Question

To show: The expression \(|a \times b{|^2} = |a{|^2}|b{|^2} - {(a \cdot b)^2}\).

Short answer

The expression \(|{\rm{a}} \times {\rm{b}}{|^2} = |{\rm{a}}{|^2}|\;{\rm{b}}{|^2} - {({\rm{a}} \cdot {\rm{b}})^2}\) is proved.

Step-by-step solution

Formula to express  \(a \cdot b\] in terms of \(\theta \]

The expression to find\({\rm{a}} \cdot {\rm{b}}\]in terms of\(\theta \].

\({\rm{a}} \cdot {\rm{b}} = |{\rm{a}}||{\rm{b}}|\cos \theta \]

Here,

\(|a|\]is the magnitude of a vector,

\(|{\rm{b}}|\]is the magnitude of\({\bf{b}}\]vector, and

\(\theta \]is the angle between vectors\({\bf{a}}\]and\({\bf{b}}\].

Rearrange the equation.

\(\cos \theta = \frac{{{\rm{a}} \cdot {\rm{b}}}}{{|{\rm{a}}||{\rm{b}}|}}\)

Calculation to show that the expression \(|a \times b{|^2} = |a{|^2}|b{|^2} - {(a \cdot b)^2}\]

Write the expression to find \(|{\rm{a}} \times {\rm{b}}|\) in terms of \(\theta \) as follows.

\(|{\rm{a}} \times {\rm{b}}| = |{\rm{a}}||{\rm{b}}|\sin \theta \)

Take square on both sides.

Substitute \(\frac{{{\rm{a}} \cdot {\rm{b}}}}{{|{\rm{a}}|{\rm{bb|}}}}\) for \(\cos \theta \),

\(\begin{array}{l}|{\rm{a}} \times {\rm{b}}{|^2} = |{\rm{a}}{|^2}|\;{\rm{b}}{|^2} - {\left( {|{\rm{a}}||{\rm{b}}|\left( {\frac{{{\rm{a}} \cdot {\rm{b}}}}{{|{\rm{a}}||{\rm{b}}|}}} \right)} \right]^2}\\ = |{\rm{a}}{|^2}|\;{\rm{b}}{|^2} - {({\rm{a}} \cdot {\rm{b}})^2}\end{array}\)

Thus, the expression \(|{\rm{a}} \times {\rm{b}}{|^2} = |{\rm{a}}{|^2}|\;{\rm{b}}{|^2} - {({\rm{a}} \cdot {\rm{b}})^2}\) is proved.