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Let $S$ be the sphere $$ x^2 + y^2 + z^2 = 14$$

I need help finding:

A. Tangent plane to $S$ at the point $P(1, 2, 3)$.
B. Distance from $Q(3, 2, 1)$ to the above tangent plane.

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3 Answers 3

up vote 2 down vote accepted

1) $f(x,y,z)=x^2+y^2+z^2-14$ $$\frac{\partial f}{\partial x} (x,y,z)=2x,\frac{\partial f}{\partial y} (x,y,z)=2y,\frac{\partial f}{\partial z} (x,y,z)=2z.$$ The vector $$\overrightarrow{n}=\overrightarrow{grad} f (P)= (2,4,6)$$ is a normal vector for the plane. A point $M(x,y,z)$ is in the plane if and only if : $$\overrightarrow{PM}. \overrightarrow{n} = 0$$ thus $$2(x-1)+4(y-2)+6(z-3)=0$$ thus: $$\Pi_P: \quad \quad 2x+4y+6z-28=0$$

2) $$d(Q,\Pi_P)=\frac{|2x_Q+4y_Q+6z_Q-28|}{\sqrt{2^2+4^2+6^2}}= \frac{2\sqrt{14}}{7}$$

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First, find the gradient $\nabla f$. Then the equation of the tangent plane to the surface is $${f_x}(1,2,3) \cdot (x - 1) + {f_y}(1,2,3) \cdot (y - 2) + {f_z}(1,2,3) \cdot (z - 3) = 0$$ where $ \cdot $ denotes the dot product. You can then find the distance (using the Euclidean norm) from $Q$ to any point on the plane.

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Or, we can use a bit of geometry.

Using the fact that the normal of the tangent plane to the given sphere will pass through it's center, (0,0,0).

We get the normal vector of the plane as

i+2j+3k. (Vector Joining point of tangency to center of sphere).

Then equation of plane can be written as:

x(1)+y(2)+z(3)=p. Where 'p' is some scalar.

As the plane passes through (1,2,3); We get p=14, Hence the tangent plane:- x+2y+3z-14=0. For finding the distance, simply use the distance formula!

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