Question

If $\overrightarrow{a,}\overrightarrow{b,}\overrightarrow{c}$are three non-coplanar mutually perpendicular unit vectors, then $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right],$ is
(a) ± 1
(b) 0
(c) −2
(d) 2

Answer 1

$$\begin{gathered} \left(\mathrm{a}\right)\pm 1 \\ \mathrm{We}\mathrm{have} \\ \left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right] \\ =\left(\overrightarrow{a}\times \overrightarrow{b}\right).\overrightarrow{c} \\ =\left|\overrightarrow{a}\times \overrightarrow{b}\right|\left|\overrightarrow{c}\right|\cos 0^{°}\mathrm{or}\left|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right|\left|\overrightarrow{\mathrm{c}}\right|\mathrm{co}\mathrm{s}{180}^{°}\left(\because \overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b},}\overrightarrow{\mathrm{c}}\mathrm{are}\mathrm{perpendicular}\mathrm{to}\mathrm{each}\mathrm{other}\right) \\ =\left|\overrightarrow{a}\times \overrightarrow{b}\right|\mathrm{or}-\left|\overrightarrow{a}\times \overrightarrow{b}\right|\left(\because \left|\overrightarrow{\mathrm{c}}\right|=1,\mathrm{co}\mathrm{s}{0}^{°}=1\mathrm{and}\cos {180}^{°}=-1\right) \\ =\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{b}\right|\sin {90}^{°}\mathrm{or}-\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{b}\right|\sin {90}^{°}\left(\because \stackrel{}{\overrightarrow{a}}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\overrightarrow{b}\right) \\ =1\mathrm{or}-1\left(\because \left|\overrightarrow{\mathrm{a}}\right|=1\mathrm{and}\left|\overrightarrow{\mathrm{b}}\right|=1\right) \\ =\pm 1 \end{gathered}$$