bar{a}, bar{b}, bar{c} are three non-coplanar | vedclass

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(B) Given that $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular vectors of magnitude $K$, we have $\bar{a} \cdot \bar{b} = \bar{b} \cdot \bar{c}

Vedclass · Accepted Answer

$\frac{\bar{a}+\bar{b}+\bar{c}}{2}$

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(B) Given that $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular vectors of magnitude $K$, we have $\bar{a} \cdot \bar{b} = \bar{b} \cdot \bar{c} = \bar{c} \cdot \bar{a} = 0$ and $|\bar{a}| = |\bar{b}| = |\bar{c}| = K$, so $\bar{a} \cdot \bar{a} = \bar{b} \cdot \bar{b} = \bar{c} \cdot \bar{c} = K^2$. Using the vector triple product formula $\bar{u} 	imes (\bar{v} 	imes \bar{w}) = (\bar{u} \cdot \bar{w})\bar{v} - (\bar{u} \cdot \bar{v})\bar{w}$, the given equation becomes: $(\bar{a} \cdot \bar{a})(\bar{r}-\bar{b}) - (\bar{a} \cdot (\bar{r}-\bar{b}))\bar{a} + (\bar{b} \cdot \bar{b})(\bar{r}-\bar{c}) - (\bar{b} \cdot (\bar{r}-\bar{c}))\bar{b} + (\bar{c} \cdot \bar{c})(\bar{r}-\bar{a}) - (\bar{c} \cdot (\bar{r}-\bar{a}))\bar{c} = \bar{0}$. Substituting the dot products: $K^2(\bar{r}-\bar{b}) - (\bar{a} \cdot \bar{r})\bar{a} + K^2(\bar{r}-\bar{c}) - (\bar{b} \cdot \bar{r})\bar{b} + K^2(\bar{r}-\bar{a}) - (\bar{c} \cdot \bar{r})\bar{c} = \bar{0}$. $3K^2\bar{r} - K^2(\bar{a}+\bar{b}+\bar{c}) - ((\bar{a} \cdot \bar{r})\bar{a} + (\bar{b} \cdot \bar{r})\bar{b} + (\bar{c} \cdot \bar{r})\bar{c}) = \bar{0}$. Since $\bar{a}, \bar{b}, \bar{c}$ are orthogonal, any vector $\bar{r}$ can be written as $\bar{r} = \frac{\bar{a} \cdot \bar{r}}{K^2}\bar{a} + \frac{\bar{b} \cdot \bar{r}}{K^2}\bar{b} + \frac{\bar{c} \cdot \bar{r}}{K^2}\bar{c}$. Thus, $(\bar{a} \cdot \bar{r})\bar{a} + (\bar{b} \cdot \bar{r})\bar{b} + (\bar{c} \cdot \bar{r})\bar{c} = K^2\bar{r}$. Substituting this back: $3K^2\bar{r} - K^2(\bar{a}+\bar{b}+\bar{c}) - K^2\bar{r} = \bar{0}$. $2K^2\bar{r} = K^2(\bar{a}+\bar{b}+\bar{c})$. $\bar{r} = \frac{\bar{a}+\bar{b}+\bar{c}}{2}$.

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