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 June 10th, 2017, 03:37 AM #1 Member   Joined: May 2017 From: Slovenia Posts: 89 Thanks: 0 Tangent plane to surface The questions says to find tangent plane to the surface x^2-y^2-3z=0 that goes through A(0,0-1) and is parallel to line : (0,0,0)+t(2,1,2). I'm really stuck here.
 June 10th, 2017, 04:03 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Well, lets see. The gradient to the surface, at point $(x_0, y_0, z_0)$ is $<2x_0, -2y_0, -3>$ so the tangent plane to the surface at that point is $2x_0(x- x_0)- 2y_0(y- y_0)- 3(z- z_0)= 0$. In order that it contain the point A(0, 0, -1) we must have $-2x_0^3+ 2y_0^2+ 3z_0= 0$. The line x= 3t, y= t, z= 2t has "direction vector" <2, 1, 2>. In order that the plane be parallel to that <2, 1, 2> must be perpendicular to the normal line to the plane: we must have $2(2x_0)+ 1(-2y_0)+ 2(-3)= 4x_0- 2y_0- 6= 0$. And, of course, since $(x_0, y_0, z_0)$ is a point on the surface, it must satisfy the equation of the surface. We must have $x_0^2- y_0^2- 3z_0= 0$. Solve the three equations $-2x_0^3+ 2y_0^2+ 3z_0= 0$ $2(2x_0)+ 1(-2y_0)+ 2(-3)= 4x_0- 2y_0- 6= 0$ and $x_0^2- y_0^2- 3z_0= 0$ for $x_0$, $y_0$, and $z_0$. Thanks from sarajoveska
June 10th, 2017, 05:46 AM   #3
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Quote:
 Originally Posted by Country Boy Well, lets see. The gradient to the surface, at point $(x_0, y_0, z_0)$ is $<2x_0, -2y_0, -3>$ so the tangent plane to the surface at that point is $2x_0(x- x_0)- 2y_0(y- y_0)- 3(z- z_0)= 0$. In order that it contain the point A(0, 0, -1) we must have $-2x_0^3+ 2y_0^2+ 3z_0= 0$. The line x= 3t, y= t, z= 2t has "direction vector" <2, 1, 2>. In order that the plane be parallel to that <2, 1, 2> must be perpendicular to the normal line to the plane: we must have $2(2x_0)+ 1(-2y_0)+ 2(-3)= 4x_0- 2y_0- 6= 0$. And, of course, since $(x_0, y_0, z_0)$ is a point on the surface, it must satisfy the equation of the surface. We must have $x_0^2- y_0^2- 3z_0= 0$. Solve the three equations $-2x_0^3+ 2y_0^2+ 3z_0= 0$ $2(2x_0)+ 1(-2y_0)+ 2(-3)= 4x_0- 2y_0- 6= 0$ and $x_0^2- y_0^2- 3z_0= 0$ for $x_0$, $y_0$, and $z_0$.
Okay, I think I got it. Thank you

June 10th, 2017, 11:26 AM   #4
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Joined: Jan 2015
From: Alabama

Posts: 3,264
Thanks: 902

Quote:
 Originally Posted by Country Boy Well, lets see. The gradient to the surface, at point $(x_0, y_0, z_0)$ is $<2x_0, -2y_0, -3>$ so the tangent plane to the surface at that point is $2x_0(x- x_0)- 2y_0(y- y_0)- 3(z- z_0)= 0$. In order that it contain the point A(0, 0, -1) we must have $-2x_0^3+ 2y_0^2+ 3z_0= 0$.
This is wrong. I meant to say $-2x_0+ 2y_0- 3(-1- z_0)= -2x_0+ 2y_0+ 3z_0+ 3= 0$

Quote:
 The line x= 3t, y= t, z= 2t has "direction vector" <2, 1, 2>. In order that the plane be parallel to that <2, 1, 2> must be perpendicular to the normal line to the plane: we must have $2(2x_0)+ 1(-2y_0)+ 2(-3)= 4x_0- 2y_0- 6= 0$. And, of course, since $(x_0, y_0, z_0)$ is a point on the surface, it must satisfy the equation of the surface. We must have $x_0^2- y_0^2- 3z_0= 0$. Solve the three equations $-2x_0^3+ 2y_0^2+ 3z_0= 0$ $2(2x_0)+ 1(-2y_0)+ 2(-3)= 4x_0- 2y_0- 6= 0$ and $x_0^2- y_0^2- 3z_0= 0$ for $x_0$, $y_0$, and $z_0$.

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