A unit vector coplanar with →i+→j+2→k and →i+2→j+→k, and perpendicular to →i+→j+→k, is
A
1√2(−→j+→k)
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B
1√2(→k−→i)
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C
1√2(→i−→k)
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D
1√2(→j−→k)
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Solution
The correct options are A1√2(−→j+→k) D1√2(→j−→k) Let the required vector be p→i+q→j+r→k According to the condition of coplanarity, we have ∣∣
∣∣112121pqr∣∣
∣∣=0 ⇒2r−q−1(r−p)+2(q−2p)=0 ⇒2r−q−r+p+2q−4p=0 or −3p+q+r=0
The vector is also perpendicular to →i+→j+→k and so, p+q+r=0