Find the equation of the plane tangent to the sphere (x - 1)^2 + (y + 2)^2 + (z - 1)^2 = 6 at th

Name: Find the equation of the plane tangent to the sphere (x - 1)^2 + (y + 2)^2 + (z - 1)^2 = 6 at the point (2, -1, 0). [Hint: A normal to this plane is the radius vector of the sphere.]
Uploaded: 2023-10-27T11:15:29-08:00
Duration: 3 min 37 s
Channel: Adi S
Description: Find the equation of the plane tangent to the sphere (x - 1)^2 + (y + 2)^2 + (z - 1)^2 = 6 at the point (2, -1, 0). [Hint: A normal to this plane is the radius vector of the sphere.]

Question

Kathleen Carty · Accepted Answer

1. We know that the equation of the sphere is $(x-1)^2 + (y+2)^2 + (z-1)^2 = 6$.

Question

Find the equation of the plane tangent to the sphere (x - 1)^2 + (y + 2)^2 + (z - 1)^2 = 6 at the point (2, -1, 0). [Hint: A normal to this plane is the radius vector of the sphere.]

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