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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.

yathish
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    $(5,0,5)$ is not in the sphere. Maybe one of the fives is $2$? – ajotatxe Oct 06 '17 at 15:20
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    Also, does the problem actually say "along a great circle of the sphere"? If that is the case, then the line through the two points must be a diameter of the sphere. If (4, 3, 2) is to be rotated into (2, 0, 5) (or (5, 0, 2)) then find the equation of the plane containing those two points and (0, 0, 0), the center of the sphere. Take the diameter to be perpendicular to that plane. – user247327 Oct 06 '17 at 15:27
  • @user247327 It need not be a diameter. You're assuming the point is rotated through a full 180 degrees in which case the two points will become the diameter. – yathish Oct 06 '17 at 15:30
  • @ajotatxe Pretty sure there's no typo. The answer (but not solution) and the question can be found on 91st page here: https://archive.org/details/analyticalsolidg032493mbp – yathish Oct 06 '17 at 15:31
  • I am sure that want @ajotatxe is saying is that there is a typo in the book. Rotating a sphere around a diameter will not change the radius of the sphere. But just to check, use formula at https://en.wikipedia.org/wiki/Rotation_matrix#Rotation_matrix_from_axis_and_angle to calculate the rotation matrix, and rotate the (4,-3,2) point – Andrei Oct 06 '17 at 16:10
  • @Andrei yup! It's a typo. Sorry. Now I know how to find the answer. Updated the question here. – yathish Oct 06 '17 at 16:18
  • @user247327 You should post this as an answer. – amd Oct 06 '17 at 20:47

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

yathish
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