Determining if a parametric curve intersects a plane at a right angle

Question

Given a plane described by $$ x + 8y + 12z = 162 $$ and a curve described by $$ \begin{align} x &= t\\ y &= t^2\\ z &= t^3 \end{align} $$ I want to find out if the curve intersects the plane at a right angle.

I started by substituting $x$, $y$, and $z$ into the equation for the curve and solving for $t$ which would find the time at which the intersection occurs. After this, I'm not entirely sure how I would go about solving it. I need a vector that describes the direction that the curve would be pointing at that time, so that I could then take the cross product with the normal vector of the plane, $\langle1,8,12\rangle$. Another issue with this method is that it requires the solution to a cubic equation which is not trivial on paper. Is there a better method to go about solving this?

Answer 1

Once you have found the time $t_0$ of intersection the derivative with respect to time at $t_0$ will well you the direction is is moving. Since $$r(t)=\begin{pmatrix}t\\t^2\\t^3\end{pmatrix}\text{ we have }r'(t)=\begin{pmatrix}1\\2t\\3t^2\end{pmatrix}$$ and now it's a matter of plugging in the intersection point.

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But since you only want to check if it's orthogonal you could work in reverse and find the times $t$ at which the directional vector is perpendicular to the plane (this will give you a quadratic equation) and check if these are intersection points.

Answer 2

The normal vector to the plane is [1,8,12]. The tangent vector to the curve is $[1,2t,3t^2]$.Note that neither of these vectors is the $\mathbf 0$ vector. So the curve intersects the plane at right angles iff these vectors are scalar multiples of each other iff they are linearly dependent iff their cross-product is $\mathbf 0$ at the particular value of $t$ where they intersect.