# Normal Plane help

• Jun 4th 2010, 11:19 PM
Warrenx
Normal Plane help
I'm studying for a final, and on my last midterm there was this question:

At what point on the following curve C is the normal plane parallel to 6x + 6y -8z =1, for C: <t^3, 3t, t^4>?

What do I do for this?

Thanks,

Warren
• Jun 4th 2010, 11:42 PM
ANDS!
The normal plane is the plane at a point P on the curve C that consists of all lines that are orthogonal to the tangent vector. Because of that property, the tangent vector is a perfect vector to chose as the "normal vector" of our normal plane. . .(Headbang) Hopefully you got all that.

Therefore, to find the tangent vector you simply take the derivative of your curve C, and then plug in your point P:

$\displaystyle T(t) = <3t^{2}, 3, 4t^{3}>$

We know the normal vector to our normal plane must be parallel to the plane 6x+6y-8z=1, which means our normal vector should be <6, 6, -8>.

See if you can take where it from here.
• Jun 5th 2010, 01:34 AM
Warrenx
Quote:

Originally Posted by ANDS!
The normal plane is the plane at a point P on the curve C that consists of all lines that are orthogonal to the tangent vector. Because of that property, the tangent vector is a perfect vector to chose as the "normal vector" of our normal plane. . .(Headbang) Hopefully you got all that.

Therefore, to find the tangent vector you simply take the derivative of your curve C, and then plug in your point P:

$\displaystyle T(t) = <3t^{2}, 3, 4t^{3}>$

We know the normal vector to our normal plane must be parallel to the plane 6x+6y-8z=1, which means our normal vector should be <6, 6, -8>.

See if you can take where it from here.

Ok, so do I just relate the parametric equations to the normal vector of the plane? I don't get how that would work with the 3 though, er... I have no idea what I am doing (Headbang).
• Jun 5th 2010, 07:27 AM
ANDS!
Here is a hint: How does <6,6,-8> relate to <3,3,-4>?
• Jun 5th 2010, 11:24 AM
zzzoak
ANDS! wants you to solve the equation

$\displaystyle T(t)=(3t^2,3,4t^3)=A(6,6,-8)$
where A=const.

• Jun 5th 2010, 11:54 AM
Warrenx
Quote:

Originally Posted by zzzoak
ANDS! wants you to solve the equation

$\displaystyle T(t)=(3t^2,3,4t^3)=A(6,6,-8)$
where A=const.

Ok, so now that i do that out and I get A=1/2, so then would I plug that into the vector and solve for the (t) of each one respectively, but I am confused on alot of things, I know that the normal plane and the given plane would be parallel when their normal vectors were parallel, and the normal vector of a normal plane is just the Tangent, but I don't conceptually see how I am supposed to find a tangent along a whole space curve that mirrors <6,6,-8> unless I say like: <6,6,-8> = r'(t)/[r'(t)]. But when I solve that I get a really funky number, for example:

r'(t) = 3(t^2)i + (3)j + 4(t^3)k

[r'(t)] = sqrt[9+9+16] = sqrt[34]

so I end up with

<6> = $\displaystyle 3/\sqrt(34)(3t^2)i$
<6> = (3)j
<-8> = $\displaystyle 4/\sqrt(34)(4t^3)k$

So I still don't get what I am doing wrong :(
• Jun 5th 2010, 12:33 PM
ANDS!
Err. My hind was that the vector <6,6,-8> is just a scaled version of <3,3,-4>, and we can use <3,3,-4> instead of <6,6,-8> (or you can equivalently just multiply T(t) by 2 to get $\displaystyle T(t)=<6t^{2},6,4t^{3}>$. It doesn't matter which way you do it.

Or course none of this matters iy you do not know WHY you are doing these things. You know that the vector function T(t) at the point P will be orthgonal to your normal plane, as would ANY vector that is parallel to the vector T(t) at point P. <3,3,-4> is parallel to <6,6,-8>, so you can use <3,3,-4> instead since that 3 matches up with the 3 in T(t).

But now what do you do? It is asking for a POINT, not a vector. We'll we know the T(t) is parallel to <3,3,-4> or $\displaystyle <3t^{2},3,2t^{3}>=<3,3,-4>$. See if you can take it from there.
• Jun 5th 2010, 01:20 PM
Warrenx
Quote:

Originally Posted by ANDS!
Err. My hind was that the vector <6,6,-8> is just a scaled version of <3,3,-4>, and we can use <3,3,-4> instead of <6,6,-8> (or you can equivalently just multiply T(t) by 2 to get $\displaystyle T(t)=<6t^{2},6,4t^{3}>$. It doesn't matter which way you do it.

Or course none of this matters iy you do not know WHY you are doing these things. You know that the vector function T(t) at the point P will be orthgonal to your normal plane, as would ANY vector that is parallel to the vector T(t) at point P. <3,3,-4> is parallel to <6,6,-8>, so you can use <3,3,-4> instead since that 3 matches up with the 3 in T(t).

But now what do you do? It is asking for a POINT, not a vector. We'll we know the T(t) is parallel to <3,3,-4> or $\displaystyle <3t^{2},3,2t^{3}>=<3,3,-4>$. See if you can take it from there.

O! ok I get it now, turns out I have the whole concept wrong, so <$\displaystyle 3t^2, 3, 4t^3>$ is just the position with respect to velocity, so f(t) =(x,y,z). and f'(t) = (x,y,z) with respect to velocity. So then all <$\displaystyle 3t^2, 3, 4t^3>$ is saying is that, at some x=3t^2 its velocity = 3, y=3 is constant, and at some z=4t^3 its velocity is -4. If this is right I think I get this now. All this after establishing that c is continuous, what plane is parallel to what plane, etc.