# Can lines in 3D be parametrized by the same variable?

• inknit
In summary, the two lines described by the equations x = 11+3t y = 7+t z = 9+2t and x=-6+4t y=-2+3t z=-7+5t do not intersect due to their linear dependence. To determine if they intersect, one of the variables can be changed to a different parameter and the system can be solved for both parameters.

#### inknit

For example.

L1: x = 11+3t y = 7+t z = 9+2t

L2: x=-6+4t y=-2+3t z=-7+5t

I was given this problem, and technically these lines don't intersect, right?

inknit said:
For example.

L1: x = 11+3t y = 7+t z = 9+2t

L2: x=-6+4t y=-2+3t z=-7+5t

I was given this problem, and technically these lines don't intersect, right?

The only way both of these lines can be described by the same variable is if the two lines are linearly dependent.

If this is not the case, you need two variables one for each line.

Alright so, they don't intersect correct? B/c if you replace the variable in L1 with let's say "s" they intersect at some point.

inknit said:
Alright so, they don't intersect correct? B/c if you replace the variable in L1 with let's say "s" they intersect at some point.

I didn't say they don't intersect. If you want to check what happens solve the linear system L1 = L2.

inknit said:
For example.

L1: x = 11+3t y = 7+t z = 9+2t

L2: x=-6+4t y=-2+3t z=-7+5t

I was given this problem, and technically these lines don't intersect, right?
They don't happen to intersect (and not just "technically") but not because the use the same parameter. A parameter has no meaning outside the equation itself. So as not to confuse yourself, it would be better to change one of them to, say, "s". To determine if they intersect, try to solve x= 11+ 3t= -6+ 4s, y= 7+ t= -2+ 3s, z= 9+ 2t= -7+ 5s for s and t.