Find point d on the line l closest to point c

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  • #1
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Homework Statement


Find point d on the line l closest to the point c (1,1,7). Point c is on the end of a vector who's origin (1,1,2) is on line l. There is an imaginary line that connects point c to point d. This imaginary line is perpendicular to the line l. This problem is relating to vectors and just before this problem I found the parametric equation of the line l so I guess Im going to need that to work this problem.


Homework Equations


parametric equation for l x=2+t u=3+2t z=5+3t
the angle between the vector with point c and line l is 36.7 degrees
c=(1,1,7)




The Attempt at a Solution

I think there is a specific equation for this but I am not sure. I thought about it for awhile but am still confused.
 

Answers and Replies

  • #2
ehild
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It would be nice to see a drawing.

Do you know how to get the projection of a vector onto the direction of an other vector and how it is related to the dot product of those vectors?

ehild
 
  • #4
ehild
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Thanks for the picture. You see the right triangle. The vectors enclose the angle of 36.7 °.

The closest distance between a fixed point and any point of a line is the length of the perpendicular light segment drawn from point C to the line. What is this length, knowing the hypotenuse of the right triangle?

ehild
 
  • #5
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Yes I know right triangles in two dimensions however this is in three dimensions. Thats what is confusing this for me.
 
  • #6
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I know that the magnitude of the vector with the point c is 5. So using cos(36.7)=x/5
I get x=4 this is the length along line l. However this is in three dimensions so I thought that there was something else I had to do.
 
  • #7
ehild
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You need the length of the side opposite to the angle. It is the distance of point C from the line.

The problem is in three dimensions, but the line and the vector define a plane.

ehild
 
  • #8
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oh sorry, the length between the vector and line l is 3.
 
  • #9
ehild
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So the distance of point C from the line is 3. I just re-read the problem, and it asks the position of the point of the line which is closest to C. You gave the distance to the closest point on the line from point (1,1,2), it was 4. Now you need to find the coordinates of the closest point.
Hint: how far do you reach along the line if you change the parameter t by 1? What is the change of t to the closest point?

ehild
 
Last edited:
  • #10
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idk if you change it by one I guess you reach along the line one. I just don't know this stuff because we haven't covered it yet in class yet we have homework over it.
 
  • #11
ehild
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The direction vector is [itex]\vec{t}[/itex]=(1,2,3) (from the parametric equation). Its length is √14=3.74. If you move by one [itex]\vec{t}[/itex], your displacement is 3.74. The distance (OP) along the line is 4. How much does the parameter t change along this distance? You need to find the coordinates of P. Both O and P fulfil the parametric equation of the line. What is t at point O? What is t at point P?

ehild
 

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  • #12
HallsofIvy
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Looks to me like you need the cosine of that angle.
 
  • #13
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The point I picked on the line l for the parametric equation was not point (1,1,2)... So whatever point you pick on the line for your parametric equation is that your t=0 point?
If I were to pick point (1,1,2) on the line l for the parametric equation then it becomes x=1+t u=1+2t z=2+3t then t would =1.0695 to move 4. so x=2.0695 u=3.139 z=5.209
edit: I chose the point (1,1,2) because that is the coordinates for point O.
 
  • #14
ehild
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Superb!!! Correct!!!:smile:

ehild
 
  • #15
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Thank you for your assistance:biggrin:
 

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