_1=\left(\begin{matrix}1&-5&0\\0&1&-1\\-1&3&2\\\end{matrix}\right)$
 1 -5 0 0 1 -1 -1 3 2
=0 r=2

$M_2=\left(\begin{matrix}1&-5&0&1\\0&1&-1&2\\-1&3&2&-5\\\end{matrix}\right)$
 1 -5 1 0 1 2 -1 3 -5
$\neq\;0$ r'=3

The line and plane are parallel. There is no intersection.

State the relationship between the plane
 : 4x-4y+2z=-2 and the line 5x-y+5z=5 -2x+2y-z=1
1. Point Intersection
2. No Intersection
3. Line Intersection