Calculus 3 help -- Is the gradient of a plane the normal to the plane?

[Mathematicsss]

Homework Statement

Is the gradient of a plane, the normal to the plane? If so, why?

Homework Equations

No idea, just a question that popped up in my head

eon of plane: n(x-x1)+n(y-y1)+n(z-z1)

The Attempt at a Solution

I found the partial derivative of each, and got the normal.[/B]

 

[LCKurtz]

Homework Statement

Is the gradient of a plane, the normal to the plane? If so, why?

Homework Equations

No idea, just a question that popped up in my head

eon of plane: n(x-x1)+n(y-y1)+n(z-z1)

The Attempt at a Solution

I found the partial derivative of each, and got the normal.[/B]

That is a general property of gradients. The gradient of $f(x,y,z)=C$ is normal to the surface given by that equation.

 

[Tio Barnabe]

It's relatively easy to show and I remembering being taught about it in a Calculus 2 class.

Let $f : \mathbb{R}^3 \longrightarrow \mathbb{R}$, parametrize the corresponding region in $\mathbb{R}^3$ by $t \in [a,b] \longmapsto (x(t),y(t),z(t)) \equiv r(t)$ and consider the cases where $f(x(t),y(t),z(t))$ is a constant $c$ for some $t = T$. Then $f(x(T),y(T), z(T)) = c$ and the corresponding points are $r(T)$. Now take the total derivative of $f(x(t),y(t),z(t))$ and evaluate it when $t = T$:

$$\frac{df(x(t),y(t),z(t)}{dt}|_T = \frac{\partial f}{\partial x}|_{r(T)} \frac{dx}{dt}|_{T} + \frac{\partial f}{\partial y}|_{r(T)} \frac{dy}{dt}|_{T} + \frac{\partial f}{\partial z}|_{r(T)} \frac{dz}{dt}|_{T} = \frac{dc}{dt}|_{T} = 0$$ This is in vector notation $\nabla f(x,y,z)|_{r(T)} \cdot r'(t)|_{T} = 0$, which shows that grad of $f$ is orthogonal to the tangent vector to any surface $f(x(t),y(t),z(t)) = c$.

In general these surfaces will not form in $\mathbb{R}^3$ what you probably have in mind as a plane, but special cases could do.

 

[Mark44]

eon of plane: n(x-x1)+n(y-y1)+n(z-z1)

Minor point, but this is not an equation. An equation has an "equals" symbol (=) between two expressions.

 

[LCKurtz]

And the coefficients of the plane likely wouldn't all be the same.

 

[CollinsArg]

If I'm not mistaken, the gradien of an equation is normal to the level of superficies/curves of that function, (that is, a dimension less than the original function), but the gradient of a function is not normal to the function where it comes from, as a gradient will be one dimension smaller than it.
So, the answer is no, but it is normal to the stright lines you can get from that plane (in a lower dimension).

 

[LCKurtz]

If I'm not mistaken, the gradien of an equation is normal to the level of superficies/curves of that function, (that is, a dimension less than the original function), but the gradient of a function is not normal to the function where it comes from, as a gradient will be one dimension smaller than it.
So, the answer is no, but it is normal to the stright lines you can get from that plane (in a lower dimension).

I'm not sure if you are replying to my post #2, but your point that there is sloppy notation in this thread is well taken. Technically, we shouldn't talk about the gradient of an equation as you did in your first line and as I did in post #2. We should talk about the gradient of a function, for example the gradient of $f(x,y,z)$. What you get is a vector field normal to the level surfaces given by $f(x,y,z) = C$. And we shouldn't say the gradient is normal to the function $f(x,y,z)$ either.