Find a point on the line closest to another point

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A plane is defined that passes through point P(3,1,4) and is orthogonal to the line given by the symmetric equation. The equation of the plane is derived using the direction vector (2,-7,1). The discussion focuses on finding the closest point on the line to point P, with suggestions to visualize the problem through drawing. Participants emphasize the importance of understanding the geometric relationships between the plane, the line, and point P to solve the problem effectively. Visual aids are recommended to clarify the spatial relationships involved.
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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?
 
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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
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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