start with what you know. take a point in R^3, and calculate it's distance from p, and then it's distance from q. if they are equal, what do you get?
I need some help with these two questions, does anyone know where to start?
Show that the points in R^3 equidistant from two fixed points p and q form a
plane, and find its equation.
Let p + tv, t in R be the parametric description of a line l and let q be a point
not on l. Show that the points (in R^3) which lie on a line through p + tv and q
(some variable t) make up a plane. What is its equation?
i can get the distance of two points using
sqrt((a1-b1)^2+(a2-b2)^2+(a3-b3)^2)
So lets say g is the point i take from R^3, and if g is equidistant to p and q, if distance from g to p = distance from g to q, but i don't know how to use this though.
Well i can visualize this in my mind, so if you have 2 fixed points, and take an infinite number of equidistant points, then make a plane connecting those points, not p and q, it will make a plane, since u used all equidistant points from two fixed points, but I dont know how to show it mathematically.
so M = ((p1-q1) / 2,(p2-q2) / 2,(p3-q3) / 2)
PQ = (q1-p1, q2-p2, q3-p3)
so RM = (x*(p1-q1) / 2,y*(p2-q2) / 2,z*(p3-q3) / 2)
so PQ dot RM =
(q1-p1, q2-p2, q3-p3) dot (x*(p1-q1) / 2,y*(p2-q2) / 2,z*(p3-q3) / 2)=0
which is
(q1-p1)*x*(p1-q1)/2 + (q2-p2)*y*(p2-q2)/2 + (q3-p3)*z*(p3-q3)/2=0
=
[(q1-p1)*x*(p1-q1) + (q2-p2)*y*(p2-q2) + (q3-p3)*z*(p3-q3)]/2=0
(q1-p1)*x*(p1-q1) + (q2-p2)*y*(p2-q2) + (q3-p3)*z*(p3-q3) = 0
is this right?