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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 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 . 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 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  June 10th, 2017, 11:26 AM   #4
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Quote:
 Originally Posted by Country Boy 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 . Tags plane, surface, tangent Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Addez123 Calculus 4 November 12th, 2016 06:51 AM xl5899 Calculus 17 November 1st, 2016 10:28 AM harley05 Calculus 1 May 11th, 2014 01:23 PM r-soy Physics 4 April 23rd, 2012 08:27 AM milko Algebra 1 August 3rd, 2008 07:14 PM

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