Thread: parametric equation on line in 3d

1. parametric equation on line in 3d

find parametric equation for the line with the following properties. the lines passes through the origin, it is contained in the plane 2x+3y-z=0. and is orthogonal to the vector v=<4,1,2>.

2. Hello, ggw!

Find parametric equation for the line with the following properties:
the line passes through the origin, it is contained in the plane 2x + 3y - z = 0.
and is orthogonal to the vector v = <4,1,2>
The parametric equations of a line has this form:

. . x .= .x1 + at
. . y .= .y1 + bt
. . z .= .z1 + ct

where (x1,y1,z1) is a point on the line and <a,b,c> is its direction vector.

The line passes through the origin: (x1,y1,z1) .= .(0,0,0)
. . We need to find its direction vector: <a,b,c>

The line is in the plane 2x + 3y - z = 0.
Hence, the line is orthogonal to the normal vector of the plane: <2,3,-1>
. . and we have: .<a,b,c>·<2,3,-1> .= .0 . . 2a + 3b - c .= .0 . [1]

The line is orthogonal to <4,1,2>.
. . So we have: .<a,b,c>·<4,1,2> .= .0 . . 4a + b + 2c .= .0 . [2]

Multiply [1] by 2: .4a + 6b - 2c .= .0
. . . Subtract [2]: .4a + b + 2c .= .0

And we have: .5b - 4c .= .0 . . b = 4c/5

Substitute into [1]: .2a + 3(-4c/5) - c .= .0 . . a = -7c/10

The direction vector of the line is: .<-7c/10, 4c/5, c>

Let c = 10 and we have: .v .= .<-7, 8, 10>

Therefore, the equations of the line are:

. . x .= .-7t
. . y .= . 8t
. . z .= .10t