# If a,b and c are unit vectors then $\left| a-b{{|}^{2}} \right.+|b-c{{|}^{2}}+|c-a{{|}^{2}}$ does not exceed

A. 4

B. 9

C. 8

D. 6

Answer

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**Hint:**We are given a, b, and c as unit vectors, and we must find the value that the given equation does not exceed. A unit vector is a vector with a magnitude of one. To answer this question, we use the whole square formula, and we get our desired answer by entering the values into the formulas and solving the equations.

**Formula Used**:

${{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}=2({{a}^{2}}+{{b}^{2}}+{{c}^{2}})-2(a.b+b.c+c.a)$

**Complete step by step solution:**

We are given that a , b and c are the unit vectors , then

$\left| a-b{{|}^{2}} \right.+|b-c{{|}^{2}}+|c-a{{|}^{2}}$-------------------------(1)

We know the formula of ${{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}=2({{a}^{2}}+{{b}^{2}}+{{c}^{2}})-2(a.b+b.c+c.a)$

Put the above values in equation (1) and we get

$\left| a-b{{|}^{2}} \right.+|b-c{{|}^{2}}+|c-a{{|}^{2}}$ = $2({{a}^{2}}+{{b}^{2}}+{{c}^{2}})-2(a.b+b.c+c.a)$

By putting the values in above equation, we get

$2({{a}^{2}}+{{b}^{2}}+{{c}^{2}})-2(a.b+b.c+c.a)$ = 2$\times 3 – 2 (a.b + b.c + c.a )$ ----------- (2)

By simplifying $2 (a.b + b.c + c.a ) = \left\{ {{(a+b+c)}^{2}}-{{a}^{2}}-{{b}^{2}}-{{c}^{2}} \right\}$

By putting the values in equation (2), we get

$2({{a}^{2}}+{{b}^{2}}+{{c}^{2}})-2(a.b+b.c+c.a)$ = 6 – $\left\{ {{(a+b+c)}^{2}}-{{a}^{2}}-{{b}^{2}}-{{c}^{2}} \right\}$

By solving the above equation, we get 9 - $|a+b+c{{|}^{2}}\le 9$

Hence, value of $\left| a-b{{|}^{2}} \right.+|b-c{{|}^{2}}+|c-a{{|}^{2}}$ does not exceed 9

**Option ‘B’ is correct**

**Note:**Vector units have both direction and magnitude. However, sometimes one is interested only in direction and not the magnitude. Vectors are often considered unit length. These unit vectors are generally used to represent direction, with a scalar coefficient providing the magnitude.

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