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दिखाएँ कि बिंदुएँ A,B तथा C जिनके स्थिति सदिश क्रमश: `-2hati + 3hatj + 5hatk, hati + 2hatj + 3hatk` तथा `7hati - hatk` हैं, सरीख हैं|

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`vec(AB) =vec(OB)-vec(OA)`, जहाँ O मूल बिंदु हैं|
`=(hati + 2hatj + 3hatk) - (-2hati + 3hatj + 5hatk)`
`=3hati -hatj - 2hatk`
साथ ही `vec(AC) = (7hati -hatk)-(-2hati + 3hatj + 5hatk) = 9hati - 3hatj -6 hatk = 3(3hati - hatj -2hatk)= 3vec(AB)`
इस प्रकार, `vec(AC) = 3vec(AB)`
`rArr` सदिश `vec(AB)` तथा `vec(AC)` समान्तर हैं|
लेकिन AB और AC में एक उभयनिष्ट बिंदु A हैं,
इसलिए बिंदुएँ A,B तथा C सरीख हैं|

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