Let →a=^i+2^j+^k,→b=^i−^j+^k and →c=^i+^j−^k . A vector in the plane of →a and →b, where projection on →c is 1√3, is
A
4^i−^j+4^k
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B
3^i+^j−3^k
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C
2^i+^j
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D
4^i+^j−4^k
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Solution
The correct option is A4^i−^j+4^k A vector in the plane of →a and →b is →a+λ→b=(^i+2^j+^k)+λ(^i−^j+^k) =(1+λ)^i+(2−λ)^j+(1+λ)^k If projection on →c is 1√3 ⇒(→a+λ→b).→c∣∣→c∣∣=±1√3 ⇒((1+λ)^i+(2−λ)^j+(1+λ)^k)(^i+^j−^k)√1+1+1=±1√3 ⇒1+λ+2−λ−1−λ√3=±1√3 ⇒2−λ=±1 ⇒λ=2±1∴λ=1 or 3 If λ=1,→a+λ→b=(1+1)^i+(2−1)^j+(1+1)^k =2^i+^j+2^k If λ=3,→a+λ→b=(1+3)^i+(2−3)^j+(1+3)^k =4^i−^j+4^k