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let veca=hati+hatj+sqrt2hatk vecb=b1hat...

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`

A

4

B

`sqrt(22)`

C

`sqrt(32)`

D

6

Text Solution

Verified by Experts

The correct Answer is:
D
`(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`
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