If vec{a}, vec{b}, vec{c} are unit vectors,th | vedclass

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(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.We know that $|\vec{x} - \vec{y}|^2 = |\ve

Vedclass · Accepted Answer

$9$

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(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.We know that $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$.Expanding the expression:$|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = (|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}) + (|\vec{b}|^2 + |\vec{c}|^2 - 2\vec{b} \cdot \vec{c}) + (|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{c} \cdot \vec{a})$$= 2(|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2) - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$$= 2(1 + 1 + 1) - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.We also know that $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0$.So,$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = |\vec{a} + \vec{b} + \vec{c}|^2 - 3$.Substituting this back:$6 - (|\vec{a} + \vec{b} + \vec{c}|^2 - 3) = 9 - |\vec{a} + \vec{b} + \vec{c}|^2$.Since $|\vec{a} + \vec{b} + \vec{c}|^2 \geq 0$,the maximum value is $9$ when $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,then what is the maximum value of $|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2$?

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