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pallavi pradeep bhardwaj Grade: 12
        

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

7 years ago

Answers : (1)

Tapasranjan Das
19 Points
										

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


 

7 years ago
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