Help me Find an equation of a plane

Question

With respect to the origin O, the lines l and m have vector equations r = 2i + k + λ(i − j + 2k) and r = 2j + 6k + µ(i + 2j − 2k) respectively. Find the equation of the plane which is parallel to l and contains m, giving your answer in the
form ax + by + cß = d.

Answer 1

I will use an ordered triplet (  ,   ,  ) notation to describe vectors.   The direction of line l  is specified by

(1 , - 1, 2)   and the direction of line my by ( 1 , 2, -2) 

A direction perpendicular to both of these is the cross product 

 ( 1 , -1, 2 ) x  (1 ,2, -2 ) =  ( -2 , 4, 3  )  This is the direction of the normal to the desired plane 

Thus the equation of the desired plane is  -2 x + 4 y+ 3 z =   λ     , for some value of λ.

To find the value of λ,  one need only select a single point on the line m to work with.

Selecting μ = 0   gives the point (0, 2 ,6).     Substituting x = 0 , y = 2 , z = 6 into the equation for the plane 

results in λ = 26.

Answer 2

v = <1, -1, 2> is parallel to the plane 

 

w = <1, 2, -2> is parallel to the plane

 

So, v x w = <-2, 4, 3> is perpendicular to the plane

 

If μ = 0, then P = (0, 2, 6) is a point on line m.

 

Let (x, y, z) be a point on the plane.

 

Then the vector <x-0, y-2, z-6> lies in the plane.

 

Equation of the plane:  <x, y-2, z-6> · <-2, 4, 3> = 0

 

                                  -2x + 4(y-2) + 3(z-6) = 0

 

                                  -2x + 4y + 3z = 26