# Vectors and Planes

• Feb 13th 2013, 03:29 AM
carla1985
Vectors and Planes
I apologize if this isnt in the right place, my actual subject is mechanics. I've know how to do some of this question but stuck on other parts.

Consider the set of points (
x,y,z)R3 lying in the plane x+y+z=1.
(i) Write the equation for this plane in the vector form

r · nˆ = d,
where d R is a constant and nˆ is a unit vector.
(ii) What is the distance of this plane from the origin (0, 0, 0)?
(iii) State the vector equation of the line that passes through the point(1, 2, 2) and is parallel to the vector nˆ .
(iv) Find the point where this line intersects the plane r · nˆ = d.
(v) Hence find the distance between the point (1
,2,2) and the plane
x + y + z = 1.

For (i) I have the plane equation as r=a+tn^ where a is a point from (x,y,z) and t is a constant. But I'm confused over the form they want it in.

Thanks x

• Feb 13th 2013, 04:10 AM
Plato
Re: Vectors and Planes
Quote:

Originally Posted by carla1985
Consider the set of points (x,y,z)∈R3 lying in the plane x+y+z=1.
(i) Write the equation for this plane in the vector form
r · nˆ = d,
where d ∈ R is a constant and nˆ is a unit vector.
(ii) What is the distance of this plane from the origin (0, 0, 0)?
(iii) State the vector equation of the line that passes through the point(1, 2, 2) and is parallel to the vector nˆ .
(iv) Find the point where this line intersects the plane r · nˆ = d.
(v) Hence find the distance between the point (1,2,2) and the plane
x + y + z = 1.

For (i) I have the plane equation as r=a+tn^ where a is a point from (x,y,z) and t is a constant. But I'm confused over the form they want it in.

Let $R=$ and $P$ be any point on the plane $\Pi: x+y+z=1$. The normal of $\Pi$ is $N=<1,1,1>$.

Now $\Pi:~ N\cdot (R-P)=0$.

DISTANCE: If the point $Q\notin \Pi$ then the distance from $Q$ to $\Pi$ is $\mathcal{D}(\Pi;Q)=\frac{|\overrightarrow {PQ} \cdot N|}{\|N\|}$.
• Feb 13th 2013, 10:59 AM
MINOANMAN
Re: Vectors and Planes
Carla 1985
I believe this is a typical question of A2 IG for grade 12 students.
1. The point (0,0,1) is a point of the plane x+y+z=1
therefore the equation of the plane is the one discribed by Plato ..where R (x,y,z) and P(0,0,1) and n(1,1,1).

2. distance (0,0,0) from the plane =1/sqrroot(1^2+1^2+1^2)=1/sqrroot(3).

3. r=(1i+2j+3k)+t(1i+1j+1k) this is the vector equation of the line passing through the point (1,2,3) and // to vector n.

4. the parametric equations of the line are: x=1+t , y=2+t , z=3+t . Substitute these into the equation of the plane x+y+z=1 to find that t = -5/3
this value of t gives x=1-5/3 ,y=2-5/3 and z=3 -5/3 these are the coordinates of the point that the plane and the line meet.

5 this is easy d=4/sqrroot(3) .

MINOAS
• Feb 13th 2013, 01:23 PM
carla1985
Re: Vectors and Planes
I still dont get the first bit, I got the general equation that plato gave but dont see what p is as I dont have a point?
• Feb 13th 2013, 01:41 PM
Plato
Re: Vectors and Planes
Quote:

Originally Posted by carla1985
I still dont get the first bit, I got the general equation that plato gave but dont see what p is as I dont have a point?

You are free to pick any point on the plane, $x+y+z=1$.
Here are some possibilities:
$(1,2-2),~(0,1,0),~(0.1,0.3,0.6),~(5,-6,2)$
Any of those co be used for $P$.
• Feb 13th 2013, 01:44 PM
jakncoke
Re: Vectors and Planes
Just pick any point on the plane, say P = (1,0,0) is a point on the plane 1^2 + 0 + 0 = 1 .

Now If any other point, call it X = (x,y,z) lies on our plane, then X - P is a vector or (x-1,y-0,z-0) = (x-1,y,z) is a vector that is parallel to our plane. So if we found a perpendicular vector to our plane, call it v, then V * (X-P) = 0 (The dot product of perp vectors is zero). This is a way we can write planes, namely find a vector parallel to our plane and a vector perpendicular to our plane and dot product of them equals 0.

so if you have a perpendicular vector V= (a,b,c) and a vector parallel to our plane got by subtracting point X = (x,y,z) from a known point P = ( $x_0,y_0,z_0$) X-P = ( $x-x_0$, $y-y_0$, $z-z_0$) then $a*(x-x_0) + b*(y-y_0) + c*(z-z_0) = a*x - a*x_0 + b*y - b*y_0 + c*z - c*z_0$ since $(x_0,y_0,z_0)$ is a known point,(all of the component are constants), you get an equation of the form $a*x+b*y+c*z = k$ (k is all the constant terms added together and brought to the right).

so since your equan is x + y + z = 1 or 1*x + 1*y + 1*z = 1, you have your perp vector (1,1,1). and X - P (from earlier) = (x-1,y-0,z-0) Perp*(X-P) = x-1+y+z = 0
or x + y + z = 1(You get your original eqn back).
• Feb 13th 2013, 03:16 PM
carla1985
Re: Vectors and Planes
:o so my plane equation literally is (r-p).n=0 I was trying to overcomplicate it and get an equation with numbers and components etc, jakncoke, that really helped me understand it, thankyou :D

I've got as far as number 4 now, when we did the other examples we used a different method. we substituted the equation of the line into the equation of the plane. I've got so far through but have ended up with 2 variables which i dont think is right.

I got:
((i+2j+3k)+tn-p).n=0
tn.n+((i+2j+3k)-p).n=0
t=-((i+2j+3k)-p).n
which is the format we had out answer in in the other questions but without the p. i get the feeling iv missed something somewhere :/
• Feb 16th 2013, 02:53 PM
HallsofIvy
Re: Vectors and Planes
Look closely at the equation x+ y+ z= x(1)+ y(1)+ z(1)= 1. Doesn't that immediately remind you of the vector product $\cdot<1, 1, 1>= 1$?