Question

Determine whether the lines L1:x=18+6t,y=7+3t,z=13+3t and L2:x=−14+7ty=−12+5tz=−8+6t intersect, are skew, or are parallel. If they intersect, determine the point of intersection; if not leave the remaining answer blanks empty.

Answer 1

Answer:

skew lines

Step-by-step explanation:

we are given 2 lines in parametric form as

L1:x=18+6t,y=7+3t,z=13+3t and L2:x=−14+7ty=−12+5tz=−8+6t$L1:x=18+6t,y=7+3t,z=13+3t \\ L2:x=-14+7t,y=-12+5t,z=-8+6t$

If the lines intersect then the two points must be equal for one value of t.

Let us try equating x,y and z coordinate.

$18+6t = -14+7t\\t=32$

when we equate y coordinate we get

$7+3t =-12+5t\\2t =19\\t =9.5$

Since we get two different t we find that these two lines cannot intersect.

Comparing direction ratios we have

I line has direction ratios as (6,3,3) and second line (7,5,6)

These two are not proportional and hence not parallel

So these lines are skew lines