Tangent Plane

  • Thread starter manenbu
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  • #1
manenbu
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Homework Statement



Find the tangent plane to the surface x2 + 2y2 + 3z2 = 21, which is parallel to the plane x+4y+6z=0

Homework Equations



Gradients.

The Attempt at a Solution


Here is my solution:
So gradient of the surface is (2x, 4y, 6z). The normal of the given plane is (1,4,6). Equating both gives us x=0.5, y=1, z=1.
So this is the point where the plane meets the surface.
Finding the equation:
1(x-0.5)+4(y-1)+6(z-1) = x - 0.5 + 4y - 4 + 6z - 6
Equation for the plane is: x + 4y + 6z = 10.5

The answer given is x + 4y + 6z = ±21. Where did I go wrong? What did I forget to multiply? And why the two solutions (positive and negative)?

Thanks!
 

Answers and Replies

  • #2
36,424
8,405

Homework Statement



Find the tangent plane to the surface x2 + 2y2 + 3z2 = 21, which is parallel to the plane x+4y+6z=0

Homework Equations



Gradients.

The Attempt at a Solution


Here is my solution:
So gradient of the surface is (2x, 4y, 6z). The normal of the given plane is (1,4,6). Equating both gives us x=0.5, y=1, z=1.
So this is the point where the plane meets the surface.
The point (.5, 1, 1) is not a point on the quadric surface, so can't be the point where the plane intersects the surface.
Finding the equation:
1(x-0.5)+4(y-1)+6(z-1) = x - 0.5 + 4y - 4 + 6z - 6
Equation for the plane is: x + 4y + 6z = 10.5

The answer given is x + 4y + 6z = ±21. Where did I go wrong? What did I forget to multiply? And why the two solutions (positive and negative)?

Thanks!
The reason for the two equations for planes is that there are two points on the quadric surface that satisfy the given conditions. The surface is an ellipsoid.
 
  • #3
manenbu
103
0
So how should I do it? I tried finding the point - but no success/
 
  • #4
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
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In your calculation, you used the following step:
If two vectors are parallel, then they are equal.​
Do you believe this to be a theorem? If not, then what statement is a theorem? And how does your calculation change if you use that statement instead?
 
  • #5
manenbu
103
0
If two vectors are parallel, then they have the same direction (unit vector). Or maybe, one is a linear combination of the other.
I still can't see though how this relates to my problem.
 
  • #6
36,424
8,405
If two vectors are parallel, then they have the same direction (unit vector). Or maybe, one is a linear combination of the other.
I still can't see though how this relates to my problem.
If two vectors are parallel, then one is a scalar multiple of the other.
 
  • #7
manenbu
103
0
ok got it!
so the gradient was (2x, 4y, 6z). The normal to the plane is (1,4,6).
both lines are parallel, so:
a(2x,4y,6z)=(1,4,6)
from here we get:
x=0.5a y=a z=a
we need to find the a which satisfies the ellipsoid equation:
x2 + 2y2 + 3z2 = 21
(0.5a)2 + 2a2 +3a2 = 21
21a2 = 84
a2 = 4
a = ±2
so we got 2 sets of points where the required plane will intersect:
one is (1, 2, 2) the other is -(1, 2, 2) - makes sense - two opposite sides of the ellipsoid.
finding the equation of the plane is easy now:
1(x-1)+4(x-2)+6(z-2) and 1(x+1)+4(x+2)+6(z+2)
which gives the two required planes.

thanks everyone!
 

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