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 November 10th, 2016, 10:00 AM #1 Member   Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2 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.
 November 11th, 2016, 04:39 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 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.
November 11th, 2016, 02:59 PM   #3
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 Originally Posted by Country Boy 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?

 November 12th, 2016, 07:14 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 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.
 November 12th, 2016, 07:51 AM #5 Member   Joined: Oct 2016 From: Sverige Posts: 32 Thanks: 2 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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