( 717, #57 ) Find parametric equations for the line through the point ( 0, 1, 2 ) that is parallel to the plane x + y + z = 2 and perpendicular to the line x = 1 + t, y = 1 - t, z = 2t.

Solution: The parallel line through the point has the vector equation L = < 0, 1, 2 > + t < a, b, c >, and we have to determine a, b, and c.

( 1 ) L is parallel to the plane x + y + z = 2

L is perpendicular to n = < 1, 1, 1 >, the normal to the plane.

L and n are perpendicular

Thus a + b + c = 0.

( 2 ) L and the line < 1, 1, 0 > + t < 1, -1, 2 > are perpendicular

Thus a - b + 2c = 0

We have two equations with three variables:

a + b = -c

a - b = -2c

Solving for a and b in terms of c :

2a = -3c a = -3c/2

2b = c b = c/2

Take c = -2. Then < a, b, c > = < 3, -1, -2 >

Thus the vector equation of the line is : L = < 0, 1, 2 > + t < 3, -1, -2 > and the parametric equations are:

x = 3t, y = 1 - t, z = 2 - 2t

by: nl