Let →a=2^i+^j+^k,→b=^i+2^j−^k and a unit vector →c be coplanar. If →c is perpendicular to →a, then →c=
A
1√2(−^j+^k)
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B
1√3(−^i−^j−^k)
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C
1√5(^i−2^j)
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D
1√5(−^i−^j−^k)
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Solution
The correct option is A1√2(−^j+^k) Since ¯¯c is coplanar with ¯¯¯a and ¯¯b, it can be written as a linear combination of ¯¯¯a and ¯¯b. Thus let ¯¯c=α¯¯¯a+β¯¯b. Given that ¯¯c.¯¯¯a=0, so 6α+3β=0. Thus, β=−2α. ∴¯¯c=α¯¯¯a−2α¯¯b=α(−3¯j+3¯¯¯k) ∴ the unit vector required =1√2(−¯j+¯¯¯k)