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Equation of a plane (General and Point-Normal Form) |
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| Nov1-08, 02:48 PM | #1 |
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Equation of a plane (General and Point-Normal Form)
Hey all,
I'm trying to figure out an example from my Linear Algebra Book: The question is as follows: Find the equation of the plane passing through the points P1(1,2,-1), P2(2,3,1) and P3(3,-1,2) It goes on to say it must satisfy this system, which is fine: 1 + 2b - c + d = 0 2a + 3b + c + d = 0 3a - b + 2c + d = 0 Then it just says, "solving for this system gives [tex]a=\frac{9}{16}t, b=-\frac{1}{16}t, c= \frac{5}{16}t, d=t[/tex] " To solve this system, would one include the d's in the matrix or leave them out? What method would you recommend for solving this? (and then I can give it a shot) Thanks in advance! |
| Nov3-08, 05:22 AM | #2 |
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You forgot to mention that you are assuming the equation of the plane (or any plane) can be written ax+ by+ cz+ d= 0! I wondered where a, b, c, and d came from!
Yes, of course you have to include d! You have to include all the variables. The equations you give are equivalent to 2b- c+ d= -1 2a+ 3b+ c+ d= 0 3a- b+ 2c+ d= 0 The augmented matrix for that system of equations is [tex]\left[\begin{array}{ccccc}0 & 2 & -1 & 1 & -1 \\ 2 & 3 & 1 & 1 & 0 \\ 3 & -1 & 1 & 1 & 0\end{array}\right][/tex] |
| Nov3-08, 02:16 PM | #3 |
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Am I correct in thinking the Gaussian elimination would be the best choice, or is there a more efficient way I'm missing?
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| Nov3-08, 05:58 PM | #4 |
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Equation of a plane (General and Point-Normal Form)
I recommend you instead use the point-normal form instead.
To find the normal, you can take the cross product n = (p2-p1)X(p3-p2). Then the condition that the normal is perpendicular to vectors on the plane means that (x-c)*n = 0 for points x and c on the plane. Here is a diagram: http://img259.imageshack.us/img259/7...malformpy3.png |
| Nov4-08, 10:22 AM | #5 |
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Awesome, Maze! But where does the 1/16 come from then?
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| Nov4-08, 10:46 AM | #6 |
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Have you tried
1) Solving the system of equations? 2) Reducing the matrix I gave you? 3) Taking the cross product as Maze suggested? Doing any of those things will show you where the "1/16" came from. The point is that you have three equations in 4 unknown coefficients. You can solve for any three in terms of the other one. Here they chose to solve for a, b, c in terms of d and use d as the parametr. Actually, it doesn't matter. If you define s to be "t/16", then, a= 9s, b= -s, c= 5s and d= 16s. |
| Nov4-08, 10:50 AM | #7 |
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Yes! Doing the cross product I obtained (9,1,-5), lacking the denominator I need.
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| Nov4-08, 11:30 AM | #8 |
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Hmm. heres what i get
p1=(1,2,1), p2=(2,3,1), p3 = (3,-1,2) p2-p1=(1,1,0), p3-p2=(1,-4,1) n = (p2-p1)X(p3-p2) = i(1-0) + j(0-1) + k(-4-1) = (1,-1,-5) 0 = n*(x-p1) = (1,-1,-5)*(x1-1, x2-2, x3-1) = x1-1-x2+2-5*x3+5 0 = x1 - x2 - 5*x3 + 6 This checks out with the 3 points, so it should be good. |
| Nov4-08, 12:40 PM | #9 |
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Hey Maze,
sorry, I'm not following your last two steps. What I did is set it up as a 2x3 matrix and solved for the three components. I used the easy method of "crossing" out a column and then finding the three determinants. This gives me the three numbers from before, but not the 1/16th... |
| Nov4-08, 01:04 PM | #10 |
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Maybe it will be easier to help if you can post your work so far.
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| Nov4-08, 01:28 PM | #11 |
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Sure,
I'm just using a trick from my linear algebra book: taking the cross product of a 2x3 1 1 0 1 -4 1 The trick is, to get the x portion, cross out the first column and take the determinant. For the y, cross out the second column, for the z, cross out the third. The three determinants I get are 9, 1, -5 |
| Nov4-08, 02:00 PM | #12 |
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Recheck the determinants,
[tex]det(\begin{matrix}1 & 0 \\ -4 & 1\end{matrix}) = 1*1-(-4*0) = 1[/tex] |
| Nov4-08, 02:13 PM | #13 |
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Wait, sorry - I accidentally used your value. The first row should be 1 1 2.
(P1 is 1,2, -1) Using that, I get the values I showed. |
| Nov4-08, 02:34 PM | #14 |
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Yeah ok with p1 = (1,2,-1) that looks good. So whats the problem then?
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| Nov4-08, 08:39 PM | #15 |
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Is there a problem with doing it my way, as I do not get the 1/16th still... or am I forgetting something?
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| Nov4-08, 09:34 PM | #16 |
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Oh, i see. Just recall
ax + by + c = 0 is the same as ax/16 + bx/16 + c/16 = 0 |
| Nov4-08, 09:46 PM | #17 |
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I'm confused...
when I wrote the solutions with 1/16th in my first post - that came from an unsolved answer in the book. I do not know how its obtained solving for my determinants,etc. |
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