Perpendicular plane?

Question

Say you have a point, such as (x,y,z) and a line with the equation x=4-2t, y=6+10t, and z=6+2t. How would you find the equation for the plane that passes through the point, and is perpendicular to the line?

Answer 1

It's problematic to call the point (x,y,z) since those are the variables in the equations. Let's call the point P(x1, y1, z1).

The parametric equation of the line is:
x = -2t
y = 6 + 10t
z = 6 + 2t

The normal vector to the plane n, is the same as the directional vector to the line.

n = <-2, 10, 2>

Any non-zero multiple of n will also be a directional vector to the line. Divide by -2.

n = <1, -5, -1>

With the normal vector n and the point P(x1, y1, z1) we can write the equation of the plane.

1(x - x1) - 5(y - y1) - 1(z - z1) = 0
x - 5y - z - x1 + 5y1 + z1 = 0

Answer 2

Eliminating parameter t from the equation of line, it can be written as
( x - 4 ) / -2 = ( y - 6 ) / 10 = ( z - 6 ) / 2

Hence the direction of the line is given by ( -2, 10, 2 ).

Let us denote the given point as ( x', y', z' ) to avoid confusion with variables x, y, z.

As the plane is perpendicular to the line, direction of its normal is the direction of the line.

Therefore, its equation can be written as -2x + 10y +2z = k.

As it passes through the point ( x', y', z' ), k = -2x' + 10y' + 2z'.

Hence, equation of the plane is -2x + 10y + 2z = -2x' + 10y' + 2z'

or, x - 5y - z = x' - 5y' - z'