The value of such that scalar product of the vectors (i+j+k) with the unit vectors parallel to the sum of the vectors (2i+4j-5k) and bi+2j+3k) is 1 is

hi pallavi,

(2i+4j-5k)+(bi+2j+3k)=(2+b)i+6j-2k=A(say)

now let xi+yj+zk be a unit vector parallel to A

therefore (x²+y²+z^2)1/2=1   =>x²+y²+z²=1 (i) and

xi+yj+zk=λ((2+b)i+6j-2k),λ be any constant

now x=λ(2+b) ,y=6λ ,z=-2λ

again,(xi+yj+zk).(i+j+k)=1

=>x+y+z=1

=>λ(2+b+6-2)=1

=>b+6=1/λ

again x²⁺y²⁺z²=1 [from (i)]

=>λ²((2+b)²+36+4)=1

=>(2+b)²+40=1/λ²

                         =(b+6)²

=>40=2b+2

=>b=19