Given a¯=cos⁡θi¯+sin⁡θj¯,b¯=−sin⁡θi¯+cos⁡θj¯ and c¯=k¯. The unit vector which makes equalangles with a¯,b¯,c¯ is

|a¯|=|b¯|=|c¯|=1So get a¯+b¯+c¯|a¯+b¯+c¯|Here a¯+b¯+c¯=(cos⁡θ−sin⁡θ)i¯+(sin⁡θ+cos⁡θ)j¯+k¯ has magnitude3.

Given a¯=cos⁡θi¯+sin⁡θj¯,b¯=−sin⁡θi¯+cos⁡θj¯ and c¯=k¯. The unit vector which makes equalangles with a¯,b¯,c¯ is

|a¯|=|b¯|=|c¯|=1So get a¯+b¯+c¯|a¯+b¯+c¯|Here a¯+b¯+c¯=(cos⁡θ−sin⁡θ)i¯+(sin⁡θ+cos⁡θ)j¯+k¯ has magnitude3.

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Given $\bar{a} = \cos θ \bar{i} + \sin θ \bar{j}, \bar{b} = - \sin θ \bar{i} + \cos θ \bar{j} and \bar{c} = \bar{k}$ . The unit vector which makes equal
angles with $\bar{a}, \bar{b}, \bar{c}$ is

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$(\bar{a} + \bar{b} + \bar{c}) \sqrt{3}$

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

None

$\frac{(\bar{a} + \bar{b} + \bar{c})}{\sqrt{3}}$

answer is C.

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Detailed Solution

$| \bar{a} | = | \bar{b} | = | \bar{c} | = 1$
So get $\frac{\bar{a} + \bar{b} + \bar{c}}{| \bar{a} + \bar{b} + \bar{c} |}$
Here $\bar{a} + \bar{b} + \bar{c} = (\cos θ - \sin θ) \bar{i} + (\sin θ + \cos θ) \bar{j} + \bar{k}$ has magnitude $\sqrt{3}$ .