Find an equation for the plane containing the point (1, 1, 1) and the line (x, y, z) = (1 + 2t, 3 − t, t).

Step-by-step explanation:

The line can be written as

Since the line is contained in the plane, if we take any two points of this line they will lie in the plane.

Give values  for t =0 and 1

We get two points (x,y,z) as (1,3,0) and (3,2,1)

Now we have another non collinearpoint (1,1,1) (given)

The plane equation can be written using these 3 points using the determinant.

x-x1 y-y1 z-z1

x2-x1 y2-y2 z2-z1

x3-x1 y3-y1 z3-z1    =0

Substitute the values form the points.

We get

Expand the determinant as

(x-1)(-1+2)-(y-3)(2-0)+z(-4-0) =0

x-1-2y+6-4z =0

x-2y-4z+5 =0

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square