Question

Find an equation for the plane consisting of all points that are equidistant from the given points.
(-6, 1, 2), (4, 3, 6).

Answer 1

I don't know how to do this myself, but I found a similar problem here http://answers.yahoo.com/question/index?qid=20080123003055AABx40u

Answer 2

ok here is what ya do. Find a vector between the two points (-6,1,2) and (4,-3,6) which gives you (tip minus tail) (10,2,4). Now you find the length of this line (sqrt(10^2+2^2+4^2) which gives you 2 sqrt(30). Now halfway between the two points will give you sqrt(30). Now find a unit vector in the direction which will be the vector devided by it's length (5/srt(30), 1/sqrt(30), 2/sqrt(30)) this is a unit vector that lies along your line. Multiply this by your half length which basically gives you (5,1,2) this is a point that lies on the plane that you need. NoNow you have two things 1) a point 2) a normal vectro to the plane which is the unit vector i told you (5/srt(30), 1/sqrt(30), 2/sqrt(30)). Now plug these into the equation of a plane. A(x-x0)+B(y-yo)+c(z-z0)=0 where ABC come from your normal vector and Xoyozo come from your point

Answer 3

note that this can be negated because there are two normal vectors which have the same path but an opposite sense or direction (+,-)

Answer 4

not sure if they use positive or negative signs in the plane equation...... just use the one they give you in your textbook