A unit vector d→ is equally inclined at an angle a with the vectors⁡a→=cos⁡θi^+sin⁡θj^,b→=−sin⁡θi^+cos⁡θi^,c→=k^Then α =

|a→|=|b→|=|c→|=1⇒ d→⋅a→=d→⋅b→=d→⋅c→=cos⁡α,c→=k^d→⋅(a→−k^)=0,d→⋅(b→−k^)=0∴ d→ is along (a→−k^)×(b→−k→)=i^j^k^cos⁡θsin⁡θ−1−sin⁡θcos⁡θ−1=(cos⁡θ−sin⁡θ)i^+(cos⁡θ+sin⁡θ)j^+k^∴d→=(cos⁡θ−sin⁡θ)i^+(cos⁡θ+sin⁡θ)j^+k^(cos⁡θ−sin⁡θ)2+(cos⁡θ+sin⁡θ)2+1=13[(cos⁡θ−sin⁡θ)i^+(cos⁡θ+sin⁡θ)j^+k^]⇒cos⁡α=d→⋅k^=13

A unit vector d→ is equally inclined at an angle a with the vectors⁡a→=cos⁡θi^+sin⁡θj^,b→=−sin⁡θi^+cos⁡θi^,c→=k^Then α =

|a→|=|b→|=|c→|=1⇒ d→⋅a→=d→⋅b→=d→⋅c→=cos⁡α,c→=k^d→⋅(a→−k^)=0,d→⋅(b→−k^)=0∴ d→ is along (a→−k^)×(b→−k→)=i^j^k^cos⁡θsin⁡θ−1−sin⁡θcos⁡θ−1=(cos⁡θ−sin⁡θ)i^+(cos⁡θ+sin⁡θ)j^+k^∴d→=(cos⁡θ−sin⁡θ)i^+(cos⁡θ+sin⁡θ)j^+k^(cos⁡θ−sin⁡θ)2+(cos⁡θ+sin⁡θ)2+1=13[(cos⁡θ−sin⁡θ)i^+(cos⁡θ+sin⁡θ)j^+k^]⇒cos⁡α=d→⋅k^=13

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$A$ unit vector $\vec{d}$ is equally inclined at an angle a with the $vectors \vec{a} = \cos θ \hat{i} + \sin θ \hat{j}, \vec{b} = - \sin θ \hat{i} + \cos θ \hat{i}, \vec{c} = \hat{k}$
Then $α =$

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$\frac{π}{4}$

$\cos^{- 1} \frac{1}{\sqrt{3}}$

$\cos^{- 1} \frac{1}{3}$

$\frac{π}{2}$

answer is A.

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Detailed Solution

$| \vec{a} | = | \vec{b} | = | \vec{c} | = 1$

$\begin{array}{l} \Rightarrow \vec{d} \cdot \vec{a} = \vec{d} \cdot \vec{b} = \vec{d} \cdot \vec{c} = \cos α, \vec{c} = \hat{k} \\ \vec{d} \cdot (\vec{a} - \hat{k}) = 0, \vec{d} \cdot (\vec{b} - \hat{k}) = 0 \end{array}$

$∴$ $\vec{d}$ is along $(\vec{a} - \hat{k}) \times (\vec{b} - \vec{k})$

$\begin{array}{l} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \cos θ & \sin θ & - 1 \\ - \sin θ & \cos θ & - 1 \end{array}| = (\cos θ - \sin θ) \hat{i} + (\cos θ + \sin θ) \hat{j} + \hat{k} \\ ∴ \vec{d} = \frac{(\cos θ - \sin θ) \hat{i} + (\cos θ + \sin θ) \hat{j} + \hat{k}}{\sqrt{(\cos θ - \sin θ)^{2} + (\cos θ + \sin θ)^{2} + 1}} \\ = \frac{1}{\sqrt{3}} [(\cos θ - \sin θ) \hat{i} + (\cos θ + \sin θ) \hat{j} + \hat{k}] \\ \Rightarrow \cos α = \vec{d} \cdot \hat{k} = \frac{1}{\sqrt{3}} \end{array}$