Question

For any two vectors $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$, prove that ${\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2=\left|\begin{array}{ccc}\overrightarrow{a}.\overrightarrow{a}& & \overrightarrow{a}.\overrightarrow{b}\\ \overrightarrow{b}.\overrightarrow{a}& & \overrightarrow{b}.\overrightarrow{b}\end{array}\right|$.

Answer 1

$$\begin{gathered} \text{RHS=}\left|\begin{array}{cc}\overrightarrow{a}.\overrightarrow{a}& \overrightarrow{a}.\overrightarrow{b}\\ \overrightarrow{b}.\overrightarrow{a}& \overrightarrow{b}.\overrightarrow{b}\end{array}\right| \\ \text{=}\left|\begin{array}{cc}{\left|\overrightarrow{a}\right|}^2& \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \\ \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta & {\left|\overrightarrow{b}\right|}^2\end{array}\right| \\ ={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2-{\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2{\cos }^2\theta \\ ={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2\left(1-{\cos }^2\theta \right) \\ ={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2{\sin }^2\theta \\ ={\left(\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin \theta \right)}^2 \\ ={\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2 \\ =\text{LHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$