# Thread: tangent hyperplanes and tanget planes, a bit confused about dimensions

1. ## tangent hyperplanes and tanget planes, a bit confused about dimensions

I thought I understood tangent planes until I saw tangent hyperplane and 3-d contour here. So a little bit of help will be most welcome. (i) and (ii) is no problem and agreed to the textbook answer. I am a bit stuck on (iii).

Question (Calculus, Binmore and Davies, 3.9(#5*))

Let $\displaystyle f(x,y,z)=ln(x^2+y^2+z^2)$

i. Find the gradient of f at the point $\displaystyle (x,y,z)^T$

ii. Find the tangent hyperplane to the hypersurface $\displaystyle u=ln(x^2+y^2+z^2)$
where $\displaystyle (x,y,z,u)^T=(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T$.

iii. Find the normal and the tangent plane to the contour

$\displaystyle ln(x^2+y^2+z^2)=0$ at $\displaystyle (1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T$.

i. $\displaystyle \nabla{f}=\frac{1}{x^2+y^2+z^2}(2x,2y,2z)^T$

ii. To find tangent hyperplane, I want to use the formula

$\displaystyle u-U=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z)+(z-Z)f_z(x,y,z)$

$\displaystyle u-0=(x-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(y-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(z-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}$

$\displaystyle u=\frac{2}{\sqrt{3}}(x+y+z)-2$

iii. To me, the equation for the countour $\displaystyle ln(x^2+y^2+z^2)=0$ looks exactly the same as the equation to the surface in (ii) as u=0 in both cases.

I understand in (ii) the hypersurface is in 4 dimentional space, and the hyperplane is in 3 dimensions. Here the contour is in 3 dimensions (??) and the tangent plane is 2-dimentional.

???

I am not sure what to do, but I was thinking may be try to express z as the function of x and y and apply the same template as in (ii) ie

$\displaystyle z-Z=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z)$

???

2. I am not sure what to do, but I was thinking may be try to express z as the function of x and y and apply the same template as in (ii)
Actually, had a better idea: the contour is f(x,y,z)=0 so the tangent plane to the countour is

$\displaystyle (x-X)f_x+(y-Y)f_y+(z-Z)f_z=0$

so that I get $\displaystyle \frac{2}{\sqrt{3}}(x+y+z)-2=0$ and $\displaystyle x+y+z=\sqrt{3}$ as per the answer in the book.

Now I need to figure out the way to find the normal vector. BUT it's probably the coefficients of x, y and z in the above equation, right? So, the normal vector is $\displaystyle (1,1,1)^T$.

3. Originally Posted by Volga
I thought I understood tangent planes until I saw tangent hyperplane and 3-d contour here. So a little bit of help will be most welcome. (i) and (ii) is no problem and agreed to the textbook answer. I am a bit stuck on (iii).

Question (Calculus, Binmore and Davies, 3.9(#5*))

Let $\displaystyle f(x,y,z)=ln(x^2+y^2+z^2)$

i. Find the gradient of f at the point $\displaystyle (x,y,z)^T$

ii. Find the tangent hyperplane to the hypersurface $\displaystyle u=ln(x^2+y^2+z^2)$
where $\displaystyle (x,y,z,u)^T=(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T$.

iii. Find the normal and the tangent plane to the contour

$\displaystyle ln(x^2+y^2+z^2)=0$ at $\displaystyle (1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T$.

i. $\displaystyle \nabla{f}=\frac{1}{x^2+y^2+z^2}(2x,2y,2z)^T$

ii. To find tangent hyperplane, I want to use the formula

$\displaystyle u-U=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z)+(z-Z)f_z(x,y,z)$

$\displaystyle u-0=(x-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(y-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(z-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}$

$\displaystyle u=\frac{2}{\sqrt{3}}(x+y+z)-2$

iii. To me, the equation for the countour $\displaystyle ln(x^2+y^2+z^2)=0$ looks exactly the same as the equation to the surface in (ii) as u=0 in both cases.
No, they are not the same thing though the difference is subtle. Consider the lower dimension case of $\displaystyle z= x^2+ y^2- 1$. That describes a two dimensional surface in a three dimensional space. It has a tangent plane and a normal vector at any given point and, in particular at (1, 0, 0). There, the normal vector is given by $\displaystyle 2\vec{i}- \vec{k}$ and the tangent plane is given by $\displaystyle \frac{x- 1}{2}- z= 0$ or $\displaystyle z= \frac{1}{2}x- \frac{1}{2}$. We can get that normal vector by thinking of the surface $\displaystyle z= x^2+ y^2- 1$ as a "level surface" of the function $\displaystyle \phi(x, y, z)= x^2+ y^2- z= 1$ and taking the gradient: $\displaystyle \nabla \phi= 2x\vec{i}+ 2y\vec{j}- \vec{k}$, which, at (1, 0, 1), is $\displaystyle 2\vec{i}- \vec{k}$.

But the graph of $\displaystyle x^2+ y^2- 1= 0$ is a circle in the xy-plane. it is, in fact, the intersection of the xy-plane with the paraboloid $\displaystyle z= x^2+ y^2- 1$. But because it is in the xy-plane, any normal to it at a given point, say (1, 0) must be in the xy-plane which was not true of the normal vector to the parabloid at (1, 0, 0). Now, instead of thinking of the surface z= x^2+ y^2- 1[/tex] as a level surface of the function $$\phi(x, y, z)= x^2+ y^2- z[tex], we think of it as a as a "level curve" of \displaystyle z(x,y)= x^2+ y^2- 1. The (two-dimensional) gradient of that is \displaystyle \nabla z= 2x\vec{i}+ 2y\vec{j} which, at (1, 0), is \displaystyle 2\vec{i}. A tangent line would be x= 1. I understand in (ii) the hypersurface is in 4 dimentional space, and the hyperplane is in 3 dimensions. Here the contour is in 3 dimensions (??) and the tangent plane is 2-dimentional. ??? I am not sure what to do, but I was thinking may be try to express z as the function of x and y and apply the same template as in (ii) ie \displaystyle z-Z=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z) ??? 4. Now, instead of thinking of the surface z= x^2+ y^2- 1$$ as a level surface of the function $\displaystyle \phi(x, y, z)= x^2+ y^2- z$, we think of it as a as a "level curve" of $\displaystyle z=x^2+y^2-1$.
So, in other words, we take a 'slice' of it, to go one dimension lower, by fixing z as a constant, and hence we find ourselvers in xy-plane? So that the grad vector 'loses' z-derivative because z is no longer variable?

Thanks for your explanation! By the way, as a practice, if I want to calculate normal vector in my example of hypersurface in (ii), I think it will be 4x1 vector as follows:

$\displaystyle u=(\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{ \sqrt{3}},-1)(x-\frac{1}{\sqrt{3}},y-\frac{1}{\sqrt{3}},z-\frac{1}{\sqrt{3}},u)^T$ thus giving the normal vector of

$\displaystyle (\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{\s qrt{3}},-1)^T$?

Question: does the size (length) of the 'normal' vector matter? if what is important is the direction only (eg such normal vector can define a plane), then I can say the vector is $\displaystyle k*(\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{ \sqrt{3}},-1)^T$ where k>0? Or the size does matter?

5. No. The point of a "normal vector" is that it is perpendicular to the surface. Only its direction is relevant. Any multiple of a normal vector is still a normal vector. Sometimes, you might be instructed to find the unit normal vector specifically to avoid any depence upon length. Of course, there are still two unit normal vectors to a surface, in opposite directions.

There are special cases in which the length of a normal vector might be important. For example, if you are integrating a vector field over a surface, you would want the length of the normal vector to reflect the "differential of surface area". You can get that by using the "fundamental vector product" for the surface.