Thread: Vectors: Finding an equation for plane that contains the two lines

Find an equation for the plane that contains the two lines
l1: x = t + 1, y = 5t – 4, z = 4t + 3
and
l2: x = -2 – t, y = 4t – 19, z = -9 – 2t.

I rewrote this in vector form
L1 = (1,-4,3) + t(1,5,4)
L2 = (-2,-19,-9) + t(-1,4,-2)

Then i went to find the point of intersection
x = 1+s = -2-t
y = -4+5s = -19+4t
z = 3+4s = -9-2t

used x to find 's'

1+s = -2-t
s = -3-t

plugged in 's' to z

3+4(-3-t) = -9-2t
3-12-4t=-9-2t
9-4t=-9-2t
-6t=0
t = 0

plugged 't' back into x

1+s = -2-(0)
s = -3

I plugged in t into x,y and z and get the following
x = -2-(0) = -2
y = -19+4(0) = -19
z = -9-2(0) = -9

P = (-2,-19,-9)

Then i went to find the cross product
n = (1,5,4)x(-1,4,-2)
n = (-26,2,9)

Then finally went ahead to find the equation.
-26(x-(-2)) + 2(y-(-19)) + 9(z-(-9)) = 0
-26x-52 + 2y+38 + 9z+81 = 0
-26x + 2y + 9z = -67


but this is WRONG!!! what am i doing wrong? Thanks!!!

silly mistake. a friend of mine helped me out. seems i accidentally did my cross product wrong. it was supposed to be (-26,-2,9) instead.