2
$\begingroup$

I was having a problem finding the points on $z=3x^2 - 4y^2$ where vector $n=<3,2,2>$ is normal to the tangent plane.

How do we calculate the tangent plane equation without a specific point to calculate it at?

I also had an idea to take the cross product of $2$ vectors in the plane and somehow compare it to the $n$ vector but I don't know exactly how to do this. Thank you for any help!

$\endgroup$
  • $\begingroup$ I don't think that is exactly what the OP is asking! $\endgroup$ – user247327 Nov 7 '18 at 15:17
4
$\begingroup$

Let $f(x,y,z)=3x^2-4y^2-z$. Then your surface is $\bigl\{(x,y,z)\in\mathbb{R}^3\,|\,f(x,y,z)=0\bigr\}$. You are after the points $(x,y,z)$ in that surface such that $\nabla f(x,y,z)$ is a multiple of $(3,2,2)$. So, solve the system$$\left\{\begin{array}{l}6x=3\lambda\\-8y=2\lambda\\-1=2\lambda\\3x^2-4y^2-z=0.\end{array}\right.$$

$\endgroup$
  • $\begingroup$ Just wanted to clarify for others that I used the bottom-most equation after getting x and y from the top 2 equations and lambda from the third one. Thank you for the answer!! $\endgroup$ – sjfklsdafjks Nov 7 '18 at 15:55
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Nov 7 '18 at 16:06
3
$\begingroup$

Let $ f(x,y,z)=3x^2 - 4y^2 - z$ then the normal vector at $p_0(x_0,y_0,z_0)$ is $$\nabla f(p)=(f_x,f_y,f_z)_p$$ or $$\nabla f(p)=(6x_0,-8y_0,-1)$$ then $$\dfrac{\nabla f(p)}{|\nabla f(p)|}=\dfrac{\vec{n}}{|\vec{n}|}$$

$\endgroup$
  • $\begingroup$ Be aware that the signs of the vectors could differ! $\endgroup$ – weee Nov 7 '18 at 15:19
  • $\begingroup$ yes, equality with a $\pm$. $\endgroup$ – Nosrati Nov 7 '18 at 15:21
  • $\begingroup$ And then how can we calculate the tangent plane? $\endgroup$ – manooooh Nov 7 '18 at 15:21
  • $\begingroup$ $p_0=(-\frac14,-\frac18,\frac18)$. $\endgroup$ – Nosrati Nov 7 '18 at 15:23
0
$\begingroup$

The problem does not ask you to find a tangent plane! It asks you to find points where the normal vector is parallel to <3, 2, 2>. The normal vector at any point of f(x,y,z)= constant is $\nabla f$. Here $f(x, y, z)= 3x^2- 4y^2- z= 0$. Find $\nabla f$ and set it equal to <3k, 2k, 2k> for some k.

$\endgroup$
0
$\begingroup$

The other answers already covered the basics: you don’t need to find any tangent planes per se, but only points at which the normal to the surface is parallel to $n$. Since you’re working in $\mathbb R^3$, you have a bit of a short cut available: two nonzero vectors are parallel iff their cross product vanishes. Thus, you can avoid introducing another variable by stating the condition in the problem as $\nabla F\times n=0$, where $F:(x,y,z)\mapsto 3x^2-4y^2-z$. This generates three equations (only two of which are independent) to solve together with the original implicit equation of the surface.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.