Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are unit vectors such that $\overrightarrow{a}.\overrightarrow{b}=0,(\overrightarrow{a}-\overrightarrow{b}).(\overrightarrow{b}+\overrightarrow{c})=0$ and $\overrightarrow{c}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}+\omega (\overrightarrow{a}\times \overrightarrow{b}),$ where $\lambda ,\mu ,\omega$ are scalars then.

• A. ${\mu }^2+{\omega }^2=1$
• B. $\lambda +\mu =1$
• C. $(\mu +1{)}^2+{\mu }^2+{\omega }^2=1$
• D. ${\lambda }^2+{\mu }^2=1$

Answer 1

The correct option is C

$(\mu +1{)}^2+{\mu }^2+{\omega }^2=1$

$(\overline{a}-\overline{b}).(\overline{b}-\overline{c})=0\Rightarrow \overline{a}.\overline{b}+\overline{c}.(\overline{a}-\overline{b})-|\overline{c}{|}^2=0$

$(\overline{a}-\overline{b}).\overline{c}=1\Rightarrow (\overline{a}-\overline{b}).(\lambda \overrightarrow{a}+\mu \overrightarrow{b}+\omega (\overrightarrow{a}\times \overrightarrow{b}\left)\right)=1\Rightarrow \lambda -\mu =1$

$\lambda =\mu +1,\overline{a}.\overline{b}=0\Rightarrow \overline{a},\overline{b},\overline{a}\times \overline{b}$ are mutually perpendicular

$|\overline{c}|=1\Rightarrow {\lambda }^2+{\mu }^2+{\omega }^2=1\Rightarrow (\mu +1{)}^2+{\mu }^2+{\omega }^2=1$