The sphere x^2+y^2+z^2=29 rotates through the angle α about…

The sphere $x^2+y^2+z^2=29$ rotates through the angle $\alpha$ about its diameter $MN$. The point $A(4,-3,2)$ moves to $A'(5,0,-2)$ along a great circle. Find: a) the line $MN$; b) the plane of rotation of $A$; c) $\alpha$; d) the angle between that plane and $z=0$; e) the projection of the great circle onto $z=0$.

Explanation of Method

Recovering the axis, plane and angle of a rotation from one point and its image, and projecting the great circle.

Step 1: Checks

$|OA|^2=16+9+4=29$ and $|OA'|^2=25+0+4=29$, so both points are on the sphere. $\checkmark$

Step 2: a) The axis

$A$ moves along a great circle, so its plane of rotation passes through $O$ and contains $A$ and $A'$. The axis is perpendicular to that plane:

\[ \vec u\parallel\overrightarrow{OA}\times\overrightarrow{OA'}=\{4,-3,2\}\times\{5,0,-2\}=\{6,18,15\}\parallel\{2,6,5\}. \]

\[MN:\ \frac x2=\frac y6=\frac z5\]

The endpoints are $M,N=\pm\sqrt{\tfrac{29}{65}}\,(2,6,5)$.

Check: $\vec u\cdot\overrightarrow{AA'}=\{2,6,5\}\cdot\{1,3,-4\}=2+18-20=0$, so $A$ and $A'$ have the same projection on the axis. $\checkmark$

Step 3: b) Plane of rotation

\[2x+6y+5z=0\]

Checks: $A$ gives $8-18+10=0$ and $A'$ gives $10+0-10=0$. $\checkmark$

Step 4: c) Angle of rotation

The circle of rotation is centered at $O$, so

\[ \cos\alpha=\frac{\overrightarrow{OA}\cdot\overrightarrow{OA'}}{29}=\frac{20+0-4}{29}=\frac{16}{29}. \]

\[\alpha=\arccos\tfrac{16}{29}\approx56.51^\circ=56^\circ31'\]

Step 5: d) Angle with $z=0$

The angle between two planes equals the angle between their normals: \[\cos\theta=\frac{|\{2,6,5\}\cdot\{0,0,1\}|}{\sqrt{65}}=\frac5{\sqrt{65}},\qquad \tan\theta=\frac{\sqrt{40}}5 .\]

\[\theta=\arccos\tfrac5{\sqrt{65}}\approx51.67^\circ=51^\circ40'\]

Step 6: e) Projection onto $z=0$

From the plane, $z=-\tfrac{2x+6y}5$. Substituting into the sphere: $25x^2+25y^2+(2x+6y)^2=725$.

\[29x^2+24xy+61y^2=725,\quad z=0\]

Checks: $(4,-3)$ gives $464-288+549=725$, and $(5,0)$ gives $725$. $\checkmark$

This is an ellipse centered at $O$. The matrix $\begin{pmatrix}29&12\\12&61\end{pmatrix}$ has eigenvalues $25$ and $65$, so

\[ \frac{u^2}{29}+\frac{v^2}{725/65}=1 . \]

The major semi-axis is $\sqrt{29}$, along $\{3,-1\}$, the trace $x+3y=0$ of the plane. The minor semi-axis is $\sqrt{29}\cos\theta=\tfrac{5\sqrt{29}}{\sqrt{65}}\approx3.34$. A circle projects to an ellipse whose shortened axis is scaled by the cosine of the tilt.

$O$ $A$ $A'$ $N$ $M$ $\tfrac x2=\tfrac y6=\tfrac z5$

The axis $MN$ (teal) is normal to the great circle through $A$ and $A'$ (blue, in $2x+6y+5z=0$). The rotation carries $A$ along the red arc to $A'$; $\alpha=\angle AOA'=\arccos\tfrac{16}{29}$.

$A_0(4,-3)$ $A'_0(5,0)$ $\sqrt{29}$ $\tfrac{5\sqrt{29}}{\sqrt{65}}$

e) The projection onto $z=0$: the ellipse $29x^2+24xy+61y^2=725$ (blue) inside the circle $x^2+y^2=29$ (dashed). Its major semi-axis $\sqrt{29}$ lies along $x+3y=0$, the trace of the plane on $z=0$; the minor one is shortened by $\cos\theta=\tfrac5{\sqrt{65}}$.

Result

\[\boxed{\,MN:\ \frac x2=\frac y6=\frac z5\,}\]

\[\boxed{\,2x+6y+5z=0\,}\]

\[\boxed{\,\alpha=\arccos\tfrac{16}{29}\approx56.51^\circ=56^\circ31'\,}\]

\[\boxed{\,\theta=\arccos\tfrac5{\sqrt{65}}\approx51.67^\circ=51^\circ40'\,}\]

\[\boxed{\,29x^2+24xy+61y^2=725,\quad z=0\,}\]

Remark

Correction: the source writes $\theta=\arccos\tfrac5{\sqrt{65}}=\arcsin\tfrac5{\sqrt{65}}\approx38.7^\circ$. These two values are not equal. $\arcsin\tfrac5{\sqrt{65}}\approx38.3^\circ$ is the angle between the normal $\vec u$ and the plane $z=0$. The dihedral angle between the two planes is $\arccos\tfrac5{\sqrt{65}}\approx51.7^\circ$. (For $\alpha$, the source's $56.6^\circ$ is a rounding slip for $56.5^\circ$.)