Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 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) $$d(Q,\Pi_P)=\frac{|2x_Q+4y_Q+6z_Q-28|}{\sqrt{2^2+4^2+6^2}}= \frac{2\sqrt{14}}{7}$$

share|cite|improve this answer

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.

share|cite|improve this answer

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!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.