Is this line perpendicular to the plane?

[masterchiefo]

Homework Statement

Line perpendicular to a plane
Plane: z(x,y)= 0.882531*x-0.346494*y+0.383108
P1(0.5;0.1;Z)
P2(0.4;0.15;Z)

Homework Equations

The Attempt at a Solution

z(0.5,0.1) = 0.7897
z(0.4,0.15) = 0.68414

vector n=ai+bj+ck

vector P2P1 = (0.5-0.4)i + (0.1-0.15)j +(0.7897-0.68414)k

n*(P2P1) = 0.1a - 0.005b + 0.10556c
c= -0.947329(a-0.5*b)

parametric

x(t,u) = t+0.5
y(t,u)=t+0.1
z(t,u)=-0.947329(t-0.5*t)

when I put this on the graph, its not perpendicular at all... :( help

 

[Mark44]

Homework Statement

Line perpendicular to a plane
Plane: z(x,y)= 0.882531*x-0.346494*y+0.383108
P1(0.5;0.1;Z)
P2(0.4;0.15;Z)

Homework Equations

The Attempt at a Solution

z(0.5,0.1) = 0.7897
z(0.4,0.15) = 0.68414

For the first one, I get .7897241, and for the second, I get .6841463

vector n=ai+bj+ck

vector P2P1 = (0.5-0.4)i + (0.1-0.15)j +(0.7897-0.68414)k
n*(P2P1) = 0.1a - 0.005b + 0.10556c

Your 2nd coordinate is wrong -- it should be -.05, not -.005 .
Also, your plane can be written as 0.882531*x - 0.346494*y - z = .383108. A normal to this plane is <0.882531, -0.346494, -1> .

c= -0.947329(a-0.5*b)

parametric

x(t,u) = t+0.5
y(t,u)=t+0.1
z(t,u)=-0.947329(t-0.5*t)

when I put this on the graph, its not perpendicular at all... :( help

 

[Mark44]

BTW, what terrible numbers in this problem! I started by using a calculator, but then gave up and made a small Excel spreadsheet.

 

[geoffrey159]

When you have a plane of equation $ax+by+cz+d = 0$, you know that the perpendicular line is directed by $\vec u = (a,b,c)$.
So the equation of your line passing through $P_1$ is ${\cal D} = \{ P_1 + t \vec u, t\in\mathbb{R}\}$