Find a point on the line closest to another point

  • Thread starter Thread starter MarcL
  • Start date Start date
  • Tags Tags
    Line Point
Click For Summary

Homework Help Overview

The problem involves finding a point on a line that is closest to a given point in three-dimensional space. It includes two parts: determining the equation of a plane that is orthogonal to a specified line and finding the closest point on that line to the point P(3,1,4).

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equation of the plane derived from the direction vector and express uncertainty about how to proceed with finding the closest point on the line. Questions are raised about the relationship between the closest point and the plane, as well as the geometric interpretation of the problem.

Discussion Status

The discussion is ongoing, with participants sharing their attempts to derive the plane's equation and exploring the geometric implications of the problem. Some participants suggest visualizing the situation through drawings to gain better insights.

Contextual Notes

There is a mention of the potential redundancy in the relationship between the normal line and the plane, as well as considerations about the minimum distance from point P to the normal line.

MarcL
Messages
170
Reaction score
2

Homework Statement


(2 part problem) a) A plane passes through the point P(3,1,4) and is orthogona to the line (x-1)/2 = (2-y)/-7 = z-3
b) Find the point on the line closest to point (3,1,4)

Homework Equations


Symmetric equation --> (x-x1)/t = (y-y1)/t = (z-z1)/t ( anyway i think that's what it is

The Attempt at a Solution


I found the equation of the plane using the direction vector as (2,-7,1)
and used this form 2(x-3)-7(y-1)+(z-4)= 0

I can't seem to be able to go on to start b. I would think of plugging P in but that seems way too easy.

Any idea on how I can approach this?
 
Physics news on Phys.org
MarcL said:
I found the equation of the plane using the direction vector as (2,-7,1)
and used this form 2(x-3)-7(y-1)+(z-4)= 0

I can't seem to be able to go on to start b. I would think of plugging P in but that seems way too easy.

Any idea on how I can approach this?
Where will that closest point be in relation to the plane?
 
MarcL said:

The Attempt at a Solution


I found the equation of the plane using the direction vector as (2,-7,1)
Check the sign of the y component of the direction vector.

MarcL said:
and used this form 2(x-3)-7(y-1)+(z-4)= 0
I can't seem to be able to go on to start b. I would think of plugging P in but that seems way too easy.

Any idea on how I can approach this?

The distance of P from the normal line is the length of the line drawn perpendicularly to the normal...Where is that line?
 
The normal, but... it seems kinda redundant if it lies on the same line , anyway to me at least.
 
MarcL said:
The normal, but... it seems kinda redundant if it lies on the same line , anyway to me at least.
Of course, it intersects the normal, but what is the position of the line drawn from P and perpendicular to the normal, with respect to the plane? Try to draw a picture.

If you can not see it, write the distance of any point of the normal line from point P. When is it minimum, and what is that minimum distance?
 
Drawing a picture, as ehild suggested, is an excellent idea. Having an image to look at gives you insights that formulas and equations simply can't provide. This took me a couple of minutes to draw.
Plane_and_Line.png
 

Similar threads

Replies
17
Views
3K
Replies
7
Views
2K
Replies
8
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K