Let →a=^i+^j+^k, →b=^i−^j+^k and →c=^i−^j−^k be three vectors. A vector →v in the plane of →a and →b, whose projection on →c is 1√3, is given by
A
^i−3^j+3^k
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B
−3^i−3^j−^k
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C
3^i−^j+3^k
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D
^i+3^j−3^k
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Solution
The correct option is C3^i−^j+3^k A vector →v in the plane of →a and →b is →v=→a+λ→b ⟹→v=(^i+^j+^k)+λ(^i−^j+^k) Projection of →v on →c =1√3 ⟹→v.→c∣∣→c∣∣=1√3 ⟹(1+λ)−(1−λ)−(1+λ)√3=1√3 ⟹(1+λ)−(1−λ)−(1+λ)=1 λ=2 So, →v=3^i−^j+3^k