Calculus 2 Question (Will Pick Best Answer Tonight)

The lines x = 2 + 4t, y = -1 - 3t, z = 4 + t and x = 8 + 2t, y = -3 + t, z = -5t intersect at the point (6,-4,5). Find an equation of the plane that contains both lines.

A direction vector of the first line is <4,-3,1> and of the second line is <2,1,-5>
A normal vector is the cross product of the above vectors, <4,-3,1)X<2,1,-5>
=<14,22,10> or scaling down, <7,11,5>
The equation of the plane is <7,11,5>.=<7,11,5>.<6,-4,5>
giving 7x+11y+5z=23

If you are not familiar with vectors you could find a point on the first line (say t=0)
so (2,-1, 4) and on the second line (t=0) so (8,-3,0) and you now have 3 points so sub
into ax+by+cy=1 and find 3 equations for a,b and c then solve. You should get
a=7/23, b=11/23, c=5/23.
and on the second line,

i) Line L1: (x - 2)/4 = (y + 1)/(-3) = (z - 4)/1 and
Line L2 is: (x - 8)/2 = (y + 3)/1 = z/(-5)

ii) Thus L1 has direction ratios (4, -3, 1) and L2 has (2, 1, -5)

iii) Let the plane be ax + by + cz = d, whose normal is (a, b, c)

iv) As both L1 & L2 are contained in the plane, dot product of direction ratios = 0
So, 4a - 3b +c = 0 and 2a + b - 5c = 0

v) Solving these two equations, we get a = 7c/5 and b = 11c/5

Thus a:b:c = 7:11:5

vi) Thus the plane is: 7x + 11y + 5z = d
Plugging the point (6, -4, 5) in this plane equation and solving d = 23

So plane equation is: 7x + 11y + 5z = 23