Parametric Intersection of Planes P1 and P2

[andrey21]

Calculate in parametric form and describe how the planes intersect

Where:

P₁ = x-3y+5z=6
P₂ = 2x-7y+9z=2



My attempt

Put planes in matrix form:

1, -3, 5, 6
2, -7, 9, 2

Find Echelon Form

1, -3, 5, 6
0, -1, -1, -10

Z = free variable = a
So:

-y-z=-10

y = 10 - a

Sub into P₁:

x-3(10-a)+5z=6
x -30 +3a +5a =9
x+8a=36
x=36-8a

So vector parametric form of equation is:

(x,y,z) = (36,10,0) + a(-8,-1,1)

Is this correct??

 

[Mark44]

You can check for yourself. If your equation is correct, each value of a will give you a point that is on both planes.

Geometrically, your solution represents the line of intersection of the two planes.

 

[andrey21]

So you mean to do the following:

use (1,-3,5) from first plane:

sub into equation gives:

z=0+a
a=5

y=10-a
-3=10-a
a=13

 

[HallsofIvy]

No, he meant check (x,y,z) = (36,10,0) + a(-8,-1,1) in the equations of both planes.
The planes are given by x-3y+5z=6 and 2x-7y+9z=2.

Put x= 36- 8a, y= 10- a and z= a into both of those. What do you get?

 

[andrey21]

Ok sorry I got a little confused before: By substitution I get:

6=6 and 2=2 therefore equation is correct

 

[Mark44]

Ok sorry I got a little confused before: By substitution I get:

6=6 and 2=2 therefore equation is correct

No, what you have found is that the point x = 36 - 8a, y = 10 - a, z = a is a solution of both equations (meaning that this point is on both planes), regardless of the value of a.

 

[andrey21]

Ok Thank you. I do have a follow up question :

A third plane is added

P₃: 2x-6y+10z=b

Find a value for b so three planes have no common point of intersection.

Now shall I just use equation established in previous question and establish value for b from there??

 

[Mark44]

That's not what I would do. I would solve this system of equations
x - 3y + 5z = 6
2x - 7y + 9z = 2
2x - 6y + 10z = b

I would set this up as an augmented matrix, and use matrix operations to row-reduce the matrix.

 

[andrey21]

Ok so adopting your technique I obtained for the final row:

b-12=0
b=12

Therefore b=11 would mean the planes have no common point of intersection

 

[Mark44]

Right, that's one of an infinite number of possible values for which the three planes have no common intersection. Notice that if b = 12, the first and third equations are equivalent, so both equations describe the same plane.

 

[andrey21]

Thank you mark 44:

The question goes on to introduce a fourth plane:

P₄=3x-10y+az=8

Now I must find a value for a for which three planes P₁,P₂and P₄ intersect in a common straight line. Shall I just adopt same approach by putting in matrix and solving?

 

[Mark44]

Yes. Keep in mind, though, that for these three planes to intersect in a line, you shouldn't get a unique solution for your system. I haven't worked this problem, so this warning may or may not be important.

One very minor thing: Don't write P₄=3x-10y+az=8. P₄ just identifies which plane you're talking about; it isn't equal to an equation. In fact, you shouldn't say that anything "equals" an equation.

Either use a colon (:) or just a space.

 

[andrey21]

Ok so by putting in matrix and solving as before I obtain:

1 ,-3, 5, 6
0,-1,(a-15),-10
0, 0,(14-a), 0

So from this I can say:

(14-a)z=0
14z=az
a=14

Then substituting:

x= 36- 8a, y= 10- a and z= a

into P₄.

we get

8=8

 

[Mark44]

Ok so by putting in matrix and solving as before I obtain:

1 ,-3, 5, 6
0,-1,(a-15),-10
0, 0,(14-a), 0

So from this I can say:

(14-a)z=0

No need to expand (14 -a)z. If (14 - a)z = 0, then 14 - a = 0, so a = 14.

14z=az
a=14

This (above) is your answer.

The part below is the check of your work, verifying that the line is actually on the third plane.

Then substituting:

x= 36- 8a, y= 10- a and z= a

into P₄.

we get

8=8

 

[andrey21]

Yes sorry that wasn't the clearest answer at the bottom there. Now I have to identify the parametric equation of that line. This is confusing me a little as to where I get this from

 

[Mark44]

You can get the parametric equation of the line from your matrix.
If a = 14, your reduced augmented matrix looks like this:
[tex]\left[\begin{array} {ccccc} 1&0&8&|&36\\0&1&1&|&10\end{array}\right][/tex]
As a system of equations, this matrix says
x + 8z = 36
y + z = 10

We can write both x and y in terms of z, and we can let z = t to work a parameter in. This gives us
x = -8t + 36
y = -t + 10
z = t

You can also write this in vector form, with r(t) = <x, y, z> = t<-8, -1, 1> + <36, 10, 0>.

 

[andrey21]

so r(t) = <x, y, z> = t<-8, -1, 1> + <36, 10, 0> is the equation of the line, which is identical to the one I established in the first post as all three planes are on the same line.

 

[Mark44]

Or rather, the line is on all three planes.