Determine a parameterization for the line which is tangent to the curve at t=2

Question

(1) A curve is given by the function $$r(t)=(t^3 -3t^2 +2t +4)i + (13-5t)j +(t^2 -t-3)k$$

Determine a parameterization for the line which is tangent to the curve at $t=2$

I started by solving for when $t=2$, and got the vector $4i + 3j - k$

I don't know what this vector means or if this was even the correct approach. Do I need to find the vector perpendicular to this, or what should I be doing instead?

Answer 1

The tangent vector to the curve will always point in the direction of the tangent of the curve. We obtain the tangent vector by differentiating, it is given by

$$ \mathbf{T}(t) = \dot{\mathbf{r}}(t) = (3t^2 - 6t + 2)\mathbf{i} -5\mathbf{j} +(2t -1)\mathbf{k} $$

The direction of the line tangent to the curve at $t=2$ is then

$$ \mathbf{T}(2) = 2\mathbf{i} - 5\mathbf{j} + 3\mathbf{k} $$

Then the line is given by

$$ \mathbf{l}(\lambda) = \lambda \mathbf{T}(2) + \mathbf{k} $$ Where $\mathbf{k}$ is fixed by the condition that $\mathbf{l}$ intersects $\mathbf{r}$ at $t=2$, so $\mathbf{k} = \mathbf{r}(2) - 2\mathbf{T}(2)$.