Find pt of intersection of tangent lines to the foll. curve at pt t=0 and t=1/2?

Question

curve is - r(t) =

i got r'(0) = <π,3π,0>

and r'(1/2) = <0,0,-π>

what is my next step?

i forgot to add r(t)...

r(t) = <πcosπt, 3πcosπt, -πsinπt>

Answer 1

The tangent line at t=0 consists of all points of the form

r(0) + s*r'(0) = <0,0,1>+s*

where s is a parameter.

The tangent line at t=1/2 consists of all points of the form

r(1/2) + tr'(1/2) = <1,3,0> + t*<0,0,-pi>.

At an intersection, you have

<0,0,1>+s* = <1,3,0> + t*<0,0,-pi>

for some values of s and t, or

= <1,3,-1>.

That is, s*pi=1, 3*s*pi=3, and t*pi=-1.

It follows that s=1/pi, and t=-1/pi.

Therefore, the point of intersection is

<0,0,1>+(1/pi)* =
<0,0,1>+<1,3,0> = <1,3,1>.

Answer 2

r(t)=
r'(t)=< πcosπt, 3πcosπt, -πsinπt>
r'(0)=<π,3π,0>
r'(1/2) = <0,0,-π>

now, the tangent line at t=0 is given by:
x=πt
y=3πt
z=1
here t is the parameter of the line

the tangent line at t=1/2 is given by:
x=1
y=3
z=-πs
here s is the parameter of the line
so these lines intersect when
πt=1, 3πt=3 and 1=-πs
i.e.
t=1/π and s=-1/π
which corresponds to the point:
(1,3,1)