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Four vectors a→,b→,c→ and x→ satisfy the relation (a→.x→)b→=c→+x→ where b→⋅a→ ≠ 1. The value of x→ in terms of a→,b→ and c→ is equal to -

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Question

Four vectors `veca, vecb, vecc` and `vecx` satisfy the relation `(veca.vecx)vecb = vecc + vecx` where `vecb * veca` ≠ 1. The value of `vecx` in terms of `veca, vecb` and `vecc` is equal to

Options

  • `((veca.vecc)vecb - vecc(veca*vecb - 1))/((veca*vecb - 1))`

  • `vecc/(veca*vecb - 1)`

  • `(2(veca*vecc)vecb + vecc)/(veca*vecb - 1)`

  • `(2(veca*vecc)vecc + vecc)/(veca*vecb - 1)`

MCQ

Solution

`((veca.vecc)vecb - vecc(veca*vecb - 1))/((veca*vecb - 1))`

Explanation:

`(veca. vecx)vecb = vecc + vecx` ......(i)

Taking dot product with `veca`, we get

`(veca. vecx)(vecb. veca) = (vecc . veca) + (veca. vecx)`

∴ `(veca. vecx)(vecb. veca - 1) = (vecc. veca)`

∴ `(veca. vecx) = (vecc. veca)/(vecb.veca - 1)`  ......(ii)

⇒ `vecx = (vecc. veca)/(vecb . veca - 1) vecb - vecc`

= `((veca. vecc)vecb - vecc(veca. vecb - 1))/(veca. vecb - 1)` 

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