Let →a=2^i+^j−^k and →b=^i+2^j+^k be two vectors. Consider a vector →c=α→a+β→b,α,β∈R. If the projection of →c on the vector (→a+→b) is 3√2, then the minimum value of (→c−(→a×→b)).→c equals
Open in App
Solution
Let →c=α→a+β→b=(2α+β)^i+(α+2β)^j+(β−α)^k Projection →c on vector →a+→b =→c.(→a+→b)|→a+→b|=3√2 ⇒((2α+β)^i+(α+2β)^j+(β−α)^k).(3^i+3^j+0^k)|3^i+3^j+0^k|=3√2 ⇒9(α+β)=3√2×3√2 ⇒α+β=2...(i) (→a×→b)=∣∣
∣
∣∣^i^j^k21−1121∣∣
∣
∣∣=3^i−3^j+3^k (→c−→a×→b).→c=|→c|2−[→a→b→c] ∵[→a→b→c]=0 |c|2=(2α+β)2+(α+2β)2+(β−α)2=6(α2+β2+α.β) For minimum value ⇒α=β=1 Hence,(→c−→a×→b).→c=18