Orthogonal projection over an orthogonal subspace

[coltson]

Homework Statement

Being F = (1,1,-1), the orthogonal projection of (2,4,1) over the orthogonal subspace of F is:

a) (1,2,3)
b) (1/3, 7/3, 8/3)
c) (1/3, 2/3, 8/3)
d) (0,0,0)
e) (1,1,1)

The correct answer is B

Homework Equations

The Attempt at a Solution

Using the orthogonal projection formula:

(1,1,-1).(2,4,-1) * (1,1,-1) = 7/3 * (1,1,-1) = (7/3,7/3,-7/3)
(1,1,-1).(1,1-1)

I am obviously misunderstanding something, I thought that maybe orthogonal subspace ain't (1,1-1), but the answer makes me believe that I should have (1,1-1) in the formula.

 

[Buffu]

(1,1,-1).(2,4,-1) * (1,1,-1) = 7/3 * (1,1,-1) = (7/3,7/3,-7/3)
(1,1,-1).(1,1-1)

I am obviously misunderstanding something, I thought that maybe orthogonal subspace ain't (1,1-1), but the answer makes me believe that I should have (1,1-1) in the formula.

Shouldn't that be (2,4,1).

 

[Mark44]

Homework Statement

Being F = (1,1,-1), the orthogonal projection of (2,4,1) over the orthogonal subspace of F is:

The problem statement is confusing to me. I think it's asking for the projection of <2, 4, 1> onto the the orthogonal subspace of F (i.e., the plane perpendicular to F). It might be helpful to think about the geometry here.

a) (1,2,3)
b) (1/3, 7/3, 8/3)
c) (1/3, 2/3, 8/3)
d) (0,0,0)
e) (1,1,1)

The correct answer is B

Homework Equations

The Attempt at a Solution

Using the orthogonal projection formula:

(1,1,-1).(2,4,-1) * (1,1,-1) = 7/3 * (1,1,-1) = (7/3,7/3,-7/3)
(1,1,-1).(1,1-1)

I am obviously misunderstanding something, I thought that maybe orthogonal subspace ain't (1,1-1), but the answer makes me believe that I should have (1,1-1) in the formula.

 

[coltson]

The problem statement is confusing to me. I think it's asking for the projection of <2, 4, 1> onto the the orthogonal subspace of F (i.e., the plane perpendicular to F). It might be helpful to think about the geometry here.

In that case, the dot product between F and a random vector contained in the plane would be equal to zero? That means that F is the normal of the plane?

But in the end, what he probably wants? The projection of F over a vector contained in this plane that is perpendicular to F?

 

[Mark44]

In that case, the dot product between F and a random vector contained in the plane would be equal to zero? That means that F is the normal of the plane?

Yes, that's what it means to be the orthogonal subspace to F.

But in the end, what he probably wants? The projection of F over a vector contained in this plane that is perpendicular to F?

It means to find the projection of <2, 4, 1> onto the plane that is orthogonal to (perpendicular to) F = <1, 1, -1>.

Again, draw a picture.

 

[coltson]

Yes, that's what it means to be the orthogonal subspace to F.

It means to find the projection of <2, 4, 1> onto the plane that is orthogonal to (perpendicular to) F = <1, 1, -1>.

Again, draw a picture.

Well, I don't know how to draw it, so drawing doesn't help. I can imagine what I would like to draw if I had the skill, and all I see is that the projection is a vector contained in the plane, which means it should satisfy the plane equation.

 

[Mark44]

Well, I don't know how to draw it, so drawing doesn't help. I can imagine what I would like to draw if I had the skill, and all I see is that the projection is a vector contained in the plane, which means it should satisfy the plane equation.

You can't do something like this?

Not labelled, but the projection of the vector onto the plane is the one heading off to the right and down.

 

[coltson]

You can't do something like this?
View attachment 218065
Not labelled, but the projection of the vector onto the plane is the one heading off to the right and down.

I do not understand. There is three vectors and three labels, how can it not be labelled? As far as I understand, it is the one with the label "Ortho F". I also do not understand what do you mean by "heading down". Doesn't it has to be contained in the plane? How can it be heading down, since it implies that it would cross the plane?

 

[Mark44]

I do not understand. There is three vectors and three labels, how can it not be labelled? As far as I understand, it is the one with the label "Ortho F". I also do not understand what do you mean by "heading down". Doesn't it has to be contained in the plane? How can it be heading down, since it implies that it would cross the plane?

What I've labelled "Orth F" is the orthogonal subspace of F; i.e. the plane.
Here's another picture that explains things better.

Anyway, what I asked before was why can't you draw something like this? It took me about 2 minutes to create the first drawing, and another minute to update it to the above drawing. Using pencil and paper, you could probably create a simple drawing like this in about the same amount of time.
What I'm calling "Proj <2, 4, 1>" is the heavy line segment that is the projection of <2, 4, 1> onto the plane. This projection is perpendicular to F (= <1, 1, -1>) and lies in the plane. Using simple trig (including the dot product), you should be able to find the vector I'm showing with the heavy line.

Clear?

 

[coltson]

Shouldn't that be (2,4,1).

Here is. A simple typo. Should have realized that before.

Anyway, what I asked before was why can't you draw something like this? It took me about 2 minutes to create the first drawing, and another minute to update it to the above drawing. Using pencil and paper, you could probably create a simple drawing like this in about the same amount of time.

Because I had not thought about drawing a "rotated" square as my plane representation. And about the second picture, I also did not think about the dotted line.