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प्रश्न
`bara, barb` and `barc` are three vectors such that `veca + vecb + vecc` 20, `|bara| = 1, |barb| = 2` and `|barc| = 3`. Then `bara. barb + barb.barc + bar(c.a)` is equal to
पर्याय
– 2
– 6
– 7
– 14
MCQ
उत्तर
– 7
Explanation:
`|veca + vecb + vecc|^2 = (veca + vecb + vecc).(veca + vecb + vecc)`
⇒ 0 = `|veca|^2 + |vecb|^2 + |vecc|^2 + 2(veca.vecb + vecb.vecc + vecc.veca)`
⇒ 0 = 1 + 4 + 9 + 2`(veca.vecb + vecb.vecc + vecc.veca)`
⇒ `veca.vecb + vecb.vecc + veca.vecc` = – 7
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