Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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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.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 \]

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.

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

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.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

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

Step by Step Solution

Step 1: Formula to express \(a \cdot b\] in terms of \(\theta \]

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

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

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

Step by Step Solution

Step 1: Formula to express \(a \cdot b\] in terms of \(\theta \]

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

Most popular questions for Math Textbooks

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