Question

find the intersection point of the lines{ p:p=(-2,4,6)+t(1,2,3)and {p:p=(-2,1,6)+t(3,2,1)} find the equation of the plane containing the two lines

Answer 1

from the first:
x = -2 + t
y = 4 + 2t
z = 6 + 3t

from the second line , I changed the parameter to to s, can't have t is both
x = -2 + 3s
y = 1 + 2s
z = 6 + s

equating the xs:
-2 + t = -2 + 3s
t = 3s
equating the ys
4+2t = 1+2s
2t - 2s = -3
2(3s) - 2s = -3
4s= -3 ------> s = -3/4
then t = -9/4

test for zs"
is 6+s = 6 + 3t ??
LS = 6 - 3/4 = 21/4
RS = 6 + 3(-9/4) = -3/4

The 2 lines do NOT intersect