Question

Vectors: Determine whether the lines intersect and if so find the point of intersection and the cosine of the angle of intersection.

Answer 1

I'm posting up the problems right now. Hold on...

Answer 2

okay.

Answer 3

x = 4t +2,y=3, z = -t+1

x= 2s+2, y = 2s+3, z=s+1

Answer 4

Would I have to factor each of these equations out and set them equal to zero to find their pts of intersection?

Answer 5

no, u have to equate all the x and y and z togther.
u will get a system of equations consisting of s and t.

Answer 6

x = 4t +2,y=3, z = -t+1
x= 2s+2, y = 2s+3, z=s+1

equating all the x, y and z.
U get 3 equations,
4t+2= 2s+2
3=2s+3
-t+1=s+1
Solving them, u get s=0 and t=0

That means to say, the 2 points will intersect at the point when s=0 and t =0

Intersection point x=0,y=3,z=0

Answer 7

As for cos of the angle of intersection, i forgot how to do it. im relearning on wiki gimme a while.

Answer 8

at s=t=0, wouldn't x = 2 and z=1

Answer 9

oh ya. opps. made a mistake there! thx for pointing it out =D

Answer 10

Oh okay thankyou. I think I can figure out the rest of the problem from here.

Answer 11

Anw i got a question how to u find the component vector from the eqn of a line. i learned vectors so long ago, i forgot how lol.

Answer 12

the component vector is your "slope" parts. just pull the constants off your t or s variables

Answer 13

x = 2 + 4t
y = 3 + 0t ; your anchor point is (2,3,1); your scaling vector is <4,0,-1>
z = 1 - 1t

Answer 14

the cos of the angle between them is a consequence of the law of cosines;

the length of a side is equal to the diffence of, the sum of the squares of the other 2 sides, and two times the product of the other sides and the cosine of the angle opposite.

c^2 =(a^2+b^2) - 2(ab cos(C))

when we have 2 vectors we can determine the length of all the "sides" created so we just solve for cos(C)

c^2 - (a^2+b^2)
-------------- = cos(C)
2ab

or in vector talk

|c|^2 - |a|^2 - |b|^2
------------------ = cos(C)
2|a||b|


the issue is then if relating this to the dot product of vectors a and b

\[\frac{(c_x ^2+c_y ^2+c_z ^2)-(a_x ^2+a_y ^2+a_z ^2)-(b_x ^2+b_y ^2+b_z ^2)}{2\sqrt{a_x ^2+a_y ^2+a_z ^2)}\sqrt{(b_x ^2+b_y ^2+b_z ^2)}}=cos(C)\]
but c is made of the components of a-b
\[\frac{((a-b)_x ^2+(a-b)_y ^2+()_z ^2)-(a_x ^2+a_y ^2+a_z ^2)-(b_x ^2+b_y ^2+b_z ^2)}{2\sqrt{a_x ^2+a_y ^2+a_z ^2)}\sqrt{(b_x ^2+b_y ^2+b_z ^2)}}=cos(C)\]

in the end it boils down

Answer 15

to

a.b
------ = cos(C)
|a||b|

Answer 16

how did you get 4,0,-1?

Answer 17

oh wait nevermind I see.