A) \[\vec{a}+\vec{b}+\vec{c}\]
B) \[\frac{{\vec{a}}}{\left| {\vec{a}} \right|}+\frac{{\vec{b}}}{\left| {\vec{b}} \right|}+\frac{{\vec{c}}}{\left| {\vec{c}} \right|}\]
C) \[\frac{{\vec{a}}}{{{\left| {\vec{a}} \right|}^{2}}}+\frac{{\vec{b}}}{{{\left| {\vec{b}} \right|}^{2}}}+\frac{{\vec{c}}}{{{\left| {\vec{c}} \right|}^{2}}}\]
D) \[\left| {\vec{a}} \right|\vec{a}-|\vec{b}|\vec{b}+|\vec{c}|\vec{c}\]
Correct Answer: B
Solution :
[b] \[\vec{a}.\vec{b}=\vec{b}.\vec{c}=\vec{c}.\vec{a}=0\] Let \[\vec{r}=\frac{{\vec{a}}}{\left| {\vec{a}} \right|}+\frac{{\vec{b}}}{\left| {\vec{b}} \right|}+\frac{{\vec{c}}}{\left| {\vec{c}} \right|}\] \[\cos \theta =\frac{\vec{r}.\vec{a}}{|\vec{r}|\left| {\vec{a}} \right|}=\frac{1}{|\vec{r}|}\] \[\cos \phi =\frac{\vec{r}.\vec{b}}{|\vec{r}|\left| {\vec{b}} \right|}=\frac{1}{|\vec{r}|}\] \[\cos \psi =\frac{\vec{r}.\vec{c}}{|\vec{r}|\left| {\vec{c}} \right|}=\frac{1}{|\vec{r}|}\] \[\therefore \theta =\phi =\psi \]
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