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. $x=cos\theta ,y=sin\theta ,z=cos2\theta ,$
• B. $x=sin\theta ,y=cos\theta ,z=-cos2\theta ,$
• C. $x=y=cos\theta ,z^2=cos2\theta ,$
• D. $x=y=cos\theta ,z^2=-cos2\theta ,$

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 $\vec{c}\cdot \vec{c}={\left|\vec{c}\right|}^2$
$\Rightarrow \left(\alpha \vec{a}+\beta \vec{b}+\gamma \left(\vec{a}\times \vec{b}\right)\right)$
$\left(\alpha \vec{a}+\beta \vec{b}+\gamma \left(\vec{a}\times \vec{b}\right)\right)={\left|\vec{c}\right|}^2$
$\Rightarrow 2{\alpha }^2+{\gamma }^2{\left|\vec{a}\times \vec{b}\right|}^2=1$
$\Rightarrow 2{\alpha }^2+{\beta }^2\left[{\left|\vec{a}\right|}^2{\left|\vec{b}\right|}^2-{\left(\vec{a}\cdot \vec{b}\right)}^2\right]=1$
$\Rightarrow {\alpha }^2+{\gamma }^2\left[1-0\right]=1\left[\because \vec{a}\perp \vec{b}\therefore \vec{a}\cdot \vec{b}=0\right]$
$\Rightarrow 2{\alpha }^2+{\gamma }^2=1$
$\Rightarrow {\gamma }^2=1-2x^2$
$=1-2co{s}^2\theta -cos2\theta$