Question

Let $\overrightarrow{a}$ & $\overrightarrow{b}$ be two non-zero perpendicular vectors. If a vector $\overrightarrow{x}$ satisfying the equation $\overrightarrow{x}$ x $\overrightarrow{b}$ = $\overrightarrow{a}$ is $\overrightarrow{x}$ = $\beta \overrightarrow{b}-\frac{1}{{\left|\overrightarrow{b}\right|}^2}$$\overrightarrow{a}$ x $\overrightarrow{b}$ then $\beta$ can be

Answer 1

$\overline{a}\times \overline{b}=\left|\overline{a}\right|\left|\overline{b}\right|\sin 90°=\left|\overline{a}\right|\left|\overline{b}\right|⟶\left(1\right)\overline{x}=\beta \overline{b}-\frac{1}{\left|\overline{b}\right|}(\overline{a}\times \overline{b})\overline{x}\times \overline{b}=\beta \overline{b}\times \overline{b}-\frac{1}{{\left|\overline{b}\right|}^2}(\overline{a}\times \overline{b})\times \overline{b}\overline{x}\times \overline{b}=-\frac{1}{{\left|\overline{b}\right|}^2}|\overline{a}\times \overline{b}|\left|\overline{b}\right|\overline{x}\times \overline{b}=-\frac{1}{{\left|\overline{b}\right|}^2}\left|\overline{a}\right|\left|\overline{b}\right|\left|\overline{b}\right|=-\left|\overline{a}\right|\hat{a}=-\overline{a}\overline{x}\times \overline{b}=-\overline{a}$

Since $\beta$is not used and $\overline{x}\times \overline{b}=-\overline{a}$

So, $\beta$ can be any real number.