Finding points of intersection using vectors

[jaejoon89]

Find the points of intersection of the lines...

L1: R = 2i + 3j + 3k + t(i - 2j + 5k)
L2: (x + 3) / 2 = (y + 1) / 2 = -z

(I assume the plural in points is wrong... since that would be impossible)
R, i, j, k are vectors; x, y, z are not

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x = 2 + t
y = 3 - 2t
z = 3 + 5t
for some value(s) of t. If this point also lies on L2, then (x + 3)/2 = -z, so that:
(2 + t + 3)/2 = -(3 + 5t)
5 + t = -6 - 10t
11t = -11
t = -1
For this to really be on L2, you also need to have (y + 1)/2 = -z, or:
(3 - 2t + 1)/2 = -(3 + 5t)
4 - 2t = -6 - 10t
8t = -10
t = -5/4

The two t's don't equal each other. What am I doing wrong?

 

[Mark44]

I parametrized the equations for L2 using a different parameter: w.

x = 2w - 3
y = 2w - 1
z = -w

For any point (x, y, z) to lie on L1 and L2, it must be that the x coordinates on both lines are equal as must be the y and z coordinates.

So,
2 + t = 2w + 3
3 - 2t = 2w - 1
3 + 5t = -w

These are three equations in two unknowns, an overdetermined system.

I ended up with no solution, which means (assuming my calculations are correct) that the two lines don't intersect.