ranyai12
Please Help! Parameterize the plane that contains the three points (3,−5,−2), (10,−12,−10), and (25,0,5).
r(s,t)=



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ranyai12
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I first assigned the points to P Q and R and then found PQ and PR and then put P+u<PQ>+v<PR> and I got <3+7s+22t,57s+5t,28s+7s> but got it wrong!

jim_thompson5910
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what did you get for vector PQ


ranyai12
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I got <7,7,8>

ranyai12
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Yes i looked at that and Im stll not sure what i dd wrong

jim_thompson5910
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how about PR?

ranyai12
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22,5,7

jim_thompson5910
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ok, so far, so good

jim_thompson5910
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I'm assuming you tried <3+7s+22t,57s+5t,28s+7t>and not <3+7s+22t,57s+5t,28s+7s> right?

ranyai12
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i did it twice and got it wrong twice

jim_thompson5910
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ok one sec

jim_thompson5910
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does it specify at all which variables are the free variables (like x and y or x and z)?

ranyai12
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no

jim_thompson5910
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hmm well I managed to find a calculator that will give the equation of the plane


jim_thompson5910
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that equation is
x25y+21z80 = 0
x25y+21z= 80
solve for x:
x25y+21z= 80
x = 80+25y21z
x = 8025y+21z

jim_thompson5910
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so if you let y = s and z = t, then you will get
<x, y, z> = <8025y+21z, y, z>
<x, y, z> = <8025s+21t, s, t>

jim_thompson5910
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unfortunately there are many ways to parameterize the plane

ranyai12
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can you please provide me with an example

jim_thompson5910
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what do you mean

ranyai12
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as in please show me a way o parameterize this

jim_thompson5910
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well you could take the cross product of vectors PQ and PR

jim_thompson5910
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that will give you the normal vector to the plane

ranyai12
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what would i do then

jim_thompson5910
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what do you get when you do the cross product

ranyai12
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gimme a sec

ranyai12
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89i+113j+189k

jim_thompson5910
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somewhat close, but that's not correct

ranyai12
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what is it then?

jim_thompson5910
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try it again

ranyai12
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ok