May 21st, 2012, 12:46 PM  #1 
Member Joined: Apr 2012 Posts: 78 Thanks: 3  LINES AND PLANES: needs checking
NEED CORRECTION, also this . means dot multiplication. My teacher comments: #6) you've made some errors (2 marks) # correct, they intersect at a point, but you need to find the point like you did in #7 for full marks (3 marks) 6.Determine the intersection, if any, of the planes with equations x + y – z + 12 =0 and 2x + 4y  3z + 8 = 0. The normal vectors for the two planes are (1, 1, 1) and (2, 4, 3).  These vectors are not collinear therefore the planes intersect in a line. x+yz = 12 (1) 2x+4y3z = 8(2) 3(1) + 2: x + y = 28 = x+28 Let x = t. y = t+28 Substituting in (1) One of the either answers=> t+t+28z = 12 or z = 2t+40. The parametric equations for the line of intersection are x = t, y = 28+t, z = 40+2t. 8. Give a geometrical interpretation of the intersection of the planes with equations x + y ? 3 = 0 y + z + 5 = 0 x + z + 2 = 0 N1= (1, 1, 0) N2= (0, 1, 1), N3= (1, 0, 1) N1 x N2 = ((1,1,0) x (0,1,1)) . (1,0,1) = (1,1,1) . (1,0,1) =2 (N1 x N2) . N3 ? 0 Since the triple dot product does not equal to 0, then these three planes must intersect in a single point. 
May 22nd, 2012, 05:56 PM  #2 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: LINES AND PLANES: needs checking
In solving your first problem you have "x + y = 28 = x+28". If you really wrote that on your paper, you should have lost points. You meant "x+ y= 28 so y= x+ 28" and should have said that. For the second, the first is a plane parallel to the zaxis, the second a plane parallel to the xaxis, and the third a plane parallel to the yaxis. It is very simple to solve the three equation simultaneously for x, y, and z, to get the point of intersection. Why didn't you do that? 
May 22nd, 2012, 10:48 PM  #3  
Member Joined: Apr 2012 Posts: 78 Thanks: 3  Re: LINES AND PLANES: needs checking Quote:
SUCH AS X= 1 AND Y=2 I am confused.  
May 23rd, 2012, 03:13 AM  #4  
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: LINES AND PLANES: needs checking Quote:
You have three equations with three variables (x,y,z). HallsOfIvy probably noticed a quick way to solve, but if you don't happen to notice that, use any technique you know for "solving systems of linear equations". Google that phrase if necessary. Heck, you could probably even Bing it! Anyhow, the "trick" is that if you subtract the third equation from the sum of the first two equations, you get 2y = 0.  
May 23rd, 2012, 09:47 AM  #5  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: LINES AND PLANES: needs checking Quote:
x+ y 3= 0 y+ z+ 5= 0 and x+z+ 2= 0 If you subtract the second equation from the first, you eliminate y to get x z 8= 0. Adding the third equation to that gives 2x 6= 0 so x= 3. Now put that into x z 8= 0 to solve for z and into x+y 3= 0 to solve for y.  
May 25th, 2012, 07:26 AM  #6  
Global Moderator Joined: Dec 2006 Posts: 20,968 Thanks: 2217  Quote:
 

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determine the intersection, if any, of the planes with equationsÂ x y â€“ z 12 =0Â andÂ 2x 4y  3z 8 = 0,how to solve equation like 2x 3y 12=0 and x 4y5=0
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