Determining equation of line from equidistant points in vector space

Question

I have a question that gives three points (in 3d space). The task involves finding the equation of a line where the points are equidistant to that of the three given points.

The points given were $(3,1,2)$, $(1, -1, 2)$, and $(3,-1,4)$.

How would I go about solving this problem?

So far I've used the distance formula for each individual point with some point $(x,y,z)$ on the line. But I am unsure of where to go from there?

Answer 1

You are almost done.

The distance formula gives you two equalities and after squaring both sides and cancellation of squares you get two linear equations which are two planes and the intersection is the required line.

$$ (x-3)^2+(y-1)^2+(z-2)^2=(x-1)^2+(y+1)^2+(z-2)^2 $$

simplifies to $$x+y =2$$

and $$ (x-3)^2+(y-1)^2+(z-2)^2=(x-3)^2+(y+1)^2+(z-4)^2 $$

simplifies to $$y-z=-3$$

The parametric equation of the line is $$x=t, y=2-t,z=5-t$$

Answer 2

The basic idea is to find the point in the plane through the points which is equidistant from them, ie the centroid. Then to take the normal to the plane through that point.

Label the points $A(3,1,2)$ and $B(1,-1,2)$ and $C(3,-1,4)$. The centroid is simply $(A+B+C)/3$ ie $(7/3,-1/3,8/3)$.

Notice that in all cases $x-y-z=0$. That is evidently the equation of the plane containing the points.

The normal to the plane through the origin is $(-t,t,t)$ (ie the parametric form, $t$ can take any value), so the normal through the circumcenter is the line $(7/3-t,-1/3+t,8/3+t)$ (again the parametric form).

Check: $AT^2=(3-7/3+t)^2+(1+1/3-t)^2+(2-8/3-t)^2=8/3+3t^2$

$BT^2=(1-7/3+t)^2+(-1+1/3-t)^2+(2-8/3-t)^2=8/3+3t^2$

$CT^2=(3-7/3+t)^2+(-1+1/3-t)^2+(4-8/3-t)^2=8/3+3t^2$

Answer 3

Given two points $A$ and $B$, the locus of the set of points equidistant to them is the perpendicular bisecting plane of the line segment $AB$. For your three fixed points, then, the line of equidistant points is the intersection of the bisecting planes for each of the three possible pairs of points. This gives you a system of three linear equations, only two of which are independent. If you need to express the line in some other form (incidentally, there’s no such thing as the equation of a line), solve the system.