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What is an isometry $\mathbb R^3\to \mathbb R^3 $ which maps $(3,1,2) \mapsto (2,2,2)$ and $(2,2,2) \mapsto (1,1,2)$?

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Hint: the first map can be handled by a simple translation. Once you've factored out that translation, consider the 'relative positions' of the points in the second map (i.e., relative to their corresponding points in the first map) - these would be $(2,2,2)-(3,1,2) = (-1, 1, 0)$ and $(1,1,2) - (2,2,2) = (-1, -1, 0)$; can you find a rotation that maps one to the other, or prove that there isn't one? And once you've got that, can you see how all the transformations can be composed to find a final result?

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You are given four points $a$, $b$, $a'$, $b'\in{\mathbb R}^3$ and want an isometry $T:\ {\mathbb R}^3\to{\mathbb R}^3$ such that $Ta=a'$, $Tb=b'$. As $|b'-a'|=|b-a|$ there is such an isometry, and as $b'-a'\ne b-a$ it cannot be a translation, so we shall look for a suitable rotation.

This rotation will have an axis $g$, and both $a$ and $b$ will move on circles around $g$. It follows that $a'-a=(-1,1,0)$ and $b'-b=(-1,-1,0)$ are both chords of such circles and thus are normal to $g$. Looking at the data we see that a parallel to the $x_3$-axis fulfills this requirement. (In general one would calculate $(a'-a)\times(b'-b)$.)

Therefore we are down to a two-dimensional problem: We look for a rotation $T'$ of the $(x_1,x_2)$-plane such that $T'(3,1)=(2,2)$ and $T'(2,2)=(1,1)$. Drawing a little figure one immediately sees that a counterclockwise rotation by $90^\circ$ around the point $(2,1)$ does the trick. (In general one would intersect the medians of $(3,1)$, $T'(3,1)$ and of $(2,2)$, $T'(2,2)$.)

Putting it all together we can say the following: Rotating space around the axis $$g:\quad t\mapsto(2,1,t)\qquad(-\infty<t<\infty)$$ by $90^\circ$ counterclockwise (as seen from $x_3=\infty$) will move the two given points as required.

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Thank you Professor Christian Blatter . Your answer is comprehensive and useful.- Madhav Bapat – Madhav Bapat Feb 9 '12 at 1:52
With your suggestion I have obtained the isometry function as $(x,y,z)-->(3-y, x-1, z)$ – Madhav Bapat Feb 10 '12 at 15:57

Isometries should preserve the length, but $(3,1,2)$ and $(2,2,2)$ have different length (as well as $(2,2,2)$ and $(1,1,2)$. Hence, there is no such isometry.

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Isometries only need to preserve lengths between points - translation, for instance, is an isometry. – Steven Stadnicki Feb 8 '12 at 7:32
Oh, for some reason I thought, the OP would look for a linear isometry... – Dirk Feb 8 '12 at 7:39

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