How to find a linear equation of a plane that passes through the point $(6,0,-2)$ and contains the line with parametric equations $x=4-2t$, $y=3+5t$, $z=7+4t$?
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Welcome to Mathematics Stack Exchange. Do you know how to find the equation of a plane containing $3$ points? – J. W. Tanner Sep 13 '19 at 03:20
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Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos Sep 13 '19 at 03:21
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Yes. I think i need to find a plane from the parametric equations first, then use normal line n=<a,b,c>, r=<x,y,z> and r_0=<x_0,y_0,z_0> and n \cdot (r-r_0)=0 to find the linear equation. But I have no idea how to convert the parametric equation to a plane. – user42789 Sep 13 '19 at 03:24
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Well, you can’t convert the parametric equation to a plane—it’s the equation of a line, after all. I assume that you know how to find the equation of a plane given three noncolinear points on it. Do that, using two points one the line. Those should be easy enough to find. – amd Sep 13 '19 at 03:26
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https://www.math.ucla.edu/~ronmiech/Calculus_Problems/32A/chap11/section5/717d31/717_31.html – gaurav saini Sep 13 '19 at 03:28
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Thanks. Problem solved! – user42789 Sep 13 '19 at 03:42
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you could compare your answer with mine (below) – J. W. Tanner Sep 13 '19 at 03:42
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Three points in the plane are $(6,0,-2)$, $(4,3, 7$) [on the line when $t=0$],
and $(2,8,11)$ [on the line when $t=1$].
Their differences are vectors in the plane:
$(6,0,-2)-(4,3,7)=(2,-3,-9)$ and $(6,0,-2)-(2,8,11)=(4,-8,-13)$.
The cross-product of the two vectors $(2,-3,-9)\times(4,-8,-13)=(-33,-10,-4)$
is normal to the plane, and so is $(33,10,4)$. So an equation of the plane is $33x+10y+4z=C$,
and you can figure out $C$ using any of the three points we know in the plane. (I got $C=190$.)
J. W. Tanner
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