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प्रश्न
ABCDEF is a regular hexagon. Show that `bar"AB" + bar"AC" + bar"AD" + bar"AE" + bar"AF" = 6bar"AO"`, where O is the centre of the hexagon.
बेरीज
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उत्तर
ABCDEF is a regular hexagon.
∴ `bar"AB" = bar"ED" "and" bar"AF" = bar"CD"`
∴ by the triangle law of addition of vectors,
`bar"AC" + bar"AF" = bar"AC" + bar"CD" = bar"AD"`
`bar"AE" + bar"AB" = bar"AE" + bar"ED" = bar"AD"`
∴ LHS = `bar"AB" +bar"AC" + bar"AD" + bar"AE" + bar"AF"`
`= bar"AD" + (bar"AC" + bar"AF") + (bar"AE" + bar"AB")`
`= bar"AD" + bar"AD" + bar"AD"`
`= 3bar"AD" = 3(2bar"AO")` ....[O is midpoint of AD]
`= 6 bar"AO"`.
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