Question

at what point do the curves r1= <2t-6, 5-t, t^2-4t+5> and r2=< 1-s, 3s-6, -s^2+6s-8> intersect? find the angle of intersection

Answer 1

if they intersect at some point then the solution to system obtain from corresponding components of curves is the point of intersection .

so here are i,j,k components
2t-6=1-s (set i components equal )
5-t=3s-6 (set j components equal )

solve these equations for s and t and let me know .

Answer 2

k let see

Answer 3

for s =-2t+5
s=11/3-t/3

Answer 4

and t im not sure is t=7/2-s/2 and t=3s+11

Answer 5

i need their numerical values
ok let me show you .
rearranging the terms
2t+s=7 ........1
t+3s=11 ....... 2

from equation 1
s=7-2t
put in second equation at the place of s
t+3(7-2t)=11
t+21-6t=11
-5t=-10
t=2
so s=3



t=2
s=3 is the solution to the system .

Answer 6

put t=2 and s=3 in any of the give curve equation to get the required point .

Answer 7

r1 and r2?

Answer 8

yes

Answer 9

ok r1=<-2,3,1>,r2=<-2,3,1> they look the same r1 and r2

Answer 10

they should be same it means you find the point of intersection correctly :)

Answer 11

i find with part 1 of the question?

Answer 12

i'm finish*

Answer 13

no now find the angle of intersection .

Answer 14

i need to use the dot product?

Answer 15

exactly . first take derivative of r1 and then r2 with respect to t
then put t=2 and s=3 to get two vectors . then go for dot product .
can you do this now ?

Answer 16

ok @sami-21

Answer 17

r'1=<2, -1, 2t-4>=<2,1,0>
r'2=< -1, 3, -2s+6>=<-1, 3 0>

Answer 18

one thing more take the derivative of r2 with respect to s !

Answer 19

i notice that @sami-21

Answer 20

<2,1,0>*<-1, 3 0>= 1

Answer 21

good now take dot product .

Answer 22

still we find the angle?

Answer 23

need*

Answer 24

yes
if you have two vector A and B then angle between them is given by

divide their dor product by product of the magnitude of each vector and take cos inverse to get angle
\[\Large \theta =\cos^{-1} \frac {A.B }{\left| A \right|.\left| B \right| }\]

Answer 25

50