(4) For any transformation F of a plane, a transformation G is called an inverse
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Question
(4) For any transformation F of a plane, a transformation G is called an inverse transformation of F if Prove or disprove: (a) For any non-zero vector AB of a plane, let T be the transla- tion along AB. Then there exists a rigid motion G such that (b) For any non-zero angle ZABC and any point O of a plane, let Ro be the rotation around O with |ZABC|. Then there exists a rigid motion G such that G Ro_1 c) For any line LAB of a plane, let R be the reflection across LAB. Then there exists a rigid motion G such that G R-1 (d) There exists a rigid motion G such that G I-1 where I is the identity map on a planeExplanation / Answer
a) Yes, such a inverse exists. The translation is T(x) = x +AB where AB is the vector AB
As regards vectors, there will always exist an inverse vector CD such that AB+CD=CD+AB=0
Then, the required inverse will be G(x)=x+CD
It is an inverse because G(T(x))=G(x+AB)=CD+(x+AB)=x+(CD+AB)=x+O=x
b) Yes, this is also true. The inverse function G is rotation about O by an angle 360- |/ ABC|
This is because G(T(x)) effectively rotates the plane by |/ ABC| first and then rotates the point by angle 360- |/ ABC|
So the combnined effect is to rotate the plane about O by angle (360- |/ ABC|)+( |/ ABC|) = 360
As rotation by 360 is puts everything back to where it initially was (it is equivalent to no rotation at all) we have G(T(x))=x
c) Yes, again. G=R= the same motion representing reflection across line AB
In general, we can say that reflecting a point about the same line twice will give us the original vector
G(T(x))=R(R(x))=R^2(x)=x
d) I is the identity map. We simply choose G=I so that G(T(x))=I(I(x))=I(x)=x
Thus, I also has an inverse
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