let `veca=hati_+hatj+sqrt2hatk` `vecb=b_1hati+b_2hatJ+sqrt2hatk` `vecc=5hati+hatj+sqrt2hatk` & `(veca+vecb)` is perpendicular to `vecc` and projection vector of `vecb` on `veca` is `veca` then find `|vecb|` (a) `6` (b) `sqrt(22)` (c) `sqrt(32)` (d) `11`

Options

A. 4
B. `sqrt(22)`
C. `sqrt(32)`
D. 6

Answer

D

Explanation

`(vecb.hata)a=veca" and "(veca+vecb).vecc=0`
`rArr" "||vecb.hata||=||veca||" "vecb.vecc=-veca.vecc`
`rArr" "(|v_(1)+b_(2)+2|)/(2)=2" "5b_(1)+b_(2)+2=-5-1-2`
`rArr" "|b_(1)+b_(2)+2|=4" "5b_(1)+b_(2)=-10`
`" "b_(1)+b_(2)=-6 or 2`
`b_(1)=-3 and b_(2)=5`
`||vecb||=5`