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Given that ` vec a , vec b , vec p , vec q` are four vectors such that ` vec a+ vec b=mu vec p , vec bdot vec q=0a n d( vec b)^2=1,w h e r emu` is a scalar. Then `|( vec adot vec q) vec p-( vec pdot vec q) vec a|` is equal to `2| vec pdot vec q|` b. `(1//2)| vec pdot vec q|` c. `| vec pxx vec q|` d. `| vec pdot vec q|`

A

`2|vecpvecq|`

B

`(1//2)|vecp.vecq|`

C

`|vecpxxvecq|`

D

`|vecp.vecq|`

Text Solution

Verified by Experts

The correct Answer is:
d
`veca + vecb = mu vecp " " vecb.vecq =0, |vecb|^92) =1`
`veca + vecb = muvecp`
` Rightarrow (veca + vecb) xx veca = mu vecp xx veca, vecb xx veca= muvecp xx veca`
` Rightarrow vecq xx ( vecb xx veca) = muvecq xx ( vecp xx veca) `
` Rightarrow (vecq .veca) vecb - ( vecq .vecb) veca = muvecq xx (vecp xx veca)`
`Rightarrow (vecq. veca) vecb = muvecq xx (vecp xx veca)`
`veca + vecb = mu vecp`
` Rightarrow vecq. (veca + vecb) = mu vecq. vecp`
` Rightarrow vecq. veca + vecq . vecb = muvecp .vecq`
` Rightarrow mu = (vecq.veca)/(vecp.vecq) `
`Rightarrow (vecq. veca)vecb = (vecq.veca)/(vecp.vecq) [(vecq.veca) .vecp - (vecq.vecp) veca]`
` Rightarrow | (vecq.veca) vecp -(vecq.vecp) veca| = |vecp.vecq|vecb| = |(vecp.vecq)| . |vecb|`
` Rightarrow |(vecq.veca) vecp.- (vecq.vecp) ]veca| = |vecp.vecq|`
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