Question

$\mathrm{If}\overrightarrow{a}=\hat{i}-\hat{j}\mathrm{and}\overrightarrow{b}=\hat{j}+\hat{k},\mathrm{then}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}.\overrightarrow{b}\right|}^2=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

$$\begin{gathered} \mathrm{For}\overrightarrow{a}=\hat{i}-\hat{j} \\ \overrightarrow{b}=\hat{j}+\hat{k} \\ \left|\overrightarrow{a}\right|=\sqrt{1^2+1^2}=\sqrt{2} \\ \left|\overrightarrow{b}\right|=\sqrt{1^2+1^2}=\sqrt{2} \\ \mathrm{By}\mathrm{lagrange}\text{'}\mathrm{s}\mathrm{identity}, \end{gathered}$$

$\begin{array}{rcl}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}.\overrightarrow{b}\right|}^2& =& {\left|a\right|}^2{\left|b\right|}^2\\ & =& {\left(\sqrt{2}\right)}^2{\left(\sqrt{2}\right)}^2\\ & =& 2\times 2\end{array}$

$\mathrm{i}.\mathrm{e}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}.\overrightarrow{b}\right|}^2=4$