L1 is (4,2,-3) + (1,3,-2)t
L2 is (3,7,k) + (-3,-3,4)t
4+t = 3-3n t
2+3t = 7-3n t
4+t-3 = 2+3t-7
t = 3
n = -4/9
Now we can see that
-3-2t = k+4nt
-3-2(3) = k+4(-4/9)(3)
k = -11/3
L2 = (3,7,-11/3) - 4/9 (-3,-3,4)
= (3,7,-11/3) + 4/9 (3,3,-4)
at t=3,
L1(3) = (4,2,-3) + 3(1,3,-2) = (7,11,-9)
L2(3) = (3,7,-11/3) - 4/3 (-3,-3,4) = (7,11,-9)
I spent hours on this and I still cant figure it out please help me:
Let L1 be the line passing through the points Q1=(4, 2, −3) and Q2=(5, 5, −5). Find a value of k so the line L2 passing through the point P1 = P1(3, 7, k) with direction vector →d=[−3, −3, 4]T intersects with L1.
3 answers
L2 is (3,7,k) + n(-3,-3,4)t
L2 = (3,7,-11/3) - 4/9 (-3,-3,4)t
= (3,7,-11/3) + 4/9 (3,3,-4)t
= (3,7,-11/3) + 4/9 (3,3,-4)t
3 answers