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November 10th, 2016, 10:00 AM   #1
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Find all points on surface whose tangent plane is parallel to given plane

This is a common question found online yet I just can't seem to get it right myself.

The surface:$\displaystyle x^2 + y^2 - z^2 - 14 = 0$
The plane: $\displaystyle x-2y+3z - 5 = 0$

Find all points on surface that are parallel with the plane.
I try find the gradient of the surface, which is: $\displaystyle <2x, 2y, -2z>$
This gradient should be parallel with the normal of the plane, which is $\displaystyle <1,-2,3>$

Now let's find the points that gives me this normal.
2x = 1 => x = 1
2y = -2 => y = -1
-2z = 3 => z = -3/2
So the point is:$\displaystyle (1,-1,-3/2)$

This is wrong, the answer is +-sqrt(5/14)*(1,-2,3)
Where are they pulling these numbers?

Last edited by skipjack; November 11th, 2016 at 03:26 PM.
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November 11th, 2016, 04:39 AM   #2
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You were asked to find a point on the surface where the tangent plane is parallel to the given plane, not the same as the tangent plane.

If x= -1, y= -2, z= -3/2, then $\displaystyle x^2+ y^2- z^2= 1+ 4+ 9/4= (4+ 16+ 9)/4= 29/4$ not 14. The point you found is not on the surface.

Your mistake was setting the two vectors equal instead of just parallel. Yes, a normal to the surface is of the form <2x, 2y, -2z> and a vector normal to the plane is <1, -2, 3>. So we must have <2x, 2y, -2z>= a<1, -2, 3> for some number a. We must also have $\displaystyle x^2+ y^2- z^2= 14$.

So we have the four equations
2x= a
2y= -2a
-2z= 3a
$\displaystyle x^2+ y^2- z^2= 14$
to solve for the four unknowns, x, y, z, and a.

Since you are asked only to find x, y, and z, not a, the first thing I would do is divide one equation by another to eliminate a.
Dividing the second equation by the first gives $\displaystyle \frac{y}{x}= -2$ so y= -2x. Dividing the third equation by the first gives $\displaystyle -\frac{z}{x}= 3$ so [math]z=-3x[/tex]. Put those into $\displaystyle x^2+ y^2- z^2= 14$ and solve for x.
Thanks from Addez123

Last edited by Country Boy; November 11th, 2016 at 04:49 AM.
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November 11th, 2016, 02:59 PM   #3
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Quote:
Originally Posted by Country Boy View Post
You were asked to find a point on the surface where the tangent plane is parallel to the given plane, not the same as the tangent plane.

If x= -1, y= -2, z= -3/2, then $\displaystyle x^2+ y^2- z^2= 1+ 4+ 9/4= (4+ 16+ 9)/4= 29/4$ not 14. The point you found is not on the surface.

Your mistake was setting the two vectors equal instead of just parallel. Yes, a normal to the surface is of the form <2x, 2y, -2z> and a vector normal to the plane is <1, -2, 3>. So we must have <2x, 2y, -2z>= a<1, -2, 3> for some number a. We must also have $\displaystyle x^2+ y^2- z^2= 14$.

So we have the four equations
2x= a
2y= -2a
-2z= 3a
$\displaystyle x^2+ y^2- z^2= 14$
to solve for the four unknowns, x, y, z, and a.

Since you are asked only to find x, y, and z, not a, the first thing I would do is divide one equation by another to eliminate a.
Dividing the second equation by the first gives $\displaystyle \frac{y}{x}= -2$ so y= -2x. Dividing the third equation by the first gives $\displaystyle -\frac{z}{x}= 3$ so $\displaystyle z=-3x$. Put those into $\displaystyle x^2+ y^2- z^2= 14$ and solve for x.
Your analysis is correct, but doing what you suggest leads to $\displaystyle -4x^2=14$, so x is imaginary. Could the original surface definition be incorrect?
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November 12th, 2016, 07:14 AM   #4
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I did not complete that calculation but what that result means is that there is no place on the surface where the tangent plane is parallel to the given plane.
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November 12th, 2016, 07:51 AM   #5
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I wrote it wrong here, the equation was
x^2 + y^2 + z^2 = 5

But with that help you gave me I managed to solve it and get correct answer. Thanks man!
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