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Question
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.
Solution
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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