Short Answer

Prove the property\(a \times (b + c) = a \times b + a \times c\).

The property \({\bf{a}} \times ({\bf{b}} + {\bf{c}}) = {\bf{a}} \times {\bf{b}} + {\bf{a}} \times {\bf{c}}\) is proved.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Formula used

Consider, \({\rm{a}} = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \), \({\rm{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \)and \({\rm{c}} = \left\langle {{c_1},{c_2},{c_3}} \right\rangle \)

Step 2: Calculate the cross product of \({\bf{a}} \times ({\bf{b}} + {\bf{c}})\)

Let,

\(\begin{array}{l}{\bf{a}} = {a_1}{\bf{i}} + {a_2}{\bf{j}} + {a_3}{\bf{k}}\\{\bf{b}} = {b_1}{\bf{i}} + {b_2}{\bf{j}} + {b_3}{\bf{k}}\\{\bf{c}} = {c_1}{\bf{i}} + {c_2}{\bf{j}} + {c_3}{\bf{k}}\end{array}\)

\({\bf{a}} \times ({\bf{b}} + {\bf{c}}) = \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1} + {c_1}}&{{b_2} + {c_2}}&{{b_3} + {c_3}}\end{array}} \right|\)

Using property of determinants, also written as

\( = \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right| = {\bf{a}} \times {\bf{b}} + {\bf{a}} \times {\bf{c}}\)

Hence, proved that the property,\({\bf{a}} \times ({\bf{b}} + {\bf{c}}) = {\bf{a}} \times {\bf{b}} + {\bf{a}} \times {\bf{c}}\).

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Essential Calculus: Early Transcendentals

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Short Answer

Prove the property\(a \times (b + c) = a \times b + a \times c\).

Step by Step Solution

Step 1: Formula used

Step 2: Calculate the cross product of \({\bf{a}} \times ({\bf{b}} + {\bf{c}})\)

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Prove the property\(a \times (b + c) = a \times b + a \times c\).

Step by Step Solution

Step 1: Formula used

Step 2: Calculate the cross product of \({\bf{a}} \times ({\bf{b}} + {\bf{c}})\)

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