Finding a Point on Line P Normal to Plane 2x-y+2z=-2

[MozAngeles]

Homework Statement

Find a point in which the line through P(3,2,1) normal to the plane 2x-y+2z=-2

Homework Equations

The Attempt at a Solution

n=<2,-1,2>, so x=3+2t y=2-t z=1-2t
so then i have point (3+2t,2-t,1-2t)
i plug that into the plane, solve for t, which i get to be -8,

so then (x,y,z) for t=-8 (-13,10,17) but this answer is differnet from the back of the book. I don't have any clue as to what i am doing wrong. any help would be nice thanksss :D

 

[Mark44]

Homework Statement

Find a point in which the line through P(3,2,1) normal to the plane 2x-y+2z=-2

I think some words are missing in your problem description. From the work below, it appears that you want to find the point in the plane at which a normal to the plane passes through (3, 2, 1).

Homework Equations

The Attempt at a Solution

n=<2,-1,2>, so x=3+2t y=2-t z=1-2t
so then i have point (3+2t,2-t,1-2t)
i plug that into the plane, solve for t, which i get to be -8,

so then (x,y,z) for t=-8 (-13,10,17) but this answer is differnet from the back of the book. I don't have any clue as to what i am doing wrong. any help would be nice thanksss :D

I get t = -8/9.

 

[MozAngeles]

yes i have to find where the line meets the plane, so am i right with my equations for x,y, and z. but when i plug them into the equation for the plane i keep on getting t=-8.

 

[Mark44]

Your parametric equations for the line are fine, but you have made a mistake in solving for t at the point where the line intersects the plane. Show us your work on solving for t.

 

[MozAngeles]

2(3+2t)-(2-t)+2(1-2t)=-2
6+4t-2+t+2-4t=-2
t=-8

 

[Mark44]

I didn't notice before, but you have a mistake in your parametric equation for z. It should be z = 1 + 2t. You have z = 1 - 2t.

 

[MozAngeles]

Ohhhh k. Thanks.