Mathematics - Vector Question with Solution | TestHub

$\because \left[\begin{array}{lll}\overrightarrow{\mathrm{a}}& \overrightarrow{\mathrm{b}}& \overrightarrow{\mathrm{c}}\end{array}\right]=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}|\sin \theta \cos \varphi ,\theta =\overrightarrow{\mathrm{a}}\wedge \overrightarrow{\mathrm{b}},\varphi =(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}{)}^{\wedge }\overrightarrow{\mathrm{c}}$

$$\begin{array}{rlrl}& =|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\sin \theta \cos \varphi & \le |\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\sin \theta & \le |\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|\end{array}$$

∵ angle between$\overrightarrow{\mathrm{c}}$ and$\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}$ is 0$\Rightarrow \overrightarrow{\mathrm{c}}=\lambda (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}),\lambda >0$ (right handed system)$\because \overrightarrow{\mathrm{c}}$ is unit vector

$$\begin{array}{rlrl}& \therefore \vec{c}=\frac{\vec{a}\times \vec{b}}{|\vec{a}\times \vec{b}|}& \because {\left[\begin{array}{lll}\vec{a}& \vec{b}& \vec{c}\end{array}\right]}_{\max }=|\vec{a}\times \vec{b}|=3\sqrt{3}=M& \text{ and }\vec{c}=\frac{\vec{a}\times \vec{b}}{|\vec{a}\times \vec{b}|}=\frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}\end{array}$$