If ∣∣→a∣∣=1, the projection of →r along →a is 2 and →a×→r+→b=→r, then →r=
A
12[→a−→b+→a×→b]
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B
12[2→a+→b+→a×→b]
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C
→a+→a×→b
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D
→a−→a×→b
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Solution
The correct option is C12[2→a+→b+→a×→b] Projection of →r on →a is 2 ⇒→r.→a=2 -----(1) →a×→r+→b=→r ------(2) Dot (1) with →a ⇒→a.→b=→r.→a=2 by (1) cross (1) with →a ⇒→a×(→a×→r)+→a×→b=→a×→r ⇒0+→a×→b=→a×→r ∴2→r−→b+→a×→b=2→r ⇒→r=12[2→a+→b+→a×→b]