Calc 3 Question?

Question

Ok two parts to this one, the first part i think i understand what im supposed to do, but correct me if im wrong.

Im given two parametric equations

x= 3+t, y= 3-2t, z= 5t-1

x= 3k-4, y= 5-2k, z= k+6

do these lines intersect??
Im guessing that you find the direction vectors of each and make sure they are not parallel?

Then the real question, what point do they intersect?

Answer 1

That won't suffice, since the lines might simply be askew. What you do is figure out what relationship t and k must have IF the lines intersect, and then figure out whether such values of t and k are actually consistent. For instance:

The x-coordinates of a point of intersection must be the same, so 3+t=3k-4, 3k-t=7

The y-coordinates must be the same, so:
3-2t=5-2k, 2k-2t=2

The z-coordinates must be the same, so:

5t-1=k+6, k-5t=-7

This gives you the system of equations:

3k-t=7
2k-2t=2
k-5t=-7

In matrix form:

[3, -1 | 7]
[2, -2 | 2]
[1, -5 | -7]

Solving (I transposed the rows to make the matrix step easier):

[1, -5 | -7]
[3, -1 | 7] subtract 3*row 1
[2, -2 | 2] subtract 2*row 1

[1, -5 | -7]
[0, 14 | 28] divide by 14
[0, 8 | 16] subtract 8/14 * row 2

[1, -5 | -7] Add 5*row 2
[0, 1 | 2]
[0, 0 | 0]

[1, 0 | 3]
[0, 1 | 2]
[0, 0 | 0]

So this system has a unique solution, k=3 and t=2. Substituting these into either set of parametric equations yields the point of intersection as (5, -1, 9).

Answer 2

You would need to find the values for t and k (if any) where
3+t=3k-4 and 3-2t=5-2k and 5t-1=k+6
Then solve for x, y, and z.