Finding the tangent line(s) to a curve in 3D parallel to a plane

Question

Given the curve $$r(t) = (1 − 2t)i + (t^2)j + (t^3/2)k, ~~ t > 0~.$$ Find a point on the curve at which the tangent line is parallel to the plane $$5x + y + z − 3 = 0~.$$

I have done everything including making the dot product of the normal of the plane and the direction of the tangent ($r'(t)$) equal to zero but it gives two values of $t$, are both values valid? How do I identify that?

Answer 1

The tangent vetor tothe curve is $\vec T= \frac{dx}{dt} \vec i+ \frac{dy}{dt} \vec j+ \frac{dz}{dt} \vec k$ the normal vector to the plane is $\vec N= 5\vec i+ \vec j +\vec k.$ The required points on the curve will be given by $\vec N. \vec T=0 \Rightarrow 3t^2+4t-20 =0$, so $t=2~\mbox{or}~ -10/3.$ So the required position vector of points are obtained by putting these two values of $t$ in $\vec r=(1-2t) \vec i +t^2 \vec j+t^3/2 \vec k$ we get $\vec r_1=-3 \vec i+ 4 \vec j + 4\vec k$ and $\vec r_2=(23/3) \vec i+ (100/9) \vec j -(500/27,) \vec k$.

Answer 2

By checking a necessary requirement for $t$, you've reduced the possible values of $t$ to a finite number of cases (two cases to be precise). The next step is now to check for every one of these values of $t$ whether the tangent at $r(t)$ is defined, and if so, indeed parallel to that plane.