• # question_answer Let the unit vectors a and b be perpendicular and the unit vector c be inclined at an angle q to both a and b. If $\mathbf{c}=\alpha \,\mathbf{a}+\beta \,\mathbf{b}+\gamma \,(\mathbf{a}\times \mathbf{b}),$ then [Orissa JEE 2003] A) $\alpha =\beta =\cos \theta ,\,\,{{\gamma }^{2}}=\cos \,\,2\theta$ B) $\alpha =\beta =\cos \theta ,\,\,{{\gamma }^{2}}=-\cos \,\,2\theta$ C) $\alpha =\cos \theta ,\,\,\beta =\sin \theta ,\,\,{{\gamma }^{2}}=\cos \,\,2\theta$ D) None of these

Solution :

• $\mathbf{c}=\alpha \,\mathbf{a}+\beta \,\mathbf{b}+\gamma \,(\mathbf{a}\times \mathbf{b})$$\Rightarrow \mathbf{c}\,.\,\mathbf{a}=\alpha$ and $\mathbf{c}\,.\,\mathbf{b}=\beta$
• $\Rightarrow \alpha =\beta =\cos \theta$
• Also, $1=\mathbf{c}\,.\,\mathbf{c}$,
• $\therefore \left[ \alpha \,\mathbf{a}+\beta \,\mathbf{b}+\gamma \,(\mathbf{a}\times \mathbf{b}) \right]\,.\,\left[ (\alpha \,\mathbf{a}+\beta \,\mathbf{b})+\gamma \,(\mathbf{a}\times \mathbf{b}) \right]=1$
• $\Rightarrow 2{{\alpha }^{2}}+{{\gamma }^{2}}{{(\mathbf{a}\times \mathbf{b})}^{2}}=1$,
• $\left\{ \because \,\,\alpha \,=\,\beta \right\}$
• $\Rightarrow 2{{\alpha }^{2}}+{{\gamma }^{2}}[{{\mathbf{a}}^{2}}{{\mathbf{b}}^{2}}-{{(\mathbf{a}.\mathbf{b})}^{2}}]=1\Rightarrow 2{{\alpha }^{2}}+{{\gamma }^{2}}=1$
• Hence, ${{\gamma }^{2}}=1-2{{\alpha }^{2}}=1-2\,{{\cos }^{2}}\theta =-\cos 2\theta .$

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