Question

At what point do the curves \({r_1}(t) = \left\langle {t,1 - t,3 + {t^2}} \right\rangle \) and \({r_2}(s) = \left\langle {3 - s,s - 2,{s^2}} \right\rangle \)intersect? Find their angle of intersection correct to the nearest degree.

Short answer

The point of intersection is \((1,0,4)\)and the angle between given vectors is \(\theta = {55^^\circ }\)

Step-by-step solution

Step 1: Given information

The given value is:

\({r_1}(t) = \langle 1 - t,3 + {t^2}\rangle \) and \({r_2}(s) = \left( {3 - s,s - 2,{s^2}} \right)\)

Step 2: Find the two vector functions intersect

To find where two vector functions intersect, just set each pair of components equal to each other.

\({x_1} = {x_2},{y_1} = {y_2},{z_1} = {z_2}\)

\(\begin{array}{l}{r_1}(t) = {r_2}(s)\\t = 3 - s\\1 - t = s - 2\\3 + {t^2} = {s^2}\end{array}\)

We now have a system of\(3\)equations and\(2\)unknowns. It turns out that the first two equations are identical, as the work to the left shows. We'll need to use the\({3^{rd}}\)equation to find the answer.

We can then plug\(s\)and\(t\)into the original vector functions\({r_1}(t)\)and\({r_2}(s)\)to find the point of intersection

\({\rm{1 - (3 - s) = s - 2}}\)

\({\rm{s - 2 = s - 2}}\)

\({s^2} = 3 + {(3 - s)^2}\)

\({s^2} = 3 + 9 - 6s + {s^2}\)

\({\rm{s = 2}}\)

\({\rm{t = 1}}\)

\({r_1}(1) = < 1,0,4 > = {r_2}(s)\)

Step 3: Find the tangent vectors

Now that we know\(s\)and\(t,\)we'll need to know at what angle the curves intersect.

Finding the tangent vectors to\({r_1}(t)\)and\({r_2}(s),\)and then finding the angle between those tangent vectors at the point\(\left( {1,0,4} \right).\)

\(\begin{array}{l}r_1^\prime (t) = < 1, - 1,2t > \\r_2^\prime (s) = < - 1,1,2s > \\r_1^\prime (1) = < 1, - 1,2 > \\r_2^\prime (2) = < - 1,1,4 > \end{array}\)

We know that\({\bf{a}} \cdot {\bf{b}} = |{\bf{a}}| \cdot |{\bf{b}}|\cos \theta ,\)therefore, the formula for the angle between two vectors can be calculated as,

\(\cos \theta = \frac{{r_1^\prime (1) \cdot r_2^\prime (2)}}{{\left| {r_1^\prime (1)} \right|\left| {r_2^\prime (2)} \right|}}\)

\(\cos \theta = \frac{{ - 1 + ( - 1) + 8}}{{\sqrt 6 \sqrt {18} }}\)

\(\cos \theta = \frac{6}{{6\sqrt 3 }}\)

\(\theta = {54.73^^\circ }\)

The point of intersection is \((1,0,4)\)and the angle between given vectors is \(\theta = {55^^\circ }\)