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+3y-5z=31 in which the coefficient of x is -1.

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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+3y-5z=31 in which the coefficient of x is -1.

Mathematics
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anything parallel to -1x+3y-5z=31 will be in the form -1x+3y-5z=D where D is a constant
`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
Well, we're also given the \(\vec n\) of the parallel plane

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\[\vec n = \langle -1~,~3~,~-5\rangle\]
Okay, so how do we put all the given information together to find the equation of the plane??
at this point you just plug (x,y,z) = (-5,2,-5) into -1x+3y-5z=D to find D
Found out the D = 26
Scratch that D = 36
side 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
So the final answer will be -1x+3y-5z-36 :) TYYY!!!!
-1x+3y-5z-36 = 0 or -1x+3y-5z=36
Oh 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 :\[a(x-x_0)+b(y-y_0)+c(z-z_0)=0\]
Thats how I would find it...
\[-1(x-(-5)) +3(y-2)-5(z-(-5))=0\]\[-(x+5)+3(y-2)-5(z+5)=0\]\[-x-5 +3y-6-5z-25=0\]\[-x+3y-5z=25+6+5\]\[-x+3y-5z=36\]
|dw:1444016189424:dw| So you'll have like a vector running in between planes...Haha

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