Consider the sphere S of radius 5 centred at (1, 0, 2). Find the equation of the plane tangent to S at the point (5,-3, 2).Hint : To get started, draw this situation in 2D with a circle.
The radius vector `vec r` from the center of the sphere to the point of tangency will be normal (perpendicular) to the tangent plane. This vector is
The equation of the tangent plane with normal vector `vec r` that passes through the point `(a,b,c)` is `vec r * (x-a,y-b,z-c)=0.` In this case, this equation becomes
`(4,-3,0)*(x-5,y+3,z-2)=4(x-5)-3(y+3)=0.` Simplifying gives
`4x-3y=29` as the equation of the plane.
Below is an analogous case in two dimensions. Note that the red tangent line, in this case, is vertical because the radius vector has zero y-coordinate. Similarly in three dimensions, the tangent plane is vertical because the radius vector has zero z-coordinate. Also note that the radius vector is normal to the tangent line.