If vec{a}, vec{b}, vec{c} are three vectors s | vedclass

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(C) We know that $|\vec{x}-\vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$.Expanding the given expression:$S = |\vec{a}-\vec{b}|^2+

Vedclass · Accepted Answer

$249$

Vedclass · Answer

(C) We know that $|\vec{x}-\vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$.Expanding the given expression:$S = |\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$$S = ( |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} ) + ( |\vec{b}|^2 + |\vec{c}|^2 - 2\vec{b} \cdot \vec{c} ) + ( |\vec{c}|^2 + |\vec{a}|^2 - 2\vec{c} \cdot \vec{a} )$$S = 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})$Given $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$,so $|\vec{a}|^2=9, |\vec{b}|^2=25, |\vec{c}|^2=49$.$S = 2(9+25+49) - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 166 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.We know that $|\vec{a}+\vec{b}+\vec{c}|^2 \ge 0$,which implies $|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \ge 0$.So,$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \ge -(9+25+49) = -83$.Therefore,$S = 166 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \le 166 - (-83) = 249$.Thus,the expression does not exceed $249$.

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$,then $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ does not exceed:

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