# Thread: Equation of a plane

1. ## Equation of a plane

Find the equation of the plane normal to $r(t) = $ when $t = 1$.

Sorry not sure how to display pi.

Where do I start?

2. Originally Posted by kenshinofkin
Find the equation of the plane normal to $r(t) = $ when $t = 1$.

Sorry not sure how to display pi.

Where do I start?
A vector tangent to r(t) is dr/dt. Evaluate at t = 1 and you have the normal vector to the plane.

r(1) gives the position vector of a point in the plane.

So you have a normal to the plane and a point in the plane.

The equation of a plane is ax + by + cz = d where <a, b, c> is a normal to the plane and d is found by substituting a known point into ax + by + cz = d (once you've substituted the values of a, b and c of course).

3. I'm assuming that $\textbf{r}(t)=\left$ is meant to represent the x,y and z co-ordinates (respectively).
If that is the case then, when t=1, we have $\textbf{r}(t)=\left$.
Recall the dot product - $\textbf{a}.\textbf{b}=|\textbf{a}||\textbf{b}|\cos (\theta)$, where $\theta$ is the angle between the two vectors.
Note that, if two vectors are orthogonal then $\theta=\pi/2$ and so $\textbf{a}.\textbf{b}=|\textbf{a}||\textbf{b}|\cos (\pi/2)=0$ (as the cosine of $\pi/2$ is zero).
Hence, you want to find another vector, call this $\textbf{b}=\left$ and find $\textbf{r}(1).\textbf{b}=0$.
In other, words, you are looking for solutions to the equation $eb_1+0*b_2+b_3=eb_1+b_3=0$. This implies that $b_1/b_3=-e$ and hence there are infinitely many solutions of this form

Also, as a sidenote, if t is not constant, you get a really interesting kind of geometry based on spirals. It 'looks' a little bit like making your way from the top of a trumpet to the bottom (or vice versa, depending) only going around the inside of the horn (I think that's what it's called)

Hope that helps

Keith

4. Originally Posted by mr fantastic
A vector tangent to r(t) is dr/dt. Evaluate at t = 1 and you have the normal vector to the plane.

r(1) gives the position vector of a point in the plane.

So you have a normal to the plane and a point in the plane.

The equation of a plane is ax + by + cz = d where <a, b, c> is a normal to the plane and d is found by substituting a known point into ax + by + cz = d (once you've substituted the values of a, b and c of course).
So if I understand this right, I take the the derivative of the vector and evaluate it at t=1 then evaluate just the vector its self at t=1. <a,b,c> will be the derivative of the vector evaluated at t=1 and the point I get from the vector I substutute for (x,y,z). Is this correct?

5. Originally Posted by kenshinofkin
So if I understand this right, I take the the derivative of the vector and evaluate it at t=1 then evaluate just the vector its self at t=1. <a,b,c> will be the derivative of the vector evaluated at t=1 and the point I get from the vector I substutute for (x,y,z). Is this correct?
Yes.