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A question about planes intersecting at lines

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I have a questiong about planes intersecting. I have two questions from earlier exams that I don't really understand how to solve.
The questions I am wondering about are question 11(a), (b) and © on the M11/5/MATHL/HP2/ENG/TZ1/XX paper, and question 7(b) on the M09/5/MATHL/HP2/ENG/TZ1/XX paper.
The questions are about lines intersecting at either points or lines when there is a variable (a) in the plane equation. 

 

Example 7(a):

Plane 1: x+y+2z=2

Plane 2: 2x-y+3z=2

Plane 3: 5x-y+az=5

 

Find the value of a for which the planes do not meet at a unique point.

 

When I look at the markscheme for help they seem to be using matrices, but since I am a May 2014 student I have no idea how to use matrices.

 

What am I missing here?

 

Edit: Thanks for the answer, it helped me solve some of the questions, however I'm still stuck at M11/5/MATHL/HP2/ENG/TZ1/XX Question 11(b).

Question 11(b):

Plane 1: ax+2y+z=3
Plane 2: -x+(a+1)y+3z=1
Plane 3: -2x+y+(a+2)z=k

Find the value of a such that the three planes do not meet at a point.

I tried doing the same thing here, i can eliminate z from Plane 1 and 2, x from plane 2 and 3 and y from plane 1 and 3. However then I end up with these equations:
x(-3a-1)+y(a-7)=8
y(-2a-1)+z(a-4)=k+2
x(a+4)+z(-2a-3)=-2k+3

I try to rearrange equation 1 and 2, and insert them into equation 3 and get:

(8-y(a-7))*(a+4)*(a-4)+((k-2)-y(-2a-1))*(-2a-3)*(-3a-1)=(-2k+3)*(-3a-1)*(a+4)

 

Now from what I understand; (a) has to be a value such that the equation does not have a solution for ( y ). But how do i find that from this mess of an equation? I think I'm doing something wrong.

Edited by Howiezi

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Planes do not meet at a unique point hence there is no solution to the three equations:

Solving two equations: (1s and 2nd) 3x+5z=4, solving the second and third, -3x+(3-a)z=-3

Then solving these two: 5+(3-a)z=7 hence (8-a)z=7

No solution hence: RHS= not LHS i.e. RHS=0

(8-a)=0

a=8

Got it?

Edited by superspeed49

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Use gaussian elimination (row reduction; don't worry, it looks and sounds more intimidating than it is), and you should arrive at something that ends up with [ 0x 0y kz = n] in the bottom row. k should include a linear term stated in terms of a. 

 

Then all you have to do is to consider the 3 possible cases. Either n=0, and then if k=0 you'll end up with infinite amount of solutions (planes intersecting in a line), or k=0 but n!=0 meaning that there will be no solutions (the planes will, in simple terms, form a prism with a common axis), or the last one where n is any real value and so is k, and hence there will be a single solution.

 

In the example above, by doing row reductions you'll end up with (a-8)z=-1. This means that if a=8, then z will always be zero and thus there will be no unique solution. Because n is not 0, the planes do not intersect in a line,  and if a is any other value than 8, then you'll end up with a unique solution.

Edited by Michael Wielondek
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