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.


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


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 centre, $(0,0,0).$

We get the normal vector of the plane as:

$\hat i+2\hat j+3\hat k$. (Vector joining point of tangency to centre 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!

  • $\begingroup$ I am wondering if a linear algebra explanation might be insightful here $\endgroup$
    – Maelstorm
    Mar 26 '18 at 20:32

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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