Hello, ggw!

The parametric equations of a line has this form: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>

. . 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