Rotation of sphere about one of its diameters.

Question

Find the equation of the diameter of the sphere $x^2+y^2+z^2=29$ such that a rotation about it will transfer the point $(4, -3, 2)$ to the point $(5, 0, -2)$ along a great circle of the sphere. Find also the angle through which the sphere must be so rotated.

Answer 1

Let the equation of diameter be $\frac{x}{l}=\frac{y}{m}=\frac{z}{n}$ (Note that diameter passes through origin)

Let the given points be $A (4,-3,2)$ and $B (5,0,-2)$. Now the direction cosines of this diameter will be perpendicular to the plane $OAB$ where $O$ is the origin. Because its given that the arc $AB$ forms a part of great circle of the sphere.

Thus using $l_1l_2+m_1m_2+n_1n_2=0$ perpendicularity condition we have,

$4l-3m+2n=0$ and

$5l +0m-2n=0$

Solving these two equations we get $\frac{l}{2}=\frac{m}{6}=\frac{n}{5}$

Hence required equation of diameter is $\frac{x}{2}=\frac{y}{6}=\frac{z}{5}$

The required angle of rotation is the angle between $OA$ and $OB$ i.e.,

$cos\theta=l_1l_2+m_1m_2+n_1n_2 = \frac{(4\times5)+(-3\times0)+(2\times-2)}{\sqrt(4^2+(-3)^2+2^2)\times\sqrt(5^2+0^2+(-5)^2)} = \frac{16}{29}$

Hence, $\theta=\arccos(\frac{16}{29})$