Howiezi Posted May 6, 2015 Report Share Posted May 6, 2015 (edited) 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=2Plane 2: 2x-y+3z=2Plane 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=3Plane 2: -x+(a+1)y+3z=1Plane 3: -2x+y+(a+2)z=kFind 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)=8y(-2a-1)+z(a-4)=k+2x(a+4)+z(-2a-3)=-2k+3I 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 May 6, 2015 by Howiezi Reply Link to post Share on other sites More sharing options...
superspeed49 Posted May 6, 2015 Report Share Posted May 6, 2015 (edited) 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=-3Then solving these two: 5+(3-a)z=7 hence (8-a)z=7No solution hence: RHS= not LHS i.e. RHS=0(8-a)=0a=8Got it? Edited May 6, 2015 by superspeed49 Reply Link to post Share on other sites More sharing options...
Michael Wielondek Posted May 6, 2015 Report Share Posted May 6, 2015 (edited) 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 May 6, 2015 by Michael Wielondek 1 Reply Link to post Share on other sites More sharing options...
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