Question

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

• 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

Answer: (D)

We have $\vec{c}=\alpha\,\vec{c}+\beta\,\vec{b}+\gamma\left(\vec{a}\times\vec{ b}\right)$
$\Rightarrow \vec{c}\cdot\vec{a}=\alpha$ and $\vec{c}\cdot\vec{b}=\beta$
$\Rightarrow \alpha=\beta=cos\,\theta$
Now $\vec{c}\cdot\vec{c}=\left|\vec{c}\right|^{2}$
$\Rightarrow \left(\alpha\,\vec{a}+\beta\,\vec{b}+\gamma\left(\vec{a}\times\vec{b}\right)\right)$
$\left(\alpha\,\vec{a}+\beta\,\vec{b}+\gamma\left(\vec{a}\times\vec{b}\right)\right)=\left|\vec{c}\right|^{2}$
$\Rightarrow 2\alpha^{2}+\gamma^{2}\left|\vec{a}\times\vec{b}\right|^{2}=1$
$\Rightarrow 2\alpha^{2}+\beta^{2}\left[\left|\vec{a}\right|^{2}\left|\vec{b}\right|^{2}-\left(\vec{a}\cdot\vec{b}\right)^{2}\right]=1$
$\Rightarrow \alpha^{2}+\gamma^{2}\left[1-0\right]=1\quad\left[\because \vec{a}\,\bot\,\vec{b}\,\therefore \vec{a} \cdot \vec{b}=0\right]$
$\Rightarrow 2\alpha^{2}+\gamma^{2}=1$
$\Rightarrow \gamma^{2}=1-2x^{2}$
$=1-2\,cos^{2}\,\theta-cos\,2\theta$