equation of a line in R3 (3dimensional)?

Question

Find the parametric equations for the line passsing through the origin and which is perpendicular to:
x+1=y+2=z+1

Please tell me how you did it...

Answer 1

Well, the direction vector of the line is (1,1,1).
Let us first find the plane perpendicular to the given line. The direction of its norm must be parallel to the given line, thus its direction vector to be (1,1,1). Let (x, y, z) to be an arbitrary point in that plane. since the origin (0,0,0) is also on the plane, the plane is: 1*(x - 0) + 1*(y - 0) + 1*(z - 0) = 0,
or: x + y + z = 0
The cross-section of the plane and the given line can be find by solving the 3 simultaneous equations (1 plane equation and 2 line equations): (1/3, -2/3, 1/3)
So the requested line must pass through (0,0,0) and (1/3, -2/3, 1/3). Thus (1, -2, 1) can be thought as the direction vector or the velocity vector (consider some thing is moving from the origin to the point (1/3, -2/3, 1/3) and beyond). Let t to be the time or the parameter, the requested parametric equations are:
x = t
y = -2t
z = t