If a - b- - c - are three vectors such that a -b -c -0 - and a -2- b -3- c -5- then the value of a - b -b - c - c - a - isa 0b 1c -19d 38

Given: a #8594;=2,b #8594;=3,c #8594;=5 a #8594;+b #8594;+c #8594;=0 #8594; a #8594;+b #8594;+c #8594;=0 #8594; #8658;a #8594;+b # ...

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Byju's Answer

Standard X

Mathematics

Solving a Quadratic Equation by Factorization Method

If a →, b, → ...

Question

If $\vec{a}, \vec{b,} \vec{c}$ are three vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0} and |\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 5,$ then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \cdot \vec{c} \cdot \vec{a} is$
(a) 0
(b) 1
(c) -19
(d) 38

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Solution

Given:
$|\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 5$
$\vec{a} + \vec{b} + \vec{c} = \vec{0}$

$\vec{a} + \vec{b} + \vec{c} = \vec{0} \Rightarrow {|\vec{a} + \vec{b} + \vec{c}|}^{2} = 0 \Rightarrow (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \Rightarrow \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2 \vec{a} \cdot \vec{b} + 2 \vec{b} \cdot \vec{c} + 2 \vec{c} \cdot \vec{a} = 0 \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \Rightarrow 2^{2} + 3^{2} + 5^{2} + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 (∵ |\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 5) \Rightarrow 4 + 9 + 25 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \Rightarrow 38 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \Rightarrow 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = - 38 \Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = - \frac{38}{2} \Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = - 19$

Hence, the correct option is (c).

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