Results 1 to 6 of 6

Math Help - Distance between a point and a line Question?

  1. #1
    Newbie
    Joined
    Jul 2008
    Posts
    5

    Question Distance between a point and a line Question?

    Hi everyone, i have a midterm exam coming up and i'm looking at problems in the section of the book to help myself prepare. Can anyone answer this question for me? I have attempted to answer this problem but without the answer i don't know if i'm right.

    Find the distance between the point P(1,-2,4) and the line given by the parametric equations:

    x= 2t
    y= t-3
    z= 2t+2

    I tried using the techniques given in the book but it's not for a point and line.

    Can someone do this one out for me? It's not a homework, i just need to see it done out step by step so i can follow it, along with the answer.

    Thank you guys, really desperate...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,682
    Thanks
    614
    Hello, chrisduluk!

    I don't know if you've had Calculus yet.
    . . I'll give you a non-Calculus solution.


    Find the distance between the point P(1,-2,4) and the line given by:

    . . x .= .2t,. . y .= .t - 3,. . z .= .2t + 2
    Let point Q be on the line: .Q (2t, t-3, 2t+2)
    . . . . . . . . . . . . . . . . . . . . . . . . . . . ____________________________________
    The distance from P to Q is: . d . = . √[(2t) - 1] + [(t - 3) + 2] + [(2t + 2) - 4]
    . . . . . . . . . . . . . . . . . . . . . . ___________
    . . which simplifies to: . d .= .√9t - 14t + 6 . [1]

    And we want to minimize d.


    If we minimize the square of d, we will minimize d.

    Let D .= . d^2 .= .9t - 14t + 6

    This is an up-opening parabola; its minimum point is at its vertex.
    The formula for the vertex is: .v .= .-b/2a

    . . We have: .a = 9, .b = -14

    . . The vertex is: .t .= .-(-14)/(29) .= .7/9

    . . . . . . . . . . . . . . . . . . . . . _________________ . . . . . ___
    Substitute into [1]: . d . = . √9(7/9) - 14(7/9) + 6 . = . √5/9

    . . . . . . . . . . . . . ._
    Therefore: . d .= .√5/3

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by chrisduluk View Post
    ...

    Find the distance between the point P(1,-2,4) and the line given by the parametric equations:

    x= 2t
    y= t-3
    z= 2t+2

    ...
    1. Rewrite the equation of the line:

    (x,y,z) = (0,-3,2) + t \cdot (2,1,2)

    2. Create the equation of a plane E which contains P and has the direction vector of the line as it's normal vector:

    E: (2,1,2) \cdot \left( (x,y,z)-(1,  -2, 4)\right)=0~\implies~(2,1,2)\cdot (x,y,z)-8=0

    3. Calculate the coordinates of the point of intersection between the plane E and the given line. Plug in the term of (x, y, z) from the line into the equation of the plane:

    (2,1,2)\cdot \left((0,-3,2) + t \cdot (2,1,2) \right)-8=0 ~\implies~ 1+9t-8 = 0 ~\implies~ t = \frac79

    4. The point of intersection has the coordinates S\left(\frac{14}9~,~ -\frac{20}9~,~ \frac{32}9 \right)

    5. Calculate the distance between P and S which is the absolute value of \overrightarrow{PS} = \left(\frac59~,~ -\frac29~,~ -\frac49  \right)

    6. \left| \overrightarrow{PS} \right| = \sqrt{\frac{25}{81} + \frac{4}{81} + \frac{16}{81}} = \frac19 \sqrt{45} = \frac13 \sqrt{5} \approx 0.745


    Please check my arithmetic!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    Vector Geometry

    We can use vector geometry to represent this problem.
    Suppose we have the line through the point \mathbf{q} parallel to the vector \mathbf{l}, and a point \mathbf{p} not on the line.
    Then \mathbf{r}=\mathbf{p}-\mathbf{q} is the vector from point \mathbf{q} to point \mathbf{p}. If it makes an angle θ with the line, and thus with \mathbf{l}, then the distance from the point \mathbf{p} to the line is d=|\mathbf{r}|\sin\theta. Now, recall that the magnitude of the cross product of two vectors \mathbf{a} and \mathbf{b} is given by
    |\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{  b}|\sin\theta.
    Thus,
    d=\frac{|\mathbf{r}\times\mathbf{l}|}{|\mathbf{l}|  }
    d=\frac{|(\mathbf{p}-\mathbf{q})\times\mathbf{l}|}{|\mathbf{l}|}

    Our point is \mathbf{p}=\mathbf{\hat{i}}-2\mathbf{\hat{j}}+4\mathbf{\hat{k}}, and our line is that which passes through the point \mathbf{q}=-3\mathbf{\hat{j}}+2\mathbf{\hat{k}} parallel to the vector
    \mathbf{l}=2\mathbf{\hat{i}}+\mathbf{\hat{j}}+2\ma  thbf{\hat{k}}.
    Thus
    \mathbf{r}=\mathbf{p}-\mathbf{q}=\mathbf{\hat{i}}+\mathbf{\hat{j}}+2\mat  hbf{\hat{k}},
    and
    d=\frac{|\mathbf{r}\times\mathbf{l}|}{|\mathbf{l}|  }
    =\frac{|(\mathbf{\hat{i}}+\mathbf{\hat{j}}+2\mathb  f{\hat{k}})\times(2\mathbf{\hat{i}}+\mathbf{\hat{j  }}+2\mathbf{\hat{k}})|}{|2\mathbf{\hat{i}}+\mathbf  {\hat{j}}+2\mathbf{\hat{k}}|}
    =\frac{|2\mathbf{\hat{j}}-\mathbf{\hat{k}}|}{\sqrt{2^2+1^2+2^2}}
    =\frac{\sqrt{2^2+(-1)^2}}{3}
    d=\frac{\sqrt{5}}{3}

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2009
    Posts
    69

    Question

    THIS is a response to the MHF helper's response for the first solution. I did this on a test, got the RIGHT answer and my teacher wrote "OH NO" when i said we have to minimize the square of d. Why? Then he gave me 0 points.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by Intsecxtanx View Post
    THIS is a response to the MHF helper's response for the first solution. I did this on a test, got the RIGHT answer and my teacher wrote "OH NO" when i said we have to minimize the square of d. Why? Then he gave me 0 points.
    Only that instructor can really answer that question.
    Prehaps he wanted you to use only vector methods.
    But as you can see the answers are the same.

    So ask him.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: May 16th 2011, 04:57 AM
  2. line point distance
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 7th 2010, 08:58 PM
  3. Distance between point and line
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 12th 2009, 04:26 PM
  4. distance from point to line
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 8th 2009, 02:52 PM
  5. Replies: 3
    Last Post: February 20th 2008, 10:17 AM

Search Tags


/mathhelpforum @mathhelpforum