Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


Essential Calculus: Early Transcendentals
Found in: Page 565
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

Answers without the blur.

Just sign up for free and you're in.


Short Answer

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 by Step Solution

Step 1: 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 \]


\(|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.

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

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.

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.