Question

`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

Options

• – 2

• – 6

• – 7

• – 14

Answer 1

– 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