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June 10th, 2017, 03:37 AM   #1
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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.
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June 10th, 2017, 04:03 AM   #2
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Well, lets see. The gradient to the surface, at point is so the tangent plane to the surface at that point is .

In order that it contain the point A(0, 0, -1) we must have .

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 .

And, of course, since is a point on the surface, it must satisfy the equation of the surface. We must have .


Solve the three equations

and

for , , and .
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June 10th, 2017, 05:46 AM   #3
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Quote:
Originally Posted by Country Boy View Post
Well, lets see. The gradient to the surface, at point is so the tangent plane to the surface at that point is .

In order that it contain the point A(0, 0, -1) we must have .

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 .

And, of course, since is a point on the surface, it must satisfy the equation of the surface. We must have .


Solve the three equations

and

for , , and .
Okay, I think I got it. Thank you
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June 10th, 2017, 11:26 AM   #4
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Quote:
Originally Posted by Country Boy View Post
Well, lets see. The gradient to the surface, at point is so the tangent plane to the surface at that point is .

In order that it contain the point A(0, 0, -1) we must have .
This is wrong. I meant to say

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 .

And, of course, since is a point on the surface, it must satisfy the equation of the surface. We must have .


Solve the three equations

and

for , , and .
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