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Dec 15th, 2006, 10:41 AM
#1
Thread Starter
Addicted Member
Quick linear algebra question
Hey guys,
I didn't know who to ask and I need an answer quickly :S. It should be simple, I just cant seem to remember how I did this problem.
What is the equation of the plane parallel to the vector (1, 1, -2) which passes through the points (1, 5, 18) and (4, 2, -6).
The answer is 5x - 3y + z = 8
How do I go about solving this?
Thanks for yout time!
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Dec 15th, 2006, 10:44 AM
#2
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Dec 15th, 2006, 06:03 PM
#3
Re: Quick linear algebra question
Moved. Please don't post technical questions in the General Discussion / Chit Chat forum.
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Dec 18th, 2006, 07:53 AM
#4
Re: Quick linear algebra question
 Originally Posted by INF3RN0666
Hey guys,
I didn't know who to ask and I need an answer quickly :S. It should be simple, I just cant seem to remember how I did this problem.
What is the equation of the plane parallel to the vector (1, 1, -2) which passes through the points (1, 5, 18) and (4, 2, -6).
The answer is 5x - 3y + z = 8
How do I go about solving this?
Thanks for yout time!
The equation in vector form is:
R = r + av1 + bv2
where bold means vector and a,b are scalars (numbers).
r is a vector from the coordinate origin to a point in the plane, say to (1,5,18)
v1 and v2 are any two (linear independent) vectors contained on the plane. For example, v1 can be the given vector (1,1,-2) and v2 can be the vector joining the 2 given points:
v2 = (4,2,-6) - (1,5,18) = (3,-3,-24)
So the equation becomes:
R = (x,y,z) = (1,5,18) + a(1,1,-2) + b(3,-3,-24)
Expressing each component separately:
x = 1 + a +3b
y = 5 + a -3b
z = 18 -2a -24b
Now you make substitutions to find z in terms of x and y:
a = x - 1 - 3b
y = 5 + x -1 -3b -3b = 4 + x -6b
b = (4 + x - y) / 6
z = 18 -2(x - 1 -3b) - 24b = 18 - 2x + 2 + 6b - 24b = 20 - 2x - 16(4 + x - y) / 6 = 20 - 2x -3(4 + x - y) = 20 - 2x - 12 - 3x + 3y = 8 - 5x + 3y
Or, rearranging,
5x - 3y + z = 8
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