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Let L1 be the line passing through the points Q1=(4, −1, −2) and Q2=(1, 0, −1) and let L2 be the line passing through the point...Asked by rr
Let L1 be the line passing through the points Q1=(2, −2, 5) and Q2=(−4, −6, 3) and let L2 be the line passing through the point P1=(7, 4, 0) with direction vector →d=[−1, −2, 3]T. Determine whether L1 and L2 intersect. If so, find the point of intersection Q.
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Answered by
mathhelper
L1:
r1= (2,-2,5) + s<6,4,2>
x = 2 + 6s
y = -2 + 4s
z = 5 + 2s
L2:
r2 = (7,4,0) + t<-1,-2,3>
x = 7 - t
y = 4 - 2t
z = 0 + 3t
If they intersect, then their corresponding x, y , and z values must be the
same for some constant s and t
I will set the y's and z's equal,
-2 + 4s = 4 - 2t ---> 4s + 2t = 6 ----> 2s + t = 3
5 + 2s = 3t
2s - 3t = -5
subtract those last two:
4t = 8
t = 2
sub into 2s+t=3
2s + 2 = 3
s = 1/2
then in L1, z = 5 + 2s = 5+5/2 = 15/2
in L2, z = 3t = 6
I did not get the same value, so the two lines do not intersect
better check my calculations, only had one coffee so far.
r1= (2,-2,5) + s<6,4,2>
x = 2 + 6s
y = -2 + 4s
z = 5 + 2s
L2:
r2 = (7,4,0) + t<-1,-2,3>
x = 7 - t
y = 4 - 2t
z = 0 + 3t
If they intersect, then their corresponding x, y , and z values must be the
same for some constant s and t
I will set the y's and z's equal,
-2 + 4s = 4 - 2t ---> 4s + 2t = 6 ----> 2s + t = 3
5 + 2s = 3t
2s - 3t = -5
subtract those last two:
4t = 8
t = 2
sub into 2s+t=3
2s + 2 = 3
s = 1/2
then in L1, z = 5 + 2s = 5+5/2 = 15/2
in L2, z = 3t = 6
I did not get the same value, so the two lines do not intersect
better check my calculations, only had one coffee so far.
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