] √3a1=b1−b2 ....(i) √3a2=b1+kb2.......(ii) And as given, a12+a22=‌
2
3
(b12+b22) Squaring and adding Eqs. (i) and (ii), 3a12+3a22=(b1−b2)2+(b1+kb2)2 ⇒‌‌2(b12+b22)=2b12+b22(k2+1)+2b1b2(k−1) ⇒‌‌b22(k2+1−2)+2b1b2(k−1)=0 ⇒‌‌(k−1)[b22(k+1)+2b1b2]=0 So, k=1