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Calculus IIIChapter 1: 3-Dimensional SpaceSection 1-1

The 3-D Coordinate System — Question 2

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Question 2

A sphere passes through four specified points in three-dimensional space.

Given A=(1,0,2),B=(5,0,2),C=(1,6,2),D=(1,0,8).\begin{aligned} A&=(1,0,2), & B&=(5,0,2),\\ C&=(1,6,2), & D&=(1,0,8). \end{aligned}

Tasks

  1. Find the center and radius of the sphere.

  2. Write its Cartesian equation.

  3. Explain why the four given points determine a unique sphere.

  4. Verify that all four points satisfy your equation.

Original worksheet page 1: question and worked solution for 1-1-002
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Question 2 – Solution

Strategy A sphere’s center is equidistant from all four points. Perpendicular-bisector conditions can be obtained by equating squared distances.

See the diagram in the original worksheet below.

Center Let the center be K=(h,k,ℓ)K=(h,k,\ell). From d(A,K)2=d(B,K)2d(A,K)^2=d(B,K)^2, (h−1)2+k2+(ℓ−2)2=(h−5)2+k2+(ℓ−2)2,(h-1)^2+k^2+(\ell-2)^2=(h-5)^2+k^2+(\ell-2)^2, so h=3h=3. Comparing AA with CC gives k=3k=3, and comparing AA with DD gives ℓ=5\ell=5. Hence the center is (3,3,5)(3,3,5) and r2=(3−1)2+(3−0)2+(5−2)2=22.r^2=(3-1)^2+(3-0)^2+(5-2)^2=22. Therefore (x−3)2+(y−3)2+(z−5)2=22.\boxed{(x-3)^2+(y-3)^2+(z-5)^2=22}.

Uniqueness and verification The three difference vectors from AA point independently in the xx-, yy-, and zz-directions, so their perpendicular-bisector planes meet in exactly one point. Substitution shows that each given point produces squared distance 2222.

Original worksheet page 2: question and worked solution for 1-1-002

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