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kenxkeith
18.07.2020 •
Mathematics
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.
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Ответ:
(A) The lines L1 and L2 are NOT parallel
(B) The lines DON'T intersect
(C) The lines ARE skew
Step-by-step 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.
Ответ:
60*24=1440*365.25=525960