From the following system of unknown vectors (→a and →b are given vectors)→x+→y=→a,→x×→y=→b,→x.→a=1 and →x=→a+λ(→a×→b)|→a|2. Then λ is
A
1
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B
−1
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C
2
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D
−2
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Solution
The correct option is A1 →x+→y=→a⋯(i)→x×→y=→b⋯(ii)→x.→a=1⋯(iii) |→a|2→x=→a+λ(→a×→b)⋯(iv)
Multiplying equation (i) by dot product of →a →x.→a+→y.→a=→a.→a 1+→y.→a=|→a|2⋯(v)
Now, taking cross product of equation (ii) with →a →a×(→x×→y)=→a×→b (→a.→y)→x−(→a.→x)→y=→a×→b (→a.→y)→x−(1)→y=→a×→bFrom(iii) (→a.→y)→x−(→a−→x)=→a×→bFrom(i) (→a.→y)→x+→x=→a+→a×→b ((→a.→y)+1)→x=→a+→a×→b
putting from equation(v) |→a|2→x=→a+→a×→b⋯(vi)
Comparing equation (iv),(vi) λ=1