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If {\bar  {a}},{\bar  {b}},{\bar  {c}}\, are mutually perpendicular vectors of equal magnitude,show that {\bar  {a}}+{\bar  {b}}+{\bar  {c}}\, is equally inclined to {\bar  {a}},{\bar  {b}},{\bar  {c}}\,

Given that the vectors a,b,c are mutually perpendicular to each other.

Hence{\bar  {a}}\cdot {\bar  {b}}={\bar  {b}}\cdot {\bar  {c}}={\bar  {c}}\cdot {\bar  {a}}=0\,

Also given the three of equal magnitude |a|=|b|=|c|\,

Now |a+b+c|^{2}=({\bar  {a}}+{\bar  {b}}+{\bar  {c}})^{2}=({\bar  {a}})^{2}+({\bar  {b}})^{2}+({\bar  {c}})^{2}+2({\bar  {a}}\cdot {\bar  {b}}+{\bar  {b}}\cdot {\bar  {c}}+{\bar  {c}}\cdot {\bar  {a}})=|a|^{2}+|b|^{2}+|c|^{2}+2(0)=3|a|^{2}\,

Hence |{\bar  {a}}+{\bar  {b}}+{\bar  {c}}|=3{\sqrt  {{\bar  {a}}}}\,

Let {\bar  {a}}+{\bar  {b}}+{\bar  {c}}\, make angles \alpha ,\beta ,\gamma \, respectively.

Therefore \cos \alpha ={\frac  {({\bar  {a}}+{\bar  {b}}+{\bar  {c}})\cdot {\bar  {a}}}{|a+b+c||a|}}={\frac  {|a|^{2}}{{\sqrt  {3}}|a|^{2}}}={\frac  {1}{{\sqrt  {3}}}}\,

Similarly \cos \beta ={\frac  {1}{{\sqrt  {3}}}},\cos \gamma ={\frac  {1}{{\sqrt  {3}}}}\,

Therefore \cos \alpha =\cos \beta =\cos \gamma \, implies that \alpha =\beta =\gamma \,

Hence the sum of the three vectors is equally inclined to the three individual vectors.

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