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Motion In A Plane

Question
CBSEENPH11018237

Show that there is one and only one set of resolved components in a particular direction.

Solution

Let a given vector straight R with rightwards arrow on top be resolved in the direction of two non-parallel vectors straight A with rightwards harpoon with barb upwards on top space a n d space B with rightwards harpoon with barb upwards on top
Let there be two sets of resolved components of resultant straight R with rightwards harpoon with barb upwards on top in the direction of straight A with rightwards harpoon with barb upwards on top space a n d space B with rightwards harpoon with barb upwards on top
That is, left parenthesis straight lambda subscript 1 straight A with rightwards harpoon with barb upwards on top comma space straight beta subscript 1 straight B with rightwards harpoon with barb upwards on top space right parenthesis space and space left parenthesis straight lambda subscript 2 straight A with rightwards harpoon with barb upwards on top comma space straight beta subscript 2 straight B with rightwards harpoon with barb upwards on top space right parenthesis 
Since, straight lambda subscript 1 straight A with rightwards harpoon with barb upwards on top space and space straight beta subscript 1 straight B with rightwards harpoon with barb upwards on top are components of straight R with rightwards harpoon with barb upwards on top
Therefore,
straight R with rightwards harpoon with barb upwards on top space equals space lambda subscript 1 A with rightwards harpoon with barb upwards on top space plus space beta subscript 1 B with rightwards harpoon with barb upwards on top space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis 
And also,
straight R with rightwards harpoon with barb upwards on top space equals space lambda subscript 2 A with rightwards harpoon with barb upwards on top space plus space beta subscript 2 stack B space with rightwards harpoon with barb upwards on top space space space space space space space space space space... left parenthesis 2 right parenthesis
So, from equations (1) and (2), we have, 
space space space space space space space straight lambda subscript 1 straight A with rightwards harpoon with barb upwards on top space plus space straight beta subscript 1 straight B with rightwards harpoon with barb upwards on top space equals space straight lambda subscript 2 straight A with rightwards harpoon with barb upwards on top space plus space straight beta subscript 2 straight B with rightwards harpoon with barb upwards on top
rightwards double arrow space space space left parenthesis space straight lambda subscript 1 minus space straight lambda subscript 2 right parenthesis straight A with rightwards harpoon with barb upwards space on top space equals space left parenthesis space straight beta subscript 2 space minus space straight beta subscript 1 right parenthesis straight B with rightwards harpoon with barb upwards on top space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis
Since, vector A and B are different non-parallel vectors, so equation (3) is possible only and only if, 
space space space space space space space space space space space straight lambda subscript 1 space minus space straight lambda subscript 2 space equals space straight beta subscript 2 space minus space straight beta subscript 1 space equals space 0
rightwards double arrow space space space space space space space space straight lambda subscript 1 space equals space space straight lambda subscript 2 space and space space straight beta subscript 2 space equals space straight beta subscript 1
Hence, both the sets of components are identical. There is one and only one set of resolved components in a particular direction.