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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+z2)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
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