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anonymous
 one year ago
Calc. III
Find an equation of a plane containing the line r = <5,2,5> + t<11,2,1> which is parallel to the plane 1x+3y5z=31 in which the coefficient of x is 1.
anonymous
 one year ago
Calc. III Find an equation of a plane containing the line r = <5,2,5> + t<11,2,1> which is parallel to the plane 1x+3y5z=31 in which the coefficient of x is 1.

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1anything parallel to 1x+3y5z=31 will be in the form 1x+3y5z=D where D is a constant

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1`Find an equation of a plane containing the line r = <5,2,5> + t<11,2,1>` so we know the point (x,y,z) = (5,2,5) is on the line and on this unknown plane

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, we're also given the \(\vec n\) of the parallel plane

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\vec n = \langle 1~,~3~,~5\rangle\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so how do we put all the given information together to find the equation of the plane??

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1at this point you just plug (x,y,z) = (5,2,5) into 1x+3y5z=D to find D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Found out the D = 26

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1side note: take the dot product of <11,2,1> and the normal vector to the plane that Jhannybean wrote you should get a dot product of 0 which shows that the entire line is contained in the parallel plane

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the final answer will be 1x+3y5z36 :) TYYY!!!!

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.11x+3y5z36 = 0 or 1x+3y5z=36

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I see, your \(P_0 = (5~,~2~,~5)\) , cross checking whether your normal vector and position vector = 0, we'll know if theyre orthogonal to one another. Then with \(\vec n = \langle 1~,~3~,~5\rangle \) and \(P_0 = (5~,~ 2~,~5)\) we can fidn the equation of a parallel plane by :\[a(xx_0)+b(yy_0)+c(zz_0)=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thats how I would find it...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[1(x(5)) +3(y2)5(z(5))=0\]\[(x+5)+3(y2)5(z+5)=0\]\[x5 +3y65z25=0\]\[x+3y5z=25+6+5\]\[x+3y5z=36\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444016189424:dw So you'll have like a vector running in between planes...Haha
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