STEM-Master
Calculus IIChapter 6: 3-Dimensional SpaceSection 6-1

The 3-D Coordinate System — Question 6

PDF ↗

Question 6

Describe geometrically the set of all points P=(x,y,z)P=(x,y,z) equidistant from A=(2,−1,3)A=(2,-1,3) and B=(−4,5,1)B=(-4,5,1). Derive a Cartesian equation for the set and show that the midpoint of ABAB belongs to it.

Original worksheet page 1: question and worked solution for 6-1-006
Show solutionHide solution

Question 6 – Solution

Strategy Equate squared distances. Quadratic terms cancel, revealing the perpendicular-bisector surface.

See the diagram in the original worksheet below.

Derivation Set (x−2)2+(y+1)2+(z−3)2=(x+4)2+(y−5)2+(z−1)2.(x-2)^2+(y+1)^2+(z-3)^2=(x+4)^2+(y-5)^2+(z-1)^2. After expanding and simplifying, 3x−3y+z+7=0\boxed{3x-3y+z+7=0}.

Interpretation This is the plane perpendicular to AB→=⟨−6,6,−2⟩\overrightarrow{AB}=\langle-6,6,-2\rangle through the midpoint M=(−1,2,2)M=(-1,2,2). Indeed, 3(−1)−3(2)+2+7=03(-1)-3(2)+2+7=0.

Verification Interchanging AA and BB does not change the locus, and every point in the plane has equal squared distance to the endpoints.

Original worksheet page 2: question and worked solution for 6-1-006

Original worksheet layout. Use Enlarge or open the PDF for a closer view.

← Section questionsChapter contents