Let →a=^i+^j+^k,→b=^i−^j+2^k and →c=x^i+(x−2)^j−^k If the vector →c lies in the plane of →a and →b,then x=
A
1
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B
−4
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C
−2
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D
0
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Solution
The correct option is C−2 →a,→b,→c are coplanar ∴[→a→b→c]=(→a×→b).→c=0 ⇒∣∣
∣∣1111−12x(x−2)−1∣∣
∣∣ ⇒1−2(x−2)−(−1−2x)+x−2+x=0 ⇒2x+4=0 on simplification ∴x=−2