A) \[\frac{{{a}^{2}}(\vec{b}\times \vec{c})+{{b}^{2}}(\vec{c}\times \vec{a})+{{c}^{2}}(\vec{a}\times \vec{b})}{2[\vec{a}\,\vec{b}\,\vec{c}]}\]
B) \[\frac{{{b}^{2}}(\vec{b}\times \vec{c})+{{a}^{2}}(\vec{c}\times \vec{a})+{{c}^{2}}(\vec{a}\times \vec{b})}{[\vec{a}\,\vec{b}\,\vec{c}]}\]
C) \[\frac{{{b}^{2}}(\vec{b}\times \vec{c})+{{a}^{2}}(\vec{c}\times \vec{a})+{{c}^{2}}(\vec{a}\times \vec{b})}{2[\vec{a}\,\vec{b}\,\vec{c}]}\]
D) \[\frac{{{a}^{2}}(\vec{a}\times \vec{b})+{{b}^{2}}(\vec{b}\times \hat{c})+{{c}^{2}}(\vec{c}\times \vec{a})}{2[\vec{a}\,\vec{b}\,\vec{c}]}\]
Correct Answer: A
Solution :
[a] If the centre \['P'\]is with position vector \[\vec{r},\] |
Then \[\vec{a}-\vec{r}=\overrightarrow{PA},\vec{b}-\vec{r}=\overrightarrow{PB},\vec{c}-\vec{r}=\overrightarrow{PC,}\] |
Where \[\left| \overrightarrow{PA} \right|=\left| \overrightarrow{PB} \right|\] |
\[=\left| \overrightarrow{PC} \right|=\left| \overrightarrow{OP} \right|=\left| {\vec{r}} \right|\] |
Consider \[\left| \vec{a}-\vec{r} \right|=\left| {\vec{r}} \right|\] |
\[\Rightarrow (\vec{a}-\vec{r}).(\vec{a}-\vec{r})=\vec{r}.\vec{r}\] |
\[\Rightarrow {{a}^{2}}=-2\vec{a}.\vec{r}+{{r}^{2}}={{r}^{2}}\] |
\[\Rightarrow {{a}^{2}}=2\vec{a}.\vec{r}\] Similarly, \[{{b}^{2}}=2\vec{b}.\vec{r}\] and \[{{c}^{2}}\] |
\[=2\vec{c}.\vec{r}\] |
Since, \[(\vec{b}\times \vec{c}),(\vec{c}\times \vec{a})\] and \[(\vec{a}\times \vec{b})\] are non-coplanar, then |
\[\vec{r}=x(\vec{b}\times \vec{c})+y(\vec{c}\times \vec{a})+z(\vec{a}\times \vec{b})\] |
\[\vec{a}.\,\vec{r}=x\,\vec{a}.(\vec{b}\times c)+y.0+z.0=x[\vec{a}\,\vec{b}\,\vec{c}]\] |
\[\Rightarrow x=\frac{\vec{a}.\vec{r}}{[\vec{a}\,\vec{b}\,\vec{c}]}=\frac{{{a}^{2}}}{2[\vec{a}\,\vec{b}\,\vec{c}]}\] |
Similarly, \[y=\frac{{{b}^{2}}}{2[\vec{a}\,\vec{b}\,\vec{c}]}\] and \[z=\frac{{{c}^{2}}}{2[\vec{a}\,\vec{b}\,\vec{c}]}\] |
Therefore, \[\vec{r}\] |
\[=\frac{{{a}^{2}}(\vec{b}\times \vec{c})+{{b}^{2}}(\vec{c}\times \vec{a})+{{c}^{2}}(\vec{a}\times \vec{b})}{2[\vec{a}\,\vec{b}\,\vec{c}]}\] |
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