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The scalar product of the vector a=i+j+k with a unit vector along the sum of vectors b=2i+4j−5k and c=λi+2j+3k is equal to one. Find the value of λ and hence, find the unit vector along b +c - Mathematics

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The scalar product of the vector `veca=hati+hatj+hatk` with a unit vector along the sum of vectors `vecb=2hati+4hatj−5hatk and vecc=λhati+2hatj+3hatk` is equal to one. Find the value of λ and hence, find the unit vector along `vecb +vecc`

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Solution

Given:

`veca=hati+hatj+hatk, vecb=2hati+4hatj−5hatk and vecc=λhati+2hatj+3hatk`

now 

`vecb+vecc=(2+lambda)hati+6hatj-2hatk`

Let `hatd` denote the unit vector along `vecb+vecc` Then,

`hatd=(vecb+vecc)/|vecb+vecc|`

`=>hatd=((2+lambda)hati+6hatj-2hatk)/sqrt((2+lambda)^2+(6)^2+(-2)^2)`

`=>hatd=((2+lambda)hati+6hatj-2hatk)/sqrt((2+lambda)^2+40)`

``

Also `veca.hatd=1`

`=>(hati+hatj+hatk).((2+lambda)hati+6hatj-2hatk)/sqrt((2+lambda)^2+40)=1`

`=>(hati+hatj+hatk)[(2+lambda)hati+6hatj-2hatk]=sqrt((2+lambda)^2+40)`

`=>2+lambda+6-2=sqrt(2+lambda)^2+40)`

`=>(lambda+6)^2=(2+lambda)^2+40`

`=>8lambda=8`

`=>lambda=1`

`therefore hatd=((2+lambda)hati+6hatj-2hatk)/sqrt((2+1)^2+40)=(3hati+6hatj-2hatk)/sqrt(49)`

`"i.e " hatd=1/7(3hati+6hatj-2hatk)`

Concept: Product of Two Vectors - Scalar (Or Dot) Product of Two Vectors
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