# Math Help - Equation of plane question.

1. ## Equation of plane question.

Its been a little while since I've done these ones and I'm having a little trouble wrapping my head around this one.

Question:
Find the equation of the plane perpendicular to the line tangent to r(t)=(3sint)i - (2cost)j + (t)k at t=pi/2.

Would the normal vector to the plane simply be the above vector? But then would I use (0,0,0) as a point in the plane?

2. Originally Posted by dmbocci
Its been a little while since I've done these ones and I'm having a little trouble wrapping my head around this one.

Question:
Find the equation of the plane perpendicular to the line tangent to r(t)=(3sint)i - (2cost)j + (t)k at t=pi/2.

Would the normal vector to the plane simply be the above vector? No.

But then would I use (0,0,0) as a point in the plane? No.

The line tangent to $\vec{\text{r}}(t)$ is in the direction of $\displaystyle {{d}\over{dt}}\vec{\text{r}}(t)$.

Use the position specified by $\vec{\text{r}}(\pi/2)$ as a point in the plane.

3. Originally Posted by dmbocci
Its been a little while since I've done these ones and I'm having a little trouble wrapping my head around this one.

Question:
Find the equation of the plane perpendicular to the line tangent to r(t)=(3sint)i - (2cost)j + (t)k at t=pi/2.

Would the normal vector to the plane simply be the above vector? But then would I use (0,0,0) as a point in the plane?
Dear dmbocci,

First find the unit tangent vector to the curve.

$\underline{T}=\dfrac{d\underline r}{ds}=\dfrac{d\underline r}{dt}\times\dfrac{dt}{ds}=\dfrac{\underline{r'}(t )}{\mid\underline{r'}(t)\mid}$

Then find the unit normal using,

$\dfrac{d\underline{T}}{ds}=\kappa\underline N$

The vector equation of the plane perpendicular to the tangent line would then be,

$\left(\underline{R}-\underline{r}\left(\frac{\pi}{2}\right)\right).\un derline N=0$ where $\underline R$ represent the position vector of any point in the plane.

Hope this helps you to solve the problem.

4. Thanks guys, great help!