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(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors with the same magnitude,let $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$.Usin

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$\frac{1}{2}(\vec{a}+\vec{b}+\vec{c})$

Vedclass · Answer

(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors with the same magnitude,let $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$.Using the vector triple product identity $\vec{u} 	imes (\vec{v} 	imes \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$,we expand each term:$\vec{a} 	imes \{(\vec{r}-\vec{b}) 	imes \vec{a}\} = (\vec{a} \cdot \vec{a})(\vec{r}-\vec{b}) - (\vec{a} \cdot (\vec{r}-\vec{b}))\vec{a} = k^2(\vec{r}-\vec{b}) - (\vec{a} \cdot \vec{r})\vec{a}$.Similarly,$\vec{b} 	imes \{(\vec{r}-\vec{c}) 	imes \vec{b}\} = k^2(\vec{r}-\vec{c}) - (\vec{b} \cdot \vec{r})\vec{b}$ and $\vec{c} 	imes \{(\vec{r}-\vec{a}) 	imes \vec{c}\} = k^2(\vec{r}-\vec{a}) - (\vec{c} \cdot \vec{r})\vec{c}$.Summing these,we get $k^2(3\vec{r} - (\vec{a}+\vec{b}+\vec{c})) - ((\vec{a} \cdot \vec{r})\vec{a} + (\vec{b} \cdot \vec{r})\vec{b} + (\vec{c} \cdot \vec{r})\vec{c}) = \vec{0}$.Since $\vec{r} = x\vec{a} + y\vec{b} + z\vec{c}$ where $x = \frac{\vec{r} \cdot \vec{a}}{k^2}$,$y = \frac{\vec{r} \cdot \vec{b}}{k^2}$,$z = \frac{\vec{r} \cdot \vec{c}}{k^2}$,the expression becomes $k^2(3\vec{r} - (\vec{a}+\vec{b}+\vec{c})) - k^2(x\vec{a} + y\vec{b} + z\vec{c}) = \vec{0}$.This simplifies to $3\vec{r} - (\vec{a}+\vec{b}+\vec{c}) - \vec{r} = \vec{0}$,which gives $2\vec{r} = \vec{a}+\vec{b}+\vec{c}$.Therefore,$\vec{r} = \frac{1}{2}(\vec{a}+\vec{b}+\vec{c})$.

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