Intersection of Three Planes

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MCV4U1-UNIT NINE-LESSON FOUR
Lesson Four: The Intersection of Three Planes
There are EIGHT ways that three planes can intersect.
Method: Examine the normals to narrow down the number of possibilities.
CASE I.
NO NORMALS PARALLEL
Solution is ...
-a point (Three planes intersect in a point.)
-a line (Three planes intersect in one unique line.)
-no solution (Three planes intersect in three unique lines.)
For this case you must solve the system to see what the solution is.
CASE II.
TWO NORMALS PARALLEL
Solution is...
-a line (Two parallel/coincident planes and one non parallel plane.)
-no solution (Two parallel/non-coincident planes and one non
parallel plane intersect in two unique lines.)
MCV4U1-UNIT NINE-LESSON FOUR
CASE III.
ALL NORMALS PARALLEL
Solution is...
-no solution (Three parallel non/coincident planes.)
-a plane (Three parallel/coincident planes.)
-no solution (Two parallel/coincident planes and one parallel/noncoincident plane.)
Example 1.

n1  1,2,3

n2  1,1,3

n3  2,1,6
 1 : x  2 y  3z  4  0
Solve  2 : x  y  3 z  8  0
 3 : 2 x  y  6 z  14  0
 no normals are parallel
Therefore we must solve the system to see what solution we have.
The reduced matrix is.....
1 2 3  4 
0 3 6  12  so the solution is a point.....
x, y, z  1,2,3


0 3 0 6 
Example 2.

n1  1,1,2 

n 2  3,1,14 

n3  1,2,0 
 1 : x  y  2 z  2
Solve  2 : 3 x  y  14 z  6
 3 : x  2 y  5
 no normals are parallel
MCV4U1-UNIT NINE-LESSON FOUR
Therefore we must solve the system to see what solution we have.
The reduced matrix is.....
1 1 2  2
0 1  2  3 so the solution is a line. Let z=t, a parameter and express x and y in terms of t.


0 0 0
0 
The solution is the line x, y, z    4t  1,3  2t, t .
Example 3. Solve
 1 : x  y  4z  5
 2 : 3 x  y  z  2
 3 : 5x  y  9z  1

n1  1,1,4 

n2  3,1,1
 no normals are parallel

n3  5,1,9 
Therefore we must solve the system to see what solution we have.
The reduced matrix is.....
1  1 4 5 
0  4 11 19  so there is no solution. The three planes intersect in three unique lines.


0 0 0  5
 1 : x  y  2z  1
Example 4. Solve  2 : 2 x  2 y  4 z  2
 3 : 3x  2 y  z  4

n1  1,1,2 

n2  2,2,4 

n3  3,2,1




 n2  2n1 , but n2  kn3 ; k  
  2 and  1 are parallel only. Check for coincidence.
D2  2 D1 .   2 and  1 are coincident too.
Therefore we have two parallel/coincident planes and one non parallel plane intersecting in a unique line.
1 
3 6 7
t  , t  ,t .
5 5
5 
 5
Let z=t, a parameter and solve for x and y in terms of t....to get.... x, y, z   
 1 : 2 x  y  3z  4
Example 5. Solve  2 : 2 x  y  3 z  7
 3 : 3 x  y  z  10
MCV4U1-UNIT NINE-LESSON FOUR

n1  2,1,3

n2  2,1,3

n3  3,1,1




 n1  1n2 , but n1  kn3 ; k  
  1 and  2 are parallel only. Check for coincidenc e.
D1  1D2 .   1 and  2 are non  coincident .
Therefore there is no solution. Two parallel, non-coincident planes are intersected by a third in two unique lines.
Example 6. Solve

n1  1,1,2 
 1 : x  y  2z  4
 2 : 2x  2 y  4z  8
 3 : 3 x  3 y  6 z  12

 1
1 
1
1
n2  2,2,4   n1  n2 
n3 and D1  D2 
D3
2
3
2
3

n3   3,3,6 
 The three planes are parallel and coincident .
Therefore the solution is the plane. Since all three equations are essentially the same, choose one and let z=t, y=s, t and
s are parameters.
Using  1 , solve for x in terms of s and t.
x  s  2t  4
Then the solution is the plane given by
ys
zt
Example 7.
Solve
 1 : 2 x  y  5z  4
 2 : 2 x  y  5z  7
 3 : 4 x  2 y  10 z  9

n1  2,1,5

n2  2,1,5

n3  4,2,10 
 
1
1
1
n 1  n2  n3 but D1  D2 and D1  D3 and D2  D3
2
2
2
Therefore the three planes are parallel/non-coincident. So there is no solution.
Example 8.
Solve
1 : 3x  4 y  5z  6
 2 : 6x  8y  10z  13
 3 :12x  16y  20z  26

n1  3,4,5

n
 2  6,8,10
n3  12,16,20



n 3  4n1  2n2 but D3  4 D1 and D3  2 D2 and 4 D1  2 D2
Therefore two planes are parallel/coincident (  3 and  2) and one plane is parall/non-coincident ( 1 ). So there is no
solution.


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