Question
If $$\vec a,\,\vec b$$ and $$\vec c$$ are unit vectors, then $${\left| {\vec a - \vec b} \right|^2} + {\left| {\vec b - \vec c} \right|^2} + {\left| {\vec c - \vec a} \right|^2}$$ does NOT exceed :
A.
$$4$$
B.
$$9$$
C.
$$8$$
D.
$$6$$
Answer :
$$9$$
Solution :
$$\hat a,\,\hat b,\,\hat c$$ are units vectors.
$$\eqalign{
& \therefore \hat a.\hat a = \hat b.\hat b = \hat c.\hat c = 1 \cr
& {\text{Now, }}x = {\left| {\hat a - \hat b} \right|^2} + {\left| {\hat b - \hat c} \right|^2} + {\left| {\hat c - \hat a} \right|^2} \cr
& = \hat a.\hat a + \hat b.\hat b - 2\hat a.\hat b + \hat b.\hat b + \hat c.\hat c - 2\hat b.\hat c + c.\hat c + \hat a.\hat a - 2\hat c.\hat a \cr
& = 6 - 2\left( {\hat a.\hat b + \hat b.\hat c + \hat c.\hat a} \right).....(1) \cr
& {\text{Also}}, \cr
& \Rightarrow \left| {\hat a + \hat b + \hat c} \right| \geqslant 0\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {\left| {\hat a + \hat b + \hat c} \right|^2} \geqslant 0 \cr
& \Rightarrow \hat a.\hat a + \hat b.\hat b + \hat c.\hat c + 2\left( {\hat a.\hat b + \hat b.\hat c + \hat c.\hat a} \right) \geqslant 0 \cr
& \Rightarrow 3 + 2\left( {\hat a.\hat b + \hat b.\hat c + \hat c.\hat a} \right) \geqslant 0 \cr
& \Rightarrow 2\left( {\hat a.\hat b + \hat b.\hat c + \hat c.\hat a} \right) \geqslant - 3 \cr
& \Rightarrow - 2\left( {\hat a.\hat b + \hat b.\hat c + \hat c.\hat a} \right) \leqslant 3 \cr
& \Rightarrow 6 - 2\left( {\hat a.\hat b + \hat b.\hat c + \hat c.\hat a} \right) \leqslant 9.....(2) \cr} $$
From (1) and (2), $$x \leqslant 9$$
$$\therefore \,x$$ does not exceed 9