A vector r→ is equally inclined with the vectors a→=cos⁡θi^+sin⁡θj^, b→=-sin⁡θi^+cos⁡θj^ and c→=k^, then the angle between r→ and a→ is

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MARKS Content Team · Accepted Answer

Correct answer: (C) cos^(-1)1/(√3). a→,b→,c→ are unit vectors and mutually perpendicular vectors and r→ is equally inclined to all 3 vectors ⇒r→=λa→+b→+c→ ⇒r→=λcos⁡θ-sin⁡θi^+sin⁡θ+cos⁡θj^+k^ Angle between r→ and a→ is cos⁡α=(r→⋅a→)/(r→a→)=(λcos^2⁡θ-sin⁡θ)/(cos⁡θ+sin^2⁡θ+sin⁡θcos⁡θ1×√3λ) cos⁡α=1/(√3)⇒α=cos^(-1)1/(√3)

A vector $\vec{r}$ is equally inclined with the vectors $\vec{a} = \cos θ \hat{i} + \sin θ \hat{j}, \vec{b} = - \sin θ \hat{i} + \cos θ \hat{j}$ and $\vec{c} = \hat{k},$ then the angle between $\vec{r}$ and $\vec{a}$ is

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