If vec{a}, vec{b}, text{ and } vec{c} are uni | vedclass

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(B) Given that $\vec{a}, \vec{b}, 	ext{ and } \vec{c}$ are unit vectors,we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
We know that $|\vec{x} - \v

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Given that $\vec{a}, \vec{b}, 	ext{ and } \vec{c}$ 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:
$|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = (|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}) + (|\vec{b}|^2 + |vec{c}|^2 - 2\vec{b} \cdot \vec{c}) + (|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{c} \cdot \vec{a})$.
Since $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$,this simplifies to:
$2(1+1+1) - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.
We also know that $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.
Thus,$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = |\vec{a} + \vec{b} + \vec{c}|^2 - 3$.
Substituting this back into the expression:
$6 - (|\vec{a} + \vec{b} + \vec{c}|^2 - 3) = 9 - |\vec{a} + \vec{b} + \vec{c}|^2$.
Since $|\vec{a} + \vec{b} + \vec{c}|^2 \geq 0$,the maximum value of the expression is $9$.

If $\vec{a}, \vec{b}, \text{ and } \vec{c}$ are unit vectors,then $|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2$ does not exceed:

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