Question

Let the unit vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ be perpendicular to eachs other and the unit vector $\overrightarrow{c}$ be inclined at an at an angle $\theta$ to both $\overrightarrow{a}$ and $\overrightarrow{b}$. If $x\overrightarrow{a}+y\overrightarrow{b}+z(\overrightarrow{a}\times \overrightarrow{b})$

• A. $x=\cos \theta ,y=\sin \theta ,z=\cos 2\theta$
• B. $x=\sin \theta ,y=\cos \theta ,z=-\cos 2\theta$
• C. $x=y=\cos \theta ,z^2=\cos 2\theta$
• D. $x=y=\cos \theta ,z^2=-\cos 2\theta$

Answer 1

The correct option is D $x=y=\cos \theta ,z^2=-\cos 2\theta$

$$\begin{array}{c}Accordingtoquestion,\\ Here........unitvector=\left|\overline{a}\right|=\left|\overline{b}\right|=\left|\overline{c}=1\right|\\ \overline{a}.\overline{b}=0\\ \overline{a}.\overline{c}=\cos \theta \\ \overline{b}.\overline{c}=\cos \theta \\ given,x\overline{a}+y\overline{b}+z(\overline{a}\times \overline{b})---------\left(1\right)\\ Now,\\ Takedotwith\overline{a},\\ \Rightarrow \overline{a}.\overline{c}=x.\overline{a}.\overline{a}+y\overline{b}.\overline{a}+z(\overline{a}\times \overline{b}).\overline{a}|(\overline{a}\times \overline{b})isperpendicularwithais0\\ \Rightarrow \cos \theta =x+y.\left(0\right)\\ \therefore x=\cos \theta \\ Again,\\ Takedotwith\overline{b},..........\\ \Rightarrow \overline{b}.\overline{c}=x.\overline{a}.\overline{b}+y\overline{b}.\overline{b}+z(\overline{a}\times \overline{b}).\overline{b}|(\overline{a}\times \overline{b})isperpendicularwithais0\\ \Rightarrow \overline{b}.\overline{c}=y\\ \therefore y=\cos \theta \\ Again,\\ Takedotwith\overline{c},...........\\ \Rightarrow \overline{c}.\overline{c}=x.\overline{a}.\overline{c}+y\overline{b}.\overline{c}+z(\overline{a}\times \overline{b}).\overline{c}\\ \Rightarrow 1={\cos }^2\theta +{\cos }^2\theta +z\left[\overline{a}\overline{b}\overline{c}\right]\\ wesimplify,\\ \left[\overline{a}\overline{b}\overline{c}\right]=(\overline{a}\times \overline{b}).\overline{c}|(\overline{a}\times \overline{b})isperpendicular\\ =\left|\overline{a}\right|\left|\overline{b}\right|\sin {90}^0\hat{n}.\overline{c}\\ =\hat{n}.\overline{c}\\ =\left|\hat{n}\right|.\left|\overline{c}\right|.\cos \beta |unitvector\left|\hat{n}\right|\&\left|\overline{c}\right|=1\\ \therefore \cos \beta =1\\ then,\\ \Rightarrow 1={\cos }^2\theta +{\cos }^2\theta +z\left[\overline{a}\overline{b}\overline{c}\right]\\ \Rightarrow 1=2{\cos }^2\theta +z\\ \Rightarrow z=1-2{\cos }^2\theta \\ \therefore z^2=-\cos 2\theta \\ So,thatx=y=\cos \theta and{z}^2=-\cos 2\theta \\ thereforthecorrectoptionisD.\end{array}$$