Verifying Vector Intersection: Two Questions

[Fila]

I have two questions I need to make sure if I'm doing correctly. Its vectors.


1)
For what values of a are the vectors i+3j-k and i+aj+k

i) inclined at 30 degree angle
cos@ = n1.n2 / |n1||n2|
cos^2(30) * 11(2+a^2) = ((3a)^2)
a=sqrt22


ii) perpendicular
(1,3,-1).(1,a,1)=0
1+3a-1=0
a=0


2)
Show that the lines
r=s(i+2j+3k)
r=(3i+5j+4k)+t(2i+3j+k)
intersect

I believe I should try and prove that they are not parallel. If their dot product is zero, then it means the lines are parallel. So if the answer is not zero, then I proved that the lines do intersect.
Do I just take the cross product of (1,2,3) and (2,3,1)? I want to know if I'm using the correct vectors to do the dot product.

 

[Muzza]

1) is correct.

Your reasoning for 2) is not correct. Two lines in 3 dimensions which are not parallel do not necessarily intersect. Instead, simply try working out the point of intersection.

 

[Fila]

Thanks!

How do I get started with finding the point of intersection in #2? I couldn't find this 3 dimension problem in the book.

 

[Benny]

I'll try to help get you started. You have two vector equations and two parameters. You can set up three equations, each of which have the same two unknowns, by equating the x, y and z components of each vector equation. You equate x,y and z(or i,j and k depending on how you wish to state it) components of the two vector equations because at the POI they are equal.

i: s = 3 + 2t (1)
j: 2s = 5 + 3t (2)
k: 3s = 4 + t (3)

Solve any two of the three equations simultaneously to obtain values for t and s. So for example solving (1) and (2) simultaneously will yield: s = 1 and t = -1. Does (s,t) = (1,-1) satisify equation 3? If they do not satisfy equation 3 then the lines cannot intersect. If they do satisfy equation 3 then I'll leave it up to you to figure out what that means.

 

[Fila]

I had no idea how to write the two lines into those 3 equations to solve it. Now I know what to do. Thanks!