MHB Points of intersection of two vector curves

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The curves r1(t) = (t, 2 − t, 35 + t²) and r2(s) = (7 − s, s − 5, s²) intersect at a unique point determined by solving the equations t = 7 - s, 2 - t = s - 5, and 35 + t² = s². The specific values of t and s, denoted as t0 and s0, can be calculated to find the intersection coordinates. The angle of intersection, θ, is derived from the tangent vectors T1 and T2 at the intersection point, using the formula cos(θ) = (T1·T2) / (||T1|| × ||T2||). This method provides both the intersection point and the angle of intersection, which can be rounded to the nearest degree. The discussion emphasizes the mathematical approach to finding these geometric properties of the curves.
carl123
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a) At what point do the curves r1(t) = (t, 2 − t, 35 + t2) and r2(s) = (7 − s, s − 5, s2) intersect?

(x,y,z) =

b) Find their angle of intersection, θ, correct to the nearest degree.
 
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a) To find the point(s) of intersection of two curves $r_1(t)$ and $r_2(s)$ you want to find those $t$ and $s$ with $r_1(t)=r_2(s)$; i.e. $t=7-s$, $2-t=s-5$ and $35+t^2=s^2$. For this problem, it turns out there is exactly one $t=t_0$ and $s=s_0$ that satisfy these equations. You can find $t_0$ and $s_0$.

b) The angle of intersection of the two curves is the angle between the two tangent vectors at the intersection point. So find $T_1=r_1^\prime(t_0)$ and $T_2=r_2^\prime(s_0)$. Then with $\theta$ the angle of intersection,
$$\cos(\theta)={T_1\cdot T_2\over ||T_1||\times||T_2||}$$
You can finish from here.
 

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