Normal unit vector from eqn in plane P

Smythe

Can anyone let me know how to go about this type of question?

Thank you

The equation of points in a plane P is x − y + 3z = 2

Find nˆ, a unit vector along the normal to P.

sponsoredwalk

Pages 34-36.

You can't really see the picture so here's a pic from my edition, very
similar:

Smythe

Thanks for the link. I can only seem to see page 34, unfortunately.

sponsoredwalk

Damn, I can see all of them, okay.

First some notation:

[latex] \overline{AB} \ = \ (\overline{B} \ - \ \overline{A}) \ = \ \overline{0(B - A)}[/latex]

The vector from point A to point B can be written as any of the above,
the third representation being a vector starting at the origin & finishing
at the point (B - A), if you look at the picture you'll see the exact same
thing going on with both vectors.

The idea is to look at the picture I posted above & consider the vector
[latex] (\overline{X} \ - \ \overline{P})[/latex] which is on the plane in the picture. By picking any point X on that
plane you can generate the whole plane, starting from point P.

Now, the vector [latex] (\overline{X} \ - \ \overline{P})[/latex] on the plane is the exact same as the vector [latex] \overline{0(X - P)}[/latex].
This is really just to make things simpler to work with.
I hope that isn't confusing but notice the slight difference between
[latex] (\overline{X \ - \ P})[/latex] & [latex] (\overline{X} \ - \ \overline{P})[/latex],
the first is a vector from the origin to the point (X - P) while the second
is the difference of two vectors (this is all explained in the previous pages
of the book & haven't seen it explained like this anywhere else so...).

Using the vector on the plane, translated to the origin, we take

[latex] \overline{0(X - P)} \ \cdot \overline{N} \ = \ (\overline{X} \ - \ \overline{P}) \ \cdot \overline{N} \ = \ 0[/latex]

which can pedantically be written as:

[latex] \overline{0(X - P)} \ \cdot N \ = \ (\overline{X} \ - \ \overline{P}) \ \cdot \overline{N} \ = \ (\overline{X} \ - \ \overline{P}) \ \cdot \ (\overline{P} \ + \ \overline{N} \ - \ \overline{P}) \ = \ 0 [/latex]

or just

[latex] (\overline{X} \ - \ \overline{P}) \ \cdot \overline{N} \ = \ 0[/latex]

This represents the entire equation of the plane in the picture using only
a vector on the plane & a vector normal to the plane.

From this you can derive the equation ax + by + cz = d or
a(x - x₀) + b(y - y₀) + c(z - z₀) = 0 by just working through
the algebra in [latex] (\overline{X} \ - \ \overline{P}) \ \cdot \overline{N} \ = \ 0[/latex].

Do this, find both ax + by + cz = d & a(x - x₀) + b(y - y₀) + c(z - z₀) = 0
& then start from either of these & see if you can determine
[latex] (\overline{X} \ - \ \overline{P}) \ \cdot \overline{N} \ = \ 0[/latex].

Then your question should be answered trivially, after converting the
normal vector into it's unit vector representation

(hint: [latex] \overline{X} \ = \ ||\overline{X}|| \hat{X}[/latex], but how can you even make sense of the above
discussion if N is a standard vector & not a unit vector ).

Smythe

Thanks very much for that.

I think I'll have to wait until we cover this in more detail in class, as I don't think I really understood the post.

I'll take a look at it again tomorrow when hopefully I'm less exhausted!

sponsoredwalk

Ah I'm sorry, I guess I didn't motivate that post at all nor did I explain
to you why I'd start at such an apparently random place. I've written a
post that motivates wtf I'm doing & why I'm doing it, really though I
wouldn't have done this unless I thought it was necessary, the whole
normal vector thing that your question is asking about always seemed
very strange to me until I fully understood what I've written (and
bothered me) so give it another shot when you feel up for it, hopefully
the following will clear everything up:

Look at the picture of the tilted plane containing the vector [latex] (\overline{X} \ - \ \overline{P})[/latex].
If we pick any point X = (x,y,z) on that plane we can then form a vector
starting @ the point P = (x₀,y₀,z₀) & ending at X = (x,y,z),
the vector [latex] \overline{PX} \ = \ (\overline{X} \ - \ \overline{P})[/latex].

Notice that because we're dealing with 3 dimensions we're free to pick a point
anywhere in space that we want & can then form a vector starting from
our point P to this random point X, it doesn't have to be on the plane.

But if we want to talk about those vectors that are contained in the tilted
plane, and those vectors only, this is equivalent to saying that we want to
talk about those points X = (x,y,z) (& only those points) in the tilted plane.

We also know from elementary algebra that you can describe the equation
of a line by ax + by = c, & you can describe the equation of a plane by
ax + by + cz = d, so every point X = (x,y,z) on the plane must satisfy
this specific equation of the tilted plane (whatever that is, we don't know
yet, the constants a, b, c (& d?) will determine the plane - i.e. they are the
reason why the plane is tilted the way it is in the picture).

But sticking with vectors, we need a way to be 100% certain that all the
vectors must lie in the plane, which implies that all points points X will be
contained in the plane. One way to do this is to find a vector that is
perpendicular to the plane. If we find this perpendicular vector [latex] \overline{N} \ = \ (a,b,c)[/latex] then
every vector [latex] \overline{PX} [/latex] in the plane will be perpendicular to [latex] \overline{N}[/latex].
That way we can ensure that those vectors, and only those vectors, that
lie in the plane will be chosen. So we use the dot product because the
dot product encodes the idea of perpendicularity very easily.
Remember two vectors [latex] \overline{A}[/latex] & [latex] \overline{B}[/latex] are perpendicular (orthogonal) if [latex] \overline{A} \cdot \ \overline{B} \ = 0[/latex].

Just think, if we start at P & pick a point X that isn't lying in the tilted
plane, if we take the dot product of the vector joining P & X, the vector
[latex] \overline{PX}[/latex], with the vector [latex] \overline{N}[/latex] then we wont get zero!
It's a test to ensure we get what we want!

Hopefully now you can see what's going on with the vector
[latex] (\overline{X} \ - \ \overline{P}) \ \cdot \ \overline{N} \ = 0[/latex]
in my first post, that's the important thing, everything else is a more
pedantic way to say the same thing.

But write out [latex] (\overline{X} \ - \ \overline{P}) \ \cdot \ \overline{N} \ = 0[/latex]
in it's coordinate form:

[latex] (\overline{(x,y,z)}) \ - \ \overline{(x_0,y_0,z_0)}) \ \cdot \ \overline{(a,b,c)} \ = 0[/latex]

Work it out, it should all be clear now.

As for the unit vector stuff, how to express any vector in terms of it's
associated unit vector? If so, when you are looking at the vector [latex]\overline{N}[/latex],
notice that in everything I've written the vector [latex]\overline{N}[/latex]
could be any length, it doesn't matter as long as [latex]\overline{N}[/latex] is
perpendicular to the plane. The unit vector associated to any vector is just
a scalar multiple of that vector, so it doesn't matter whether you use [latex]\overline{N}[/latex]
or it's associated unit vector [latex]\hat{N}[/latex] in
[latex] (\overline{X} \ - \ \overline{P}) \ \cdot \ \overline{N} \ = 0[/latex] .

So hopefully you should be able to understand:

1) Why we need to specify a starting point P in the plane,
2) Why we're free to pick a point X anywhere
3) The reason why we want to restrict those points X to the plane,
4) How to restrict those points in the plane by working through vectors,
5) What the vector equation of the plane is,
6) How to derive both forms of the classical equation of the plane
|||I mentioned in my last post from the vector equation of the plane,
7) How to identify the Normal vector used in the vector form of the eq. of a plane
|||simply by looking at either of the classical equations of the plane,
8) How to do points 5 through 7 by instead using the unit vector associated with the normal vector
9) How it doesn't matter that everything we do through vectors is being done
|||as if each vector was situated at the origin even though the plane most certainly
|||does not pass through the origin.

equivariant

Smythe wrote: »

Can anyone let me know how to go about this type of question?

Thank you

First, try this variaton. I'll explain why its relevant in a minute

Find a vector perpendicular to the plane x-y+3z=0

How to solve this? Here are some hints. Look at the left hand side. That is (1)x+(-1)y+(3)(z). Can we think of that as a dot product? If yes, then the equation is saying that a dot product =0. What does that mean about the two vectors?


OK, assuming that you can do the variation, what about your original problem? That has x-y+3z=2. The only difference was the constant term. Here is the key. If you change the constant term in a linear equation, that just moves the plane to a parallel plane. So any vector perpendicular to x-y+3z=0 is also perpendicular to x-y+3z=2 (and vice versa). You should think about this a bit, but it is true.

Smythe

sponsoredwalk wrote: »

Ah I'm sorry

No problem. I'd been working since 6.00am yesterday morning when I read your post, so wasn't really able to absorb it when I read it yesterday evening.

I see what you guys mean now.

I borrowed the book you linked-to from the library. It looks quite informative, so I'll give that a read as well.

Thanks very much again for your help sponsoredwalk and equivariant.