Find a system of two linear equations in three unknowns whos

Find a system of two linear equations in three unknowns whose solution set is the line consisting of scalar multiples of the vector (1,2,1).

Solution

Solution :

We know that the two linear equations would be equations of the two planes that intersect at the vector. And the normals of the planes are perpendicular to the vector.

Well there is considerable freedom of choice here: you can choose the placement of the line as long as it has direction (1,2,1) and you can choose the orientations of the planes as long as they both contain (1,2,1).
Let\'s choose a line through the origin:

(x,y,z) = k(1,2,1)

The plane equation containing this line must contain the origin so bear this in mind for now.
The normal is necessarily any vector perpendicular to (1,2,1) so let\'s make one plane contain the x-axis and the other contain the y-axis so that the normals are given by the cross products:

n1 = (1,0,0)×(1,2,1) = (0,-1,2)
n2 = (0,1,0)×(1,2,1) = (1,0,-1)

The equations of the planes are given by the fact that a vector (x,y,z) pointing to a point in the plane is, in each case, perpendicular to the normals:

(x,y,z).(0,-1,2) = 0
(x,y,z).(1,0,-1) = 0

-y + 2z = 0
x - z = 0

These are the desired equations.
Now, if you want both equations to have all 3 variables you could simply add or subtract multiples of one to or from the other (below I first add then subtract the first equation to/from the second).

x - y + z = 0
x + y - 3z = 0

Check that solving the first two gives the line, put x = z = k then y = 2k then (x,y,z) = k(1,2,1) is the line.