Find all points with same distance to (A) and (B)

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In summary, to find all points (x1,x2,x3) that have the same distance to A and B, we can first find the line they both lie on using the equation L=(3,4,-3)+t(2,1,-3). Then, we can find the plane with point C=(A+B)/2 and normal L, which can be found by brute force or by using analytic geometry. Alternatively, we can pull out the normal from L's equation and use the mid-point formula (A+B)/2 to find the equation of the plane.
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reedy
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I have two given points.

A=(-1,2,3) and B=(3,4,-3).

I want to find all points (x1,x2,x3) that have the same distance to A and B.

Attempt:

I think finding the line they both lie on is a good first step. To do this, I've drawn a vector between A and B, and found the line's equation. L=(3,4,-3)+t(2,1,-3). I've come to the conclusion that I have to find a perpendicular bisector of some sort. And I guess cross product won't work, cause I only have one line(?). What is the next step?
 
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  • #2
A perpendicular bisector would be appropriate in two dimensions, but in three dimensions we are looking for a plane with normal L.

We can find it by bruteforce:
[itex](x,y,z)[/itex] is
[tex]\sqrt{(x+1)^2+(y-2)^2+(z-3)^2}[/tex]
from A and
[tex]\sqrt{(x-3)^2+(y-4)^2+(z+3)^2}[/tex]
from B so you're looking for all triples (x,y,z) such that:
[tex]\sqrt{(x+1)^2+(y-2)^2+(z-3)^2} = \sqrt{(x-3)^2+(y-4)^2+(z+3)^2}[/tex]
Square, expand and cancel a bit and you should get the equation of a plane that describes all points equally far from A and B.

You could also have noted that [itex]C=(A+B)/2[/itex] is equally far from A and B, so you want to find the plane with point C and normal CA. From analytic geometry you should be able to find this plane (the same as before), but in my opinion this is more complicated than bruteforce.
 
  • #3
rasmhop said:
A perpendicular bisector would be appropriate in two dimensions, but in three dimensions we are looking for a plane with normal L.

Of course! What was I thinking...

But if that is true, which it is, wouldn't it be easier to just use the fact that
[PLAIN said:
http://en.wikipedia.org/wiki/Normal_vector]For[/PLAIN] a plane given by the equation ax + by + cz = d, the vector (a,b,c) is a normal.

and pull out the normal from L's equation, (2,1,-3), and establish that the plane's equation is 2x1+x2-3x3?

If so, I would need a right hand side as well. And the only point I know for sure in the plane is C.

You say that C=(A+B)/2. Why is that? Is it plain ol' division on each coordinate? (-1,2,3)+(3,4,-3)=(2,6,0) // division // (1,3,0)

Gives me: 2x1+x2-3x3=5 which is a correct answer according to the textbook. But the last step confused me.
 
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  • #4
reedy said:
and pull out the normal from L's equation, (2,1,-3), and establish that the plane's equation is 2x1+x2-3x3?
Yes it would be possible, and what you find easiest is up to you. Neither is particularly complicated.

You say that C=(A+B)/2. Why is that? Is it plain ol' division on each coordinate? (-1,2,3)+(3,4,-3)=(2,6,0) // division // (1,3,0)
Yes. We use (A+B)/2 because that is the mid-point of the line segment connecting A and B. For one way of seeing this consider the vector AB. Adding half of this to A we get half-way to B from A so:
C = A + 1/2 * AB = A + 1/2(B-A) = 1/2(A+B)
 
  • #5
rasmhop said:
C = A + 1/2 * AB = A + 1/2(B-A) = 1/2(A+B)
Ahh, perfect. Now I see it - thanks a bunch!
 

FAQ: Find all points with same distance to (A) and (B)

1. How do I calculate the distance between two points?

To calculate the distance between two points, you can use the distance formula, which is √((x2-x1)^2 + (y2-y1)^2). The x and y values represent the coordinates of each point.

2. What does it mean for two points to have the same distance to point A and B?

If two points have the same distance to point A and B, it means that they are equidistant from both points. This can also be referred to as being equidistant from a line segment.

3. Can there be more than one point that has the same distance to points A and B?

Yes, it is possible for there to be more than one point with the same distance to points A and B. In fact, there can be an infinite number of points that satisfy this condition.

4. How can I find all the points with the same distance to points A and B on a graph?

To find all the points with the same distance to points A and B on a graph, you can plot the two points A and B, and then draw a circle with its center at the midpoint between A and B. The radius of the circle will be the distance between A and B. Any point on the circumference of the circle will have the same distance to points A and B.

5. Is there a specific method for finding all the points with the same distance to points A and B?

Yes, there is a specific method for finding all the points with the same distance to points A and B. It involves using the distance formula and setting the distance equal to the radius of a circle centered at the midpoint between A and B. This will give you the equation of the circle, and any point on the circumference of this circle will have the same distance to points A and B.

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