# Finding the point where C interesects the xz-plane

1. Oct 12, 2005

### mr_coffee

Hello everyone, this problem has several steps and i'm studying for an exam, so i need to get all of them! The first part is:
Let C be the curve with equations x = 2-t^3; y = 2t-1; z = ln(t);
Find the point where C intersects xz-plane. So i said let y = 0; and i'd get
Po = (z-t^3,0,ln(t)), but i don't t hink this si right because isn't x, y, and z unit vectors? like <2-t^3,2t-1,ln(t)>?

So once i find this, i'm suppose to find the parametric equations of the tagnent line at (1,1,0); then find an equation fo the normaml plane to C at (1,1,0); I think if i can get the first part i can figure out the rest! thanks!

2. Oct 12, 2005

### sqrt(-1)

Using the given parameterisation we have a curve with position vector

$$\mathbf{\rm{r}}(t) = (2 - t^3)\mathbf{\rm{i}} + (2t - 1)\mathbf{\rm{j}} + \ln(t)\mathbf{\rm{k}}$$

Clue : If y = 0 then what value of t should you use to find the point of intersection?