# How do I find a tangent plane without a specified point?

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!

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

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$$\left\{\begin{array}{l}6x=3\lambda\\-8y=2\lambda\\-1=2\lambda\\3x^2-4y^2-z=0.\end{array}\right.$$

• 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!! – sjfklsdafjks Nov 7 '18 at 15:55
• I'm glad I could help. – José Carlos Santos Nov 7 '18 at 16:06

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 $$\dfrac{\nabla f(p)}{|\nabla f(p)|}=\dfrac{\vec{n}}{|\vec{n}|}$$

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

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.

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.