find equation of line through the point (1,-1,1) perpendicular to the line 3x=2y=z and parallel to plane x+y-z=0
Two direction vectors which are perpendicular to the required line are the normal of the given plane, <1, 1, -1> and a direction vector for the given line. A direction vector for the line 3x=2y=z is found by finding two points. I went with (0,0,0) and (2,3,6). So the direction vector is <2,3,6>.
to find a direction vector for the line we need, we take the cross product.
v = <1, 1, -1> x <2, 3, 6> = <9, -8, 1>
Write the required line out in parametric form using the given point as (x0, y0, z0).
x = x0 + at
y = y0 + bt
z = z0 + ct
x = 1 + 9t
y = -1 - 8t
z = 1 + t