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May 21st, 2012, 12:46 PM   #1
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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+y-z = -12 (1)
2x+4y-3z = -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+28-z = -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.
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May 22nd, 2012, 05:56 PM   #2
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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 z-axis, the second a plane parallel to the x-axis, and the third a plane parallel to the y-axis. 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?
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May 22nd, 2012, 10:48 PM   #3
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Re: LINES AND PLANES: needs checking

Quote:
Originally Posted by HallsofIvy
For the second, the first is a plane parallel to the z-axis, the second a plane parallel to the x-axis, and the third a plane parallel to the y-axis. 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?
SO SHOULD I PICK RANDOM X AND Y

SUCH AS X= 1 AND Y=2

I am confused.
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May 23rd, 2012, 03:13 AM   #4
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Re: LINES AND PLANES: needs checking

Quote:
Originally Posted by livestrong136
SO SHOULD I PICK RANDOM X AND Y

SUCH AS X= 1 AND Y=2

I am confused.
No, and there's no need for yelling!
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.
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May 23rd, 2012, 09:47 AM   #5
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Re: LINES AND PLANES: needs checking

Quote:
Originally Posted by livestrong136
Quote:
Originally Posted by HallsofIvy
For the second, the first is a plane parallel to the z-axis, the second a plane parallel to the x-axis, and the third a plane parallel to the y-axis. 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?
SO SHOULD I PICK RANDOM X AND Y

SUCH AS X= 1 AND Y=2

I am confused.
No, I didn't mean that I meant, as I said, solve the three equations. You have
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.
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May 25th, 2012, 07:26 AM   #6
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Quote:
Originally Posted by HallsofIvy
If you really wrote that on your paper, you should have lost points.
Not necessarily. Marking systems often just give points for valid work (even if poorly presented). Once the available marks are earned, anything else is assumed to be rough working.
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