Question

Directions: The following question has four choices, out of which one or more are correct.Let α=ai^+bj^+ck^,β=bi^+cj^+ak^ and y=ci^+aj^+bk^ be three co-planar vectors with a≠b andv=i^+j^+k^, Then v is perpendicular to

• A. (A) α
• B. (B) β
• C. (C) γ
• D. (D) None of these

Answer 1

Answer: (A), (D)

(A) α

Answer 2

Explanation:
|abcbcacab|=0a3+b3+c3−3abc=0(a+b+c)(a2+b2+c2−ab−bc−ca)=012(a+b+c)(2a2+2b2+2c2−2ab−2bc−2ca)=0⇒a+b+c=0Or a=b=cBut, a≠b, therefore,a+b+c=0Nowα.v=(ai^+bj^+ck^)⋅(i^+j^+k^)=a+b+c=0β⋅v=(bi^+cj^+ak^)⋅(i^+j^+k^)=b+c+a=0γ⋅v=(ci^+aj^+bk^)⋅(i^+j^+k^)=c+a+b=0Hence, this is the required solution.