Asked by Jale
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)) =
(Use the parameter t.)
(x(t), y(t), z(t)) =
Answers
Answered by
Damon
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 !
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 !
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