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| Question 1202170:  Determine the vector equation of the line that passes through the point A (-2, 3, 6) and is parallel to the line of intersection between the two planes
 pi 1: 2x - y + z = 0 and
 pi 2: y  + 4z = 0
 the answer is supposed to be vector r2 = (-2, 3, 6) + s (-5, -8, 2) sER
 Answer by math_tutor2020(3817)
      (Show Source): 
You can put this solution on YOUR website! The given planes are
 2x - y + z = 0
 y + 4z = 0
 They intersect along some line which I'll call L1.
 
 
 To determine the equation of a line, we need 2 points on it.
 
 To generate a point in 2D, we plug in some x value to find y. This gives an (x,y) ordered pair.
 We'll do a similar thing in 3D for an ordered triple (x,y,z)
 
 Let's plug in x = 0
 2x - y + z = 0
 2*0 - y + z = 0
 -y + z = 0
 
 We have this system
 -y + z = 0
 y + 4z = 0
 
 Apply the elimination method, or any other method of your choice, to find the solution to that system is (y,z) = (0,0)
 Coupled with x = 0 leads to the point (x,y,z) = (0,0,0) on the line L1.
 
 -------------------------------------------------------
 
 Let's try x = 1
 2x - y + z = 0
 2*1 - y + z = 0
 2 - y + z = 0
 -y + z = -2
 
 We have this system
 -y + z = -2
 y + 4z = 0
 
 Solve that system to get (y,z) = (8/5,-2/5)
 Therefore the point (1,8/5,-2/5) is also on the line L1.
 
 The two points
 (0,0,0)
 and
 (1,8/5,-2/5)
 are on the line L1
 
 
 The vector going from (0,0,0) to (1,8/5,-2/5) is < 1,8/5,-2/5 >
 This is a direction vector.
 Any parallel line will have the same direction vector or a scaled version of it.
 
 
 One possible equation of line L1 as a vector equation is:
 (x,y,z) = startPoint + s*DirectionVector
 (x,y,z) = (-2,3,6) + s*(1,8/5,-2/5)
 
 It may not be entirely clear how to go from the direction vector (1,8/5,-2/5) to (-5,-8,2), but we can introduce a scale factor.
 
 Recall that vector (x,y,z) can be scaled to k*(x,y,z) = (kx,ky,kz) for any nonzero real number k.
 Vector (x,y,z) is parallel to vector (kx,ky,kz)
 
 
 If we use k = -5, then
 k*(x,y,z) = (kx,ky,kz)
 k*(1,8/5,-2/5) = (k*1,k*8/5,k*(-2/5))
 -5*(1,8/5,-2/5) = (-5*1,-5*8/5,-5*(-2/5))
 -5*(1,8/5,-2/5) = (-5,-8,2)
 
 The vectors (1,8/5,-2/5) and (-5,-8,2) are parallel.
 They point along the same straight line.
 
 That is how we go from
 (x,y,z) = (-2,3,6) + s*(1,8/5,-2/5)
 to
 (x,y,z) = (-2,3,6) + s*(-5,-8,2)
 where "s" is any real number.
 
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