Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Prove the formula \((a \times b) \cdot (c \times d) = \left| {\begin{array}{*{20}{c}}{a \cdot c}&{b \cdot c}\\{a \cdot d}&{b \cdot d}\end{array}} \right|\).

The formula \((a \times b) \cdot (c \times d) = \left| {\begin{array}{*{20}{l}}{a \cdot c}&{b \cdot c}\\{a \cdot d}&{b \cdot d}\end{array}} \right|\) is proved.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Concept of vectors.

A vector is an object that has both a magnitude as well as a direction.

Step 2: Write out \({\bf{a}} \times {\bf{b}}\)and \({\bf{c}} \times {\bf{d}}\) in component form.

\({\bf{a}} \times {\bf{b}} = \left\langle {{a_2}{b_3} - {a_3}{b_2},{a_3}{b_1} - {a_1}{b_3},{a_1}{b_2} - {a_2}{b_1}} \right\rangle \)

\({\bf{c}} \times {\bf{d}} = \left\langle {{c_2}{d_3} - {c_3}{d_2},{c_3}{d_1} - {c_1}{d_3},{c_1}{d_2} - {c_2}{d_1}} \right\rangle \)

\(({\bf{a}} \times {\bf{b}}) \cdot ({\bf{c}} \times {\bf{d}})\)

\( = \left( {{a_2}{b_3} - {a_3}{b_2}} \right)\left( {{c_2}{d_3} - {c_3}{d_2}} \right) + \left( {{a_3}{b_1} - {a_1}{b_3}} \right)\left( {{c_3}{d_1} - {c_1}{d_3}} \right)\)

\(\quad + \left( {{a_1}{b_2} - {a_2}{b_1}} \right)\left( {{c_1}{d_2} - {c_2}{d_1}} \right)\)

\(\begin{array}{l} = {a_2}{b_3}{c_2}{d_3} - {a_2}{b_3}{c_3}{d_2} - {a_3}{b_2}{c_2}{d_3} + {a_3}{b_2}{c_3}{d_2}\\ + {a_3}{b_1}{c_3}{d_1} - {a_3}{b_1}{c_1}{d_3} - {a_1}{b_3}{c_3}{d_1} + {a_1}{b_3}{c_1}{d_3}\\ + {a_1}{b_2}{c_1}{d_2} - {a_1}{b_2}{c_2}{d_1} - {a_2}{b_1}{c_1}{d_2} + {a_2}{b_1}{c_2}{d_1}\end{array}\)

\(\begin{array}{l} = {a_1}{b_2}{c_1}{d_2} + {a_1}{b_3}{c_1}{d_3} + {a_2}{b_1}{c_2}{d_1} + {a_2}{b_3}{c_2}{d_3}\\ + {a_3}{b_1}{c_3}{d_1} + {a_3}{b_2}{c_3}{d_2} - {a_2}{b_1}{c_1}{d_2} - {a_3}{b_1}{c_1}{d_3}\\ - {a_1}{b_2}{c_2}{d_1} - {a_3}{b_2}{c_2}{d_3} - {a_1}{b_3}{c_3}{d_1} - {a_2}{b_3}{c_3}{d_2}\end{array}\)

Step 3: Write the determinant of the right side of the equation in terms of components.

\(\left| {\begin{array}{*{20}{c}}{{\bf{a}} \cdot {\bf{c}}}&{{\bf{b}} \cdot {\bf{c}}}\\{{\bf{a}} \cdot {\bf{d}}}&{{\bf{b}} \cdot {\bf{d}}}\end{array}} \right| = ({\bf{a}} \cdot {\bf{c}})({\bf{b}} \cdot {\bf{d}}) - ({\bf{b}} \cdot {\bf{c}})({\bf{a}} \cdot {\bf{d}})\)

\(\begin{array}{l} = \left( {{a_1}{c_1} + {a_2}{c_2} + {a_3}{c_3}} \right)\left( {{b_1}{d_1} + {b_2}{d_2} + {b_3}{d_3}} \right)\\ - \left( {{b_1}{c_1} + {b_2}{c_2} + {b_3}{c_3}} \right)\left( {{a_1}{d_1} + {a_2}{d_2} + {a_3}{d_3}} \right)\end{array}\)

\(\begin{array}{l} = {a_1}{c_1}{b_1}{d_1} + {a_1}{c_1}{b_2}{d_2} + {a_1}{c_1}{b_3}{d_3}\\ + {a_2}{c_2}{b_1}{d_1} + {a_2}{c_2}{b_2}{d_2} + {a_2}{c_2}{b_3}{d_3}\\ + {a_3}{c_3}{b_1}{d_1} + {a_3}{c_3}{b_2}{d_2} + {a_3}{c_3}{b_3}{d_3}\\ - {b_1}{c_1}{a_1}{d_1} - {b_1}{c_1}{a_2}{d_2} - {b_1}{c_1}{a_3}{d_3}\\ - {b_2}{c_2}{a_1}{d_1} - {b_2}{c_2}{a_2}{d_2} - {b_2}{c_2}{a_3}{d_3}\\ - {b_3}{c_3}{a_1}{d_1} - {b_3}{c_3}{a_2}{d_2} - {b_3}{c_3}{a_3}{d_3}\end{array}\)

Sort into “abcd”

\(\begin{array}{l} = {a_1}{b_1}{c_1}{d_1} + {a_1}{b_2}{c_1}{d_2} + {a_1}{b_3}{c_1}{d_3}\\ + {a_2}{b_1}{c_2}{d_1} + {a_2}{b_2}{c_2}{d_2} + {a_2}{b_3}{c_2}{d_3}\\ + {a_3}{b_1}{c_3}{d_1} + {a_3}{b_2}{c_3}{d_2} + {a_3}{b_3}{c_3}{d_3}\\ - {a_1}{b_1}{d_1}{c_1} - {a_2}{b_1}{c_1}{d_2} - {a_3}{b_1}{c_1}{d_3}\\ - {a_1}{b_2}{c_2}{d_1} - {a_2}{b_2}{c_2}{d_2} - {a_3}{b_2}{c_2}{d_3}\\ - {a_1}{b_3}{c_3}{d_1} - {a_2}{b_3}{c_3}{d_2} - {a_3}{b_3}{c_3}{d_3}\end{array}\)

only the "1111", "2222", "3333" terms cancel out.

\(\begin{array}{l} = {a_1}{b_2}{c_1}{d_2} + {a_1}{b_3}{c_1}{d_3} + {a_2}{b_1}{c_2}{d_1}\\ + {a_2}{b_3}{c_2}{d_3} + {a_3}{b_1}{c_3}{d_1} + {a_3}{b_2}{c_3}{d_2}\\ - {a_2}{b_1}{c_1}{d_2} - {a_3}{b_1}{c_1}{d_3} - {a_1}{b_2}{c_2}{d_1}\\ - {a_3}{b_2}{c_2}{d_3} - {a_1}{b_3}{c_3}{d_1} - {a_2}{b_3}{c_3}{d_2}\end{array}\)

Step 4: Write the determinant of the right side of the equation in terms of components.

\(({\bf{a}} \times {\bf{b}}) \cdot ({\bf{c}} \times {\bf{d}}) = \left| {\begin{array}{*{20}{c}}{{\bf{a}} \cdot {\bf{c}}}&{{\bf{b}} \cdot {\bf{c}}}\\{{\bf{a}} \cdot {\bf{d}}}&{{\bf{b}} \cdot {\bf{d}}}\end{array}} \right|\)

So, it’s true.

Hence, the elements of the determinant are just scalars

\(\left| {\begin{array}{*{20}{l}}{{\bf{a}} \cdot {\bf{c}}}&{{\bf{b}} \cdot {\bf{c}}}\\{{\bf{a}} \cdot {\bf{d}}}&{{\bf{b}} \cdot {\bf{d}}}\end{array}} \right| = ({\bf{a}} \cdot {\bf{c}})({\bf{b}} \cdot {\bf{d}}) - ({\bf{b}} \cdot {\bf{c}})({\bf{a}} \cdot {\bf{d}})\)

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Essential Calculus: Early Transcendentals

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Short Answer

Prove the formula \((a \times b) \cdot (c \times d) = \left| {\begin{array}{*{20}{c}}{a \cdot c}&{b \cdot c}\\{a \cdot d}&{b \cdot d}\end{array}} \right|\).

Step by Step Solution

Step 1: Concept of vectors.

Step 2: Write out \({\bf{a}} \times {\bf{b}}\)and \({\bf{c}} \times {\bf{d}}\) in component form.

Step 3: Write the determinant of the right side of the equation in terms of components.

Step 4: Write the determinant of the right side of the equation in terms of components.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Prove the formula \((a \times b) \cdot (c \times d) = \left| {\begin{array}{*{20}{c}}{a \cdot c}&{b \cdot c}\\{a \cdot d}&{b \cdot d}\end{array}} \right|\).

Step by Step Solution

Step 1: Concept of vectors.

Step 2: Write out \({\bf{a}} \times {\bf{b}}\)and \({\bf{c}} \times {\bf{d}}\) in component form.

Step 3: Write the determinant of the right side of the equation in terms of components.

Step 4: Write the determinant of the right side of the equation in terms of components.

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