Let →α=^i+^j+^k,→β=^i−^j−^k;→γ=−^i+^j−^k be three vectors. A vector →δ, in the plane of →α and →β, whose projection on →γ is 1√3, is given by
A
−^i−3^j−3^k
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B
^i−3−3^k
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C
−^i+3^j+3^k
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D
^i+3^j−3^k
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Solution
The correct option is C−^i+3^j+3^k →δ=→α+n→β =(^i+^j+^k)(^i−^j−^k) →δ=(1+n)^i+(1−n)^j+(1−n)^k →δ.→γ|→γ|=1√3⇒|−1−n+1−n−1+n|√3=1√3 |n+1|=1⇒n+1⇒n=0 or n=−2 →δ=−^i+3^j+3^k