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If `veca and vecb` are two non collinear vectors `and `vecuveca0(veca.vecb)vecb and vecv=vecaxxvecb` then `|vecv|` is (A) `|vecu|` (B) `|vecu|+|vecu.vecb|` (C) `|vecu|+|vecu.veca|` (D) none of theseA. `|vecu|`B. `|vecu|+ |vecu. Veca|`C. `|vecu| + |vecu.vecb|`D. `|vecu|+ vecu. (veca + vecb)` |
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Answer» Correct Answer - a,c we have `vecv= vecaxxvecb= |veca||vecb| sin theta hatn = sin theta hatn` where `veca and vecb` are unit vectors. Therefore, `|vecv|= sin theta` Now, `vecu = veca - (veca.vecb)vecb` `= veca -vecb cos theta ( " where " veca. Vecb = cos theta)` `|vecu|^(2) = | veca-vecb cos theta|^(2)` ` 1 + cos^(2) theta -2 cos theta . cos theta` ` =1 - cos^(2) theta = sin^(2) theta = |v|^(2)` ` Rightarrow |vecu|= |vecv|` Also , `vecu . vecb = veca. vecb - (veca.vecb) (vecb.vecb)` ` = veca.vecb-veca.vecb=0` `|vecu.vecb|=0` `|vecv|=|vecu|+ |vecu.vecb|` is also correct. |
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