Angle of intersection between two parametric curves

[ElijahRockers]

Homework Statement

This is a problem involving parametric equations.

r₁= <t,2-t,12+t²>
r₂= <6-s,s-4,s²>

At what point do the curves intersect?

Find the angle of intersection, to the nearest degree.

The Attempt at a Solution

I found the point of intersection, (2,0,16). This is when t=2 and s=4.

I found the tangent vectors.
d/dt(r1) = <1,-1,2t>
d/ds(r2) = <-1,1,2s>

I used r₁\cdotr₂ = |r1||r2|cos\theta, using the tangent vectors at t=2 and s=4, and solved for theta.. I came up with 23°, but the system tells me I'm wrong. What happened?

EDIT:: Okay... it seems like I was coming up with a different answer every time. Got it on my last try though. Gotta be more careful... for anyone wondering, the correct answer is 29°

 

[fresh_42]

Homework Statement

This is a problem involving parametric equations.

r₁= <t,2-t,12+t²>
r₂= <6-s,s-4,s²>

At what point do the curves intersect?

Find the angle of intersection, to the nearest degree.

The Attempt at a Solution

I found the point of intersection, (2,0,16). This is when t=2 and s=4.

I found the tangent vectors.
d/dt(r1) = <1,-1,2t>
d/ds(r2) = <-1,1,2s>

I used r₁\cdotr₂ = |r1||r2|cos\theta, using the tangent vectors at t=2 and s=4, and solved for theta.. I came up with 23°, but the system tells me I'm wrong. What happened?

EDIT:: Okay... it seems like I was coming up with a different answer every time. Got it on my last try though. Gotta be more careful... for anyone wondering, the correct answer is 29°

Let's see, so that once and forever not everybody has to redo the exercise. Note that $t=2$ and $s=4.$
\begin{align*}
\dot r_1 \cdot \dot r_2 &= (1,-1,4)\cdot (-1,1,8)=30=|(1,-1,4)|\cdot |(-1,1,8)|\cdot \cos\theta =\sqrt{18\cdot 66}\cos\theta \\
&=6\sqrt{33}\cos\theta \Longrightarrow \cos\theta =\dfrac{5}{\sqrt{33}} \approx 0.87\Longrightarrow \theta \approx 29.5
\end{align*}

 

[Delta2]

I found myself facing a mini confusion because the OP uses the same letters, $r_1,r_2$ for the position vectors of the curves and for the tangent(velocity) vectors of the curves. The $r_1,r_2$ we use for the dot product are the tangent vectors right?

 

[fresh_42]

I found myself facing a mini confusion because the OP uses the same letters, $r_1,r_2$ for the position vectors of the curves and for the tangent(velocity) vectors of the curves. The $r_1,r_2$ we use for the dot product are the tangent vectors right?

Right. I had forgotten the dot (corrected now). $r_j(t)$ are the parameterized curves, like a walk along the time axis $t$, and $\dfrac{d}{dt} r_j = \dot r_j$ are the tangent vectors, the velocity with which we walk, or tangentially fly from the road if suddenly friction stops working.

 

[fresh_42]

I came up with 23°, but the system tells me I'm wrong. What happened?

A golden rule for such occasions and especially in exams: Writing is faster than thinking!

This sounds paradoxical, but it isn't. If you made a mistake, then it is faster to detect. If you made none, then you do not have to juggle numbers or even more important: units! in your mind. Write it down without thinking about it. How long does it take to write 100 characters without thinking about them?