Equations of Planes — Question 5

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Question 5

The sphere (x−1)2+(y+2)2+(z−3)2=36(x-1)^2+(y+2)^2+(z-3)^2=36 contains the point T=(3,2,7)T=(3,2,7). Find the tangent plane at TT and the line normal to the sphere there.

Original worksheet page 1: question and worked solution for 6-3-005
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Question 5 – Solution

Strategy The radius from the center to TT is normal to the tangent plane.

See the diagram in the original worksheet below.

Normal direction The center is C=(1,−2,3)C=(1,-2,3) and T−C=⟨2,4,4⟩=2⟨1,2,2⟩T-C=\left\langle 2,4,4\right\rangle=2\left\langle 1,2,2\right\rangle.

Tangent plane Using normal ⟨1,2,2⟩\left\langle 1,2,2\right\rangle through TT, (x−3)+2(y−2)+2(z−7)=0,orx+2y+2z=21.\boxed{(x-3)+2(y-2)+2(z-7)=0},\quad\text{or}\quad\boxed{x+2y+2z=21}.

Normal line r→=⟨3,2,7⟩+t⟨1,2,2⟩\boxed{\vec r=\left\langle 3,2,7\right\rangle+t\left\langle 1,2,2\right\rangle}.

Verification |T−C|=6|T-C|=6, so TT is indeed on the sphere.

Original worksheet page 2: question and worked solution for 6-3-005

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