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;
^i−^j+3^k
−^i−^j−^k
^i+3^j−^k
3^i−^j+3^k
Solution:
Let →v=λα+μβ→v=(λ+μ)^i+(λ−μ)^j+(λ+μ)^kProjection of →v on →c=→v.→c∣∣→c∣∣=1√3→(λ+μ)−(λ−μ)−(λ+μ)√3=1√3→μ−λ=1orμ=λ+1→→v=(2λ+1)^i−^j+(2λ+1)^kFor λ=1,→v=3^i−^j+3^k