1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Projections on to a line

  1. Jan 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the projection, P', of the point P on the line p, the distance of P from p and the coordinates of the point R symmetric to P with respect to p, where P = [1, 2, 0], and p : X = [3, 0, 0] + t(0, 1, 0), t ∈ IR.

    *Sorry about all the P's, but this is how the question is written.
    2. Relevant equations
    https://en.wikibooks.org/wiki/Linear_Algebra/Orthogonal_Projection_Onto_a_Line

    3. The attempt at a solution
    Projection of P on to p = [P ⋅ X / X ⋅X] * X

    = [ [1,2,0] ⋅ [3,0,0] + t(0,1,0) / [3,0,0] + t(0,1,0) ⋅ [3,0,0] + t(0,1,0) ] * [3,0,0] + t(0,1,0)

    I'm a bit confused with what [3,0,0] + t(0,1,0) actually means. I've usually seen the line being equal to L = cv, where v is a vector and c is some scalar. So from this I'd assume that [3,0,0] is the vector for the line, and t(0,1,0) is the scalar.
     
  2. jcsd
  3. Jan 25, 2016 #2

    Mark44

    Staff: Mentor

    It's the sum of two vectors. The first vector, <3, 0, 0> extends from the origin to the point (3, 0, 0). The second part of the sum is a scalar multiple of the vector <0, 1, 0>. As a sum, you get a vector from the origin to some point on the line that is parallel to the y-axis and passes through the point (3, 0, 0). For example, if t = 1, the vector extends out to the point (3, 1, 0).
    That would apply only to lines that go through the origin.
    No, both are vectors -- <3, 0, 0> and t<0, 1, 0> and their sum is a vector -- you can't add a scalar to a vector.
     
  4. Jan 25, 2016 #3

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The "points" are usually represented as position vectors in some coordinate system - which tells you the distance (and direction) from some origin O. (What is the difference between the point P and the vector OP?)
    A line requires two points to define it ... eg. if L is a point on the line through P and Q then L = P +t(Q-P) for some t.
    In the equation you are used to, P=O. Off what you are used to, c=t and v=Q-P is the "vector".
    You can also define a line as the set of points equi-distant from P and Q (i.e. |LP|=|LQ|) ... try working out the equation from that.
     
  5. Jan 26, 2016 #4
    ok.

    so I can find t using the equation:
    t = [ P ⋅ p' / p' ⋅ p' ] ⋅ p'
    = ( [1,2,0] ⋅ [3,1,0] / [3,1,0]⋅ [3,1,0] ) ⋅ [3,1,0]
    = 5/10 ⋅ [3,1,0]
    t= [ 3/2 , 1/2, 0 ]

    Does this mean that [3,0,0] + 1/2 (0,1,0) = [3,1/2,0]

    Does this mean the proj P on to p' = [3,1/2,0]

    I don't think this is right because (P-p') ⋅ p' = 0 because P-p' is orthogonal to p'

    if p' = [3,1/2,0] then I need to ⋅ it with another vector that makes it = 0
     
  6. Jan 26, 2016 #5

    Mark44

    Staff: Mentor

    ???
    This makes no sense. t is a scalar -- a number.
    Also, what is p'? In post 1 you give explanations for P, P', and p, but not for p'.
    I can appreciate that you might be confused, with this problem's overuse of a single letter, differentiated only by upper and lower case, and with or without an apostrophe.
    Where did you get this formula -- t = [ P ⋅ p' / p' ⋅ p' ] ⋅ p' -- ?
    As you have written it, you are trying to find the dot product of a scalar and a vector. That doesn't make sense. You can only take the dot product of two vectors. In the brackets you have P⋅p' (a scalar -- a number) divided by p' ⋅ p' (a scalar), which gives you a scalar. So far, so good. Then you have the dot product of a scalar (in the brackets) and p'. That's not right.

    Possibly you mean for the final multiplication to be scalar multiplication rather than the dot product. If so, don't use the same symbol for dot product that you use for multiplication of a vector by a scalar.
    Yes. Here t = 1/2. It's NOT a vector, so can't possibly be equal to <3/2, 1/2, 0>
     
  7. Jan 26, 2016 #6

    Mark44

    Staff: Mentor

    Have you drawn a sketch that includes the information here?

    What does IR mean, as you have it in "t ∈ IR"?
     
  8. Jan 26, 2016 #7
    Sorry, I copied and pasted the question. t ∈ R.

    The problem I am having with this question is understanding [3,0,0] + t(0,1,0). I've calculated the projection, presuming the vector that is on the line p is [3,0,0].

    You are right in post #5 @Mark44, I have used an incorrect equation. It should have been
    t = [ P ⋅ p / p ⋅ p ]

    where
    t is a scalar
    P is the vector [1,2,0]
    p is the vector that sits on the line

    t = [1,2,0] ⋅ [3,0,0] / [3,0,0] ⋅ [3,0,0] ]
    = 3/9 = 1/3

    projection of P on to p = 1/3 * [3,0,0]
    = [1,0,0]

    To check the answer we know that the vector P - the projection dotted with the vector p = 0 because they are orthogonal to each other.

    (P - projp P) ⋅ [3,0,0] = 0
    (0,2,0) ⋅ [3,0,0] = 0

    I've attached a photo of my working. I know I probably haven't answered the question properly, because I ignored the + t(0,1,0) bit, but I'm confused with the notation and what it means. The projection of P on to p will be any multiple of the vector [3,0,0] + t(0,1,0). If [3,0,0] and (0,1,0) both sit on the line then I can create a linear combination of the two and say that the vector [3,1,0] multiplied by some scalar, t, is equal to the projection of P on to p.
     

    Attached Files:

  9. Jan 26, 2016 #8

    Mark44

    Staff: Mentor

    As I said before, this is the sum of two vectors: the first, <3, 0, 0>, is the vector from the origin to the point (3, 0, 0). I use <> to denote vectors, and () to denote points. The second vector is t<0, 1, 0> which is exactly the same as <0, t, 0>. If you add <3, 0, 0> and t<0, 1, 0> you get a vector that extends from the origin to some point on the line.

    For example, if t = 0, the vector sum is <3, 0, 0>. If t = 1/2, you get the vector <3, 0, 0> + <0, 1/2, 0> = <3, 1/2, 0>, which extends from the origin to the point (3, 1/2, 0). For each value of t, you get a different vector that points to a different point on the line.
    (3, 0, 0) is a point on the line, but the direction of the line is the vector <0, 1, 0>. In other words, the line is parallel to the y-axis.
    No. If t = 1/3, and assuming you're still working with your parametric form of the line, you get <3, 0, 0> + (1/3)<0, 1, 0> = <3, 1/3, 0>. This vector goes from the origin to the point (3, 1/3, 0).
    Your drawing would be better if you had overlaid your vectors onto an x-y coordinate system. Because of the vectors involved in this problem, in which the z-coordinate is 0 for all, you can get away with drawing everything in two dimensions.
     
  10. Jan 26, 2016 #9

    Mark44

    Staff: Mentor

    Here's a sketch of your problem. Rather than call the line p, which is very confusing, I have labeled it L. Point P(1, 2, 0) can be thought of as the head of the vector ##\vec{OP}##, which extends from the origin to point P. The line segment I show going from the origin out to line L is the vector <3, 0, 0> + 1<0, 1, 0>, where t = 1. With different values of t, you get different vectors, all of which end on line L.

    From the drawing, it should be pretty obvious what the projection of ##\vec{OP}## in the direction of L is, as well as how far P is from line L.
    Projection.png x
     
  11. Jan 27, 2016 #10
    Ok, that makes much more sense. Thank you.

    So the line can be thought of as a vector [3,0,0] but the projection we want is the projection of P onto a vector that goes from the origin to the line, and sits on the y-axis at t-intervals of y.
     
  12. Jan 27, 2016 #11

    Mark44

    Staff: Mentor

    NO!
    The vector <3, 0, 0> is parallel to the x-axis. In my updated drawing, below, the vector <3, 0, 0> is the horizontal red line segment. The line (L in my drawing) is parallel to the y-axis.
    No.
    Here is the problem statement from post #1:
    This problem is very poorly written, IMO, which makes it very difficult to understand. Some things that stand out:
    1. If you project a point (P) onto a line, you get a point. The problem seems to be asking for a vector.
    2. The way the problem is written suggests that point P is on the line (it isn't).
    3. The problem's author is very unimaginative, using P, P', and p to describe different things in the problem, especially using p as the label for a line.

    You have complained about your textbook before. Who is the author of this textbook? Maybe your complaint about the book being worthless is valid.

    Here is how I interpret the first part of this problem.
    Find the projection of the vector ##\vec{OP}## (in green) onto line L (in black).​

    The red vector, <3, 1, 0> is the sum of two other vectors (in blue), <3, 0, 0> + t<0, 1, 0>. In the drawing, t = 1. Each choice of t results in a vector that points to a different point on the line (L, in black).

    You seem to be trying to find the projection of <1, 2, 0> (in green) onto the red vector. The problem, the best I can understand it, is asking you to project the vector <1, 2, 0> onto the line L (in black).

    Projection.png
     
  13. Jan 27, 2016 #12
    I agree with the statement about the question. Too many P's! I think the question is asking for the projection of P onto the line (3,1,0) or any other [3,0,0] + t(0,1,0) though.
     
  14. Jan 27, 2016 #13

    Mark44

    Staff: Mentor

    That's not how I read the problem. It's asking for the projection onto the line I'm calling L, not onto a vector <3, 0, 0> + t<0, 1, 0>. This vector generates all of the points on line L.

    You should probably ask your teacher for his or her opinion of what this problem ias asking for.

    BTW, what textbook are you using?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Projections on to a line
  1. Projections checkß (Replies: 1)

  2. Vector projections (Replies: 10)

Loading...