Calculus with Vector Functions — Question 6

Question 6 – Solution

Strategy A tangent is parallel to a plane when its direction is orthogonal to the plane normal; it is perpendicular when its direction is parallel to the normal.

Parallel condition $\vec r'(t)=\left\langle 1,2t,3t^2-3\right\rangle$ and $\vec n=\left\langle 2,-1,1\right\rangle$. Thus $$\vec r'(t)\cdot\vec n=3t^2-2t-1=(3t+1)(t-1).$$ So $t=1$ or $t=-1/3$, giving points $\boxed{(1,1,-2)}$ and $\boxed{(-1/3,1/9,26/27)}$.

Perpendicular condition Parallelism of $\vec r'(t)$ with $\vec n$ would force scale $1/2$ from the first component, then $2t=-1/2$, but the third component at $t=-1/4$ is not $1/2$. Hence none exist.