# Thread: Find an equation of a plane normal to a given vector

1. ## Find an equation of a plane normal to a given vector

I have to find an equation for a plane normal to $r(t)=< e^t sin(\frac{\pi{}}2{t}), e^t cos(\frac{\pi{}}2{t}), t^2 >$
when $t=1$

Is the equation in the form:

$ . - r(1) = 0?$

would i take the derivative of x, y, and z?

Would this require the product rule for x, and y individually? not z of course!

here is my attempt at just the product rule of x alone:

$e^t d/dx(sin(\frac{\pi{}}2{t}) + sin(\frac{\pi{}}2{t}) . d/dx (e^t)$

would it then be?

$e^t cos(\frac{\pi{}}2{t}) + e^t sin(\frac{\pi{}}2{t})$

I guess my 2 questions are. is my generic form correct of the equation for a plane normal to the vector correct above?
and when i take the derivative do i use the product rule and did i do it correctly?

I am not sure how to get <a,b,c> and the point <x,y,z> either.

I am completely lost

Thank you

2. ## Re: Find an equation of a plane normal to a given vector

Originally Posted by icelated
I have to find an equation for a plane normal to $r(t)=< e^t sin(\frac{\pi{}}2{t}), e^t cos(\frac{\pi{}}2{t}), t^2 >$
when $t=1$
Is the equation in the form:
$ . - r(1) = 0?$

You have a great deal of that wrong.
The equation in the form:
$\nabla r(1) \cdot \left( {\left\langle {x,y,z} \right\rangle - r(1)} \right) = 0$

3. ## Re: Find an equation of a plane normal to a given vector

We haven't learned del yet.

4. ## Re: Find an equation of a plane normal to a given vector

Originally Posted by icelated
We haven't learned del yet.
You need the tangent (velocity) vector to the curve: $r'(t)$.
That is the normal of the plane. In this case it is $\left\langle {{x_t}(1),{y_t}(1),{z_t}(1)} \right\rangle$

5. ## Re: Find an equation of a plane normal to a given vector

Originally Posted by icelated
I have to find an equation for a plane normal to $r(t)=< e^t sin(\frac{\pi{}}2{t}), e^t cos(\frac{\pi{}}2{t}), t^2 >$
when $t=1$

Is the equation in the form:

$ . - r(1) = 0?$

would i take the derivative of x, y, and z?

Would this require the product rule for x, and y individually? not z of course!

here is my attempt at just the product rule of x alone:

$e^t d/dx(sin(\frac{\pi{}}2{t}) + sin(\frac{\pi{}}2{t}) . d/dx (e^t)$

would it then be?

$e^t cos(\frac{\pi{}}2{t}) + e^t sin(\frac{\pi{}}2{t})$
Not quite. The derivative of $sin(\frac{\pi}{2}t)$ is $\frac{\pi}{2}cos(\frac{\pi}{2}t)$. You forgot the $\frac{\pi}{2}$ multiplying.

I guess my 2 questions are. is my generic form correct of the equation for a plane normal to the vector correct above?
and when i take the derivative do i use the product rule and did i do it correctly?

I am not sure how to get <a,b,c> and the point <x,y,z> either.

I am completely lost

Thank you