If a - b - c - are unit vectors such that a - b - c -0 - then the value of a - b - b - c - c - a - isa 1b 3c -32d none of these

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Byju's Answer

Standard XII

Mathematics

Selecting Consecutive Terms in A.P

If a →, b →, ...

Question

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0},$ then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} is$
(a) 1
(b) 3
(c) $- \frac{3}{2}$
(d) none of these

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Solution

Given:
$\vec{a}, \vec{b}, \vec{c}$ are unit vectors
$\vec{a} + \vec{b} + \vec{c} = \vec{0}$

$\vec{a} + \vec{b} + \vec{c} = \vec{0} \Rightarrow {|\vec{a} + \vec{b} + \vec{c}|}^{2} = 0 \Rightarrow (\vec{a} + \vec{b} + \vec{c}) . (\vec{a} + \vec{b} + \vec{c}) = 0 \Rightarrow \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2 \vec{a} \cdot \vec{b} + 2 \vec{b} \cdot \vec{c} + 2 \vec{c} \cdot \vec{a} = 0 \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \Rightarrow 1 + 1 + 1 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 (∵ \vec{a}, \vec{b} and \vec{c} are unit vectors) \Rightarrow 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = - 3 \Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = - \frac{3}{2}$

Hence, the correct option is (c).

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If a→, b→, c→ are unit vectors such that a→+b→+c→=0→, then the value of a→·b→+b→·c→+c→·a→ is (a) 1 (b) 3 (c) -32 (d) none of these

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