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(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. Consider the expression $E = |\vec{a}-\ve

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. Consider the expression $E = |\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$. Expanding this,we get $E = (\vec{a}^2 + \vec{b}^2 - 2\vec{a} \cdot \vec{b}) + (\vec{b}^2 + \vec{c}^2 - 2\vec{b} \cdot \vec{c}) + (\vec{c}^2 + \vec{a}^2 - 2\vec{c} \cdot \vec{a})$. Since $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$,this simplifies to $E = 6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$. We know that $|\vec{a} + \vec{b} + \vec{c}|^2 \geq 0$,which implies $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0$. Substituting the magnitudes,$3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0$,so $2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq -3$. Thus,the maximum value of $E$ is $6 - (-3) = 9$. Therefore,$k = 9$. Finally,$k(2|\vec{a}|^2 + 3|\vec{b}|^2 - 4|\vec{c}|^2) = 9(2(1) + 3(1) - 4(1)) = 9(2 + 3 - 4) = 9(1) = 9$.

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors and the maximum value of $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ is $k$,then $k(2|\vec{a}|^2+3|\vec{b}|^2-4|\vec{c}|^2) = $

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