Question 1202170
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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.


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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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