1.

Let vec(a),vec(b), & vec(c) be three vectors such that |vec(b)| = 2|vec(a)| & |vec(c)| = 3|vec(a)|. The Angle between each pair of vectors is 60^(@) such that |vec(a) + 2vec(b) + 3vec(c)| = sqrt(21) then sqrt(7)| vec(c)| is equal to

Answer»


Solution :`|vec(a)| = k`
`|vec(b)| = 2k`
`|vec(C)| = 3k`
`|vec(a) + 2vec(b) + 3vec(c)| = sqrt(21)`
`(|vec(a) + 2vec(b) + 3vec(c)|)^(2) = 21`
`a^(2) + 4b^(2) + 9C^(2) + 2[2vec(a) cdot vec(b) + 6vec(b) cdot vec(c) + 3vec(c) cdot vec(a)] = 21`
`k^(2) + 16k^(2) + 81k^(2) + 2[(4k^2)/2 + (36k^2)/2 + (9k^2)/2] = 21`
`147k^(2)= 21 IMPLIES k = +- 1/(sqrt7)`
`|c| = 3k = 3/(sqrt7)` [magnitude is ALWAYS +ve]`


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