Question

For any two vectors a and b, P.T $|a\times b{|}^2=(a.a)(b.)-(a.b{)}^2=a^2{b}^2-(a.b{)}^2$.

Answer 1

$\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left(a\cdot b\right)b-{\left(a\cdot b\right)}^2$

$a^2{b}^2-{\left(a\cdot b\right)}^2$

$\therefore {\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2=\left(\overrightarrow{a}\times \overrightarrow{b}\right)\cdot \left(\overrightarrow{a}\times \overrightarrow{b}\right)$

$=\left(ab\sin \theta \right)\hat{n}\cdot \left(ab\sin \theta \right)\hat{n}$

$\Rightarrow \left(a^2{b}^2\right)\left[1-{\cos }^2\theta \right]=a^2{b}^2-{\left[ab\cos \theta \right]}^2$

$\Rightarrow \left(\overrightarrow{a}\cdot \overrightarrow{a}\right)\left(\overrightarrow{b}\cdot \overrightarrow{b}\right)-{\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)}^2$