devarshi-dt-logo

Question:

Let a⃗=2î+ĵ-k̂ and b⃗=î+2ĵ+k̂ be two vectors. Consider a vector C⃗=αa⃗+βb⃗,α,β∈ R. If the projection of c⃗ on the vector (a⃗+b⃗) is √(2), then the minimum value of (c⃗-(a⃗×b⃗)).c⃗ equal to.

18.00

Solution:

Correct option is A. 18.00
c→=α(2i^+j^−k^)+β(i^+2j^+k^)⇒c→=(2αa+β)i^+(α+2β)j^+(β−α)k^
c→ (a→+b→)|a→+b→|=32⇒9(α+β)=18⇒α+β=2
(c→−a→×b→).c→=(αa→+βb→−a→×b→−a→×b→ (αa→+βb→)=6α2+6αβ=6(α2−2;α+4)=min value18