Coordinates on a parametric curve

Question

A curve is defined by the parametric equation $(x, y, z)$ $=$ $(-2 + 3t, 1 + 3t^2, 2t-3t^3)$. There is a unique point $P$ on the curve with the property that the tangent line at $P$ passes through the point $(-8, 10, 8)$. What are the coordinates of $P$?

So I tried to find the equation of the tangent line first. So what I did was subtract the point $(-8, 10, 8)$ from the general equation of the line. This gave me $(-6-3t, 9-3t^2, 8-2t+3t^3)$. The derivative for the general equation of the line is $(3 , 6t, 2-9t^2)$. So I seem to understand that the tangent line is parallel, and therefore a scalar multiple, of the equation of the derivative. But I can't seem to see what the relationship is and how that can help me get the required coordinates.

Any help?

Answer 1

If the tangent line passes through a fixed point, the cross product of that line and the vector between the fixed point and the tangency point is the zero vector. Thus $$(-6-3t, 9-3t^2, 8-2t+3t^3)×(3, 6t, 2-9t^2)=(0,0,0)$$ Expanding out the LHS gives a vector whose components are, respectively, a quartic, cubic and quadratic. We only need the quadratic to derive possible values for $t$: $$-9t^2-36t-27=0$$ $$t^2+4t+3=0\qquad t=-1\lor t=-3$$ Substituting $t=-3$ into the other components of the cross product does not make them zero, but substituting $t=-1$ does, so it is the correct value and $$P=(-2 + 3t, 1 + 3t^2, 2t-3t^3)|_{t=-1}=(-5,4,1)$$

Answer 2

Note that $\vec{r'}=(3,6t,2-9t^2)$ is defined by $\vec t=\lambda\vec{r'}\text{, }\lambda \in\ \Bbb R$

So, $$\vec P+\vec t=(-8,10,8)$$$$ (-2+3t,1+3t^2,2t-3t^3)+(3\lambda,6\lambda t,2\lambda-9\lambda t^2)=(-8,10,8)$$

$$\implies-2+3t+3\lambda=-8$$ $$3(t+\lambda)=-6$$ $$t=-2-\lambda$$ Similarly, $$1+3t^2+6\lambda t=10$$and$$2t-3t^3+2\lambda-9\lambda t^2=8$$

Can you take it from here?