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Thread: Distance between lines in R^3

  1. #1
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    Question Distance between lines in R^3

    Hi,

    I'm trying to determine the distance between the following lines:
    Distance between lines in R^3-screen-shot-2018-02-02-6.35.25-pm.png

    To solve this question, I'll be using the following distance formula:
    Distance between lines in R^3-screen-shot-2018-02-02-6.36.59-pm.png

    I know that the vector 'm' would be (1,1,3), but in terms of vector 'QP' I'm not sure how to determine this. Can someone help?

    - otownsend
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  2. #2
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    Re: Distance between lines in R^3

    You realize, don't you, that because both lines have direction vector <1, 1, 3?> the two lines are parallel?

    Take any point on the first line, say take (3, 1, -2) with m= 0. The plane perpendicular to the first line, through that point, is 1(x- 3)+ 1(y- 1)+ 3(z+2)= x+ y+ 3z+ 2= 0. The second line, x= 1+ n, y= n, z= 1+ 3n, intersects the plane where 1+ n+ n+ 3(1+ 3n)+ 2= 11n+ 6= 0 so n= -6/11. So the point is x= 1- 6/11= 5/11, y= -6/11, z= 1- 18/11= -7/11. The distance between the two lines is the distance between the points (3, 1, -2) and (5/11, -6/11, -7/11).
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    Re: Distance between lines in R^3

    Quote Originally Posted by otownsend View Post
    I'm trying to determine the distance between the following lines:
    Click image for larger version. 

Name:	Screen Shot 2018-02-02 at 6.35.25 PM.png 
Views:	6 
Size:	21.2 KB 
ID:	38504
    To solve this question, I'll be using the following distance formula:
    [ATTACH=CONFIG]38505[/ATTACH
    I know that the vector 'm' would be (1,1,3), but in terms of vector 'QP' I'm not sure how to determine this. Can someone help?
    Just let $P=(3,1,-2)~\&~Q=(1,0,1)$ and as you said $m=(1,1,3)$ then apply the distance formula.
    $\overrightarrow {QP} = \left\langle {2,1, - 3} \right\rangle $
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