Finding the closest point on a plane to a point not on the plane

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Homework Help Overview

The problem involves finding the point on a specified plane that is closest to a given point not on the plane. The plane is defined by the equation 5x - 14y + 2z + 9 = 0, and the external point is Q = (-2, 15, -7).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the closest point and the normal vector to the plane. There is an exploration of the parameterization of points along the normal vector and how to relate this to the plane's equation. Questions arise regarding the validity of given answers and the interpretation of the equation for the plane.

Discussion Status

Participants are actively engaging with the problem, attempting to clarify their understanding of the relationships between points, the normal vector, and the plane's equation. Some express uncertainty about the correctness of provided answers and seek confirmation or correction from others.

Contextual Notes

There is mention of a typo in the original post that may have affected the discussion. Participants are also questioning the accuracy of an answer key in relation to their findings.

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Homework Statement


Find the point P on the plane given by 5x - 14y + 2z + 9 = 0 which is nearest to the point Q = (-2, 15, -7).


Homework Equations


Distance formulas for distance between the plane and the origin and the plane and a point not on a plane. And of course, the Cartesian equation of a plane.


The Attempt at a Solution


I thought the closest point to Q must be along the normal vector to the plane. The vector would be when P - Q is a scalar multiple of N. My problem is that the answer given was (-1/25)(5, -14, 2). I don't understand this answer since it's independent of Q. Is the answer wrong?
 
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Remember that any point on the normal joining plane P and point Q can be written as:

p(t)=(5,-14,2)t+(-2,15,-7)

Since we are interested in the value t* making p(t*) a point on the plane, we insert this expression into the equation for P, and solve for t*.

Then, we determine p(t*) explicitly.
 
I think that's similar to what I tried doing but didn't have time to write out when I was posting that. I'm not sure what you mean by the "equation for P", but I'm guessing you mean the equation for the plane? So what you're saying is, p(t) = tN + Q, P = p(t*) = (t*)N + Q. If so, that is what I tried doing.

I'll carry it through to completion now though. P = (5, -14, 2)t* + (-2, 15, -7), so x = 5t* - 2, y = -14t* + 15, and z = 2t* - 7. Plugging this into the equation for the plane gives 5(5t* - 2) -14(-14t* + 15) + 2(2t* - 7) = 25t* + 196t* + 4t* - 10 - 210 - 14 = 225t* - 234 = -9 --> 225t* = 225 --> t* = 1. Thus, P = N + Q = (3, 1, -5).

Such was the answer I had gotten earlier, yet it does not match up with the book's answer. So did I misunderstand something/do something wrong?
 
The point given as an answer does not, as far as I can see, lie on the plane.

Yours is correct.
 
Thanks a lot! For future reference, I made a typo in the OP which I have now corrected. The point in the OP is indeed on the plane, but if you are getting the same answer as me on this, then I'm sure that point is incorrect too. If not, anyone please correct this since I only reluctantly change an answer key.
 

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