Find parametric equations for the line through the point (0,2,2)that is parallel to the plane x+y+z = 2 and perpendicular to the line x=1+t, y=2−t, z=2t.

(Use the parameter t.)

(x(t), y(t), z(t)) =

perpendicular to the direction

1 i - 1 j + 2 k
Vxi + Vyj + Vzk
if perpendicular dot product is 0
Vx - Vy + 2 Vz = 0

also parallel to plane x+y+z = constant 2
so normal to the normal to that plane
1 i + 1 j + 1 k
Vxi + Vyj + Vzk
Vx + Vy + Vz = 0
so we have

Vx - Vy + 2 Vz = 0
Vx + Vy + Vz = 0

let Vx = 1 and solve for Vy and Vz (only direction, not magnitude is needed)
- Vy + 2 Vz = -1
+ Vy + 1 Vz = -1
------------------ add

3 Vz = -2
Vz = -2/3

Vy = -1 +2/3 = -1/3
so direction is
1 i -2/3 j - 1/3 k
or
3 i - 2 j - 1 k

so
line through point is
(0 , 2, 2 ) + ( 3, -2, -1) t

check my arithmetic !

To find the parametric equations for the line, we need to consider two conditions: it should be parallel to the given plane and perpendicular to the given line.

First, let's find the direction vector for the line that is parallel to the plane. To do this, we can take the coefficients of x, y, and z in the plane equation x + y + z = 2. The direction vector for the line parallel to the plane is (1, 1, 1).

Next, let's find the direction vector for the line that is perpendicular to the given line x = 1 + t, y = 2 - t, z = 2t. To do this, we can take the coefficients of t in each of the equations for x, y, and z. The direction vector for the line perpendicular to the given line is (1, -1, 2).

Since the line we are looking for is both parallel to the plane and perpendicular to the given line, its direction vector should be perpendicular to both (1, 1, 1) and (1, -1, 2). We can find this direction vector by taking the cross product of these two vectors:

(1, 1, 1) × (1, -1, 2) = (-3, 1, -2)

Now, we have the direction vector for the line we are looking for, which is (-3, 1, -2). We can use this direction vector along with the given point (0, 2, 2) to get the parametric equations for the line.

The parametric equations for the line are:
x(t) = 0 + (-3)t
y(t) = 2 + t
z(t) = 2 + (-2)t

Therefore, the parametric equations for the line are:
x(t) = -3t
y(t) = 2 + t
z(t) = 2 - 2t