# 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).

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sciencesolve | Certified Educator

You need to evaluate the directional vector of the line that links the center of sphere `(1,0,2)` and the tangency point `P (5,-3,2)` , such that:

`bar v = <5-1,-3-0,2-2> => bar v = <4,-3,0>`

You need to evaluate the magnitude of the directional vector bar v, such that:

`|bar v| = sqrt(4^2 + (-3)^2 + 0^2)`

`|bar v| = sqrt(16 + 9) => |bar v| = sqrt 25 => |bar v| = 5`

The directional vector bar v is normal to the tangent plane to the sphere, at the point `(5,-3,2)` , hence, the equation of the tangent plane is the following, such that:

`4*(x - 5) - 3(y + 3) + 0*(z - 2) = 0`

`4x - 20 - 3y - 9 = 0 => 4x - 3y - 29 = 0`

**Hence, evaluating the equation of the tangent plane to the given sphere, under the given conditions, yields **`4x - 3y - 29 = 0.`