Equations of Planes — Question 5

Question 5

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

Question 5 – Solution

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

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

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

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

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