How to show that three mutually perpendicular tangent lines can be drawn to a sphere

Question

How to show that three mutually perpendicular tangent lines can be drawn to a sphere $x^2 + y^2 +z^2 = r^2$ from any point on the sphere $2(x^2 + y^2 +z^2) = 3r^2$

Answer 1

Here I have showed you how to get the answer with full detailed working.

Answer 2

The problem is spherically symmetric, so without loss of generality we can set $r=1$ and consider only the point $P=(0,0,\sqrt\frac32)$

Simple trig gives you that the angle between anu line passing through $P$ that is tangent to the unit sphere must make an angle $\theta$ with the $z$ axis satisfying ...

$$ \sin\theta = \sqrt\frac23 \text{ and }\cos\theta = \sqrt\frac13 $$

we can describe our three tangent lines in the form ... $$\begin{eqnarray*} \vec l_1&= (0,0,\sqrt\frac32) + t \hat n_1 \\ \vec l_2&= (0,0,\sqrt\frac32) + u\hat n_2 \\ \vec l_3&= (0,0,\sqrt\frac32) + v\hat n_3 \\ \end{eqnarray*} $$ Where the $\hat n_k$ are mutually perpendicular unit vectors satisfying $\hat n_k \cdot \hat z=\sqrt\frac13$

Choose $\hat n_1$ to be in the y-z plane, it can be ...

$$\hat n_1 =(0, \sqrt\frac23 ,\sqrt\frac13 ) $$

$\hat n_{2,3} . \hat n_{1}=0$ means that they must be of the form ..

$$\hat n_{2,3} =(a, b ,-b\sqrt 2 ) $$

$\hat n_{2,3} \cdot \hat z=\sqrt\frac13$ gives you $b=-\frac13$

$|\hat n_{2,3}|=1$ gives $a=\pm \sqrt\frac 23$

$( \sqrt\frac 23, -\frac13, \frac {\sqrt2} 3) \cdot ( -\sqrt\frac 23, -\frac13, \frac {\sqrt2} 3)=0$ So all conditions are satisfied.

Answer 3

It is clear that the problem remains essentially unchanged under translations and rotations.

Take the usual $xyz$ axes in $\mathbb{R}^3$ and move a ball of radius $r$ until it touches each axis at one place.

By symmetry, the ball touches each axis at a distance $d$ from the origin, hence the centre of the ball is at $(d,d,d)$ and so the distance from the origin to the centre of the ball is $\sqrt{3} d$.

Since each axis is tangential to the ball, the radius $r$ satisfies $r^2 = d^2 + d^2$, or $d = {1 \over \sqrt{2}} r$.

Hence the distance from the origin to the centre of the ball is ${\sqrt{3 \over 2}} r$. In particular, a ball of radius ${\sqrt{3 \over 2}} r$ centred at $(d,d,d)$ has three mutually perpendicular tangent lines (the axes) that touch the $r$ sphere tangentially.