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Found in: Page 565

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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

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

See the step by step solution

## Step 2: 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.