78.6k views
3 votes
Determine whether the lines L1:x=26+7t,y=8+3t,z=18+5t and L2:x=−11+8ty=−11+5tz=−13+8t intersect, are skew, or are parallel. If they intersect, determine the point of intersection; if not leave the remaining answer blanks empty.

1 Answer

4 votes

Answer:

(A) The lines L1 and L2 are NOT parallel

(B) The lines DON'T intersect

(C) The lines ARE skew

Explanation:

L1: x=26+7t; y=8+3t; z=18+5t

L2: x=-11+8t; y=-11+5t; z=-13+8t

TEST FOR PARALLEL

The direction vector for L1 is (7,3,5)

The direction vector for L2 is (8,5,8)

Since 8 is not a multiple of 7,

5 is not a multiple of 3,

and 8 is not a multiple of 5; both direction vectors are NOT propositional.

The lines are hence unparallel lines. They do not move in the same direction.

TEST FOR INTERSECTION

We equate the functions of X, the functions of Y or the functions of Z, and substitute the value of t gotten, into another equation herein.

26 + 7t = -11 + 8t

7t - 8t = -11 - 26

-t = -37

t = 37

Substituting 37 for t in equation 3, we have

18+5(37) = -13+8(37)

18+185 = -13+296

203 = 283

Obviously, 203 is not equal to 283

These 3 equations aren't consistent, so the lines don't intersect or cross paths. Lines L1 and L2 are skew.

User Turtlefight
by
5.1k points