The linear equation of the intersection line passing through the point (- 3,2,5) and parallel to the plane x-4z = 0 and 2x-y-5z = 1

The linear equation of the intersection line passing through the point (- 3,2,5) and parallel to the plane x-4z = 0 and 2x-y-5z = 1

Let t = Z be z = t
Because x = 4Z = 4T, t = x / 4
Y = 2x-5z-1 = 8t-5t-1 = 3t-1, i.e. t = (y + 1) / 3
So the intersection of plane x-4z = 0 and plane 2x-y-5z = 1 is as follows:
x/4=(y+1)/3=z
The direction vector of the line is a = (4,3,1)
Therefore, the linear equation is as follows:
(x+3)/4=(y-2)/3=z-5

Find the linear equation which passes through the point (- 3,2, - 5) and is parallel to two planes x-4z-3 = 0 and 2x-y-5z + 1 = 0

Law one
Find out the intersection of two planes
It's x = 4Z + 3, y = 3Z + 7
The intersection line must be parallel to the straight line
So the direction vector of the line is (4,3,1)
So the line is (x + 3) / 4 = (Y-2) / 3 = Z + 5
Method 2
Find the normal vector of two planes first
They are (1,0, - 4) and (2, - 1, - 5) respectively
A line is perpendicular to two normal vectors
We can find that the direction vector is (4,3,1)
So the line is (x + 3) / 4 = (Y-2) / 3 = Z + 5
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