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Three vectors a→,b→ and c→ satisfy the condition a→+b→+c→=0→. Evaluate the quantity μ = a→.b→+b→.c→+c→.a→, if |a→| = 3, |b→| = 4 and |c→| = 2. - Mathematics

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प्रश्न

Three vectors `veca, vecb` and `vecc` satisfy the condition `veca + vecb + vecc = vec0`. Evaluate the quantity μ = `veca.vecb + vecb.vecc + vecc.veca`, if `|veca|` = 3, `|vecb|` = 4 and `|vecc|` = 2.

योग
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उत्तर

Since `veca + vecb + vecc = vec0`, we have

`veca + vecb + vecc = vec0` = 0

or `veca.veca + veca.vecb + veca.vecc` = 0

Therefore `veca. vecb + veca.vecc = -|veca|^2` = – 9  ...(1)

Again, `vecb.(veca + vecb + vecc)` = 0

or `veca.vecb + vecb.vecc = -|vecb|^2` = – 16  ...(2)

Similarly `veca.vecc + vecb.vecc` = – 4.  ...(3)

 Adding (1), (2) and (3), we have

`2(veca.vecb + vecb.vecc + veca.vecc)` = – 29

or 2μ = – 29, i.e., μ = `(-29)/2`

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