Let `a=hat(i)+hat(j)+sqrt(2)hat(k),b=b_(1)hat(i)+b_(2)hat(j)+sqrt(2)hat(k)` and `c=5hat(i)+hat(j)+sqrt(2)hat(k)` be three vectors such that the projection vector of b on a is a. If `a+b` is perpendicular to c, then `|b|` is equal to

Options

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

Answer

B

Explanation

According to given information we have the following figure

Clearly projection of b on a = `(b. a)/(|a|)`
`=((b_(1)hat(i) +b_(2)hat(j) +sqrt(2hat(k)))(hat(i)+hat(j)+ sqrt(2)hat(k)))/(sqrt(1^(2)+1^(2)+(sqrt(2))^(2)))`
`=(b_(1)+b_(2)+2)/(sqrt(4))=(b_(1)+b_(2)+2)/(2)`
but projection of b on a =`|a|`
`:. (b_(1)+b_(2)+2)/(2) = sqrt(1^(2) +1^(2) +(sqrt(2))^(2)`
`rArr (b_(1)+b_(2)+2)/(2) =2 rArr b_(1)+b_(2) =2`
Now `a+b=(hat(i)+hat(j)+sqrt(2)hat(k)) + (b_(1)hat(i)+ b_(2)hat(j) +_ sqrt(2)hat(k))`
`=(b_(1)+1)hat(i)+ (b_(2)+1)hat(j)+ 2sqrt(2)hat(k)`
`:' (a+b) bot c ` therefore (a+b) .c =0
`rArr {(b_(1) +1) hat(i) +(b_(2) +1) hat(j) +2sqrt(2)hat(k)} (5hat(i)+hat(j)+sqrt(2)hat(k)) =0`
`rArr 5(b_(b_(1) +1) (b_(2)+1) +2sqrt(2)) =0`
`rArr " " 5b_(1) + b_(2)=-10`
From Eqs (i) and (ii) `b_(1) =-3` and `b_(2) =5`
`rArr b=- 3hat(i) +5hat(j) +sqrt(2)hat(k)`
`rArr |b| =sqrt((-3)^(2) +(5)^(2)+(sqrt(2))^(2))=sqrt(36) =6`