If vec{a}, vec{b} and vec{c} are vectors with | vedclass

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(C) Given,$|\vec{a}|=2, |\vec{b}|=3$ and $|\vec{c}|=4$.We know that $|\vec{x}-\vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$.There

Vedclass · Accepted Answer

$87$

Vedclass · Answer

(C) Given,$|\vec{a}|=2, |\vec{b}|=3$ and $|\vec{c}|=4$.We know that $|\vec{x}-\vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$.Therefore,$|\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})$.We also know that $|\vec{a}+\vec{b}+\vec{c}|^2 \geq 0$,which implies $|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0$.Thus,$-2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \leq |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2$.Substituting this into our expression:$|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2 \leq 2(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2) + (|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2) = 3(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2)$.Substituting the given magnitudes:$3(2^2+3^2+4^2) = 3(4+9+16) = 3(29) = 87$.Thus,the upper bound is $87$.

If $\vec{a}, \vec{b}$ and $\vec{c}$ are vectors with magnitudes $2, 3$ and $4$ respectively,then the best upper bound of $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ among the given values is

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