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Q13E

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Essential Calculus: Early Transcendentals
Found in: Page 556
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

To show

(a) \({\rm{i}} \cdot {\rm{j}} = 0,{\rm{j}} \cdot {\rm{k}} = 0\) and \({\rm{k}} \cdot {\rm{i}} = 0\).

(b) \({\rm{i}} \cdot {\rm{i}} = 1,{\rm{j}} \cdot {\rm{j}} = 1\) and \({\rm{k}} \cdot {\rm{k}} = 1\)

(a)\({\rm{i}} \cdot {\rm{j}} = 0,{\rm{j}} \cdot {\rm{k}} = 0\) and \({\rm{k}} \cdot {\rm{i}} = 0\)is shown.

(b) \({\rm{i}} \cdot {\rm{i}} = 1,{\rm{j}} \cdot {\rm{j}} = 1\) and \({\rm{k}} \cdot {\rm{k}} = 1\)is shown.

See the step by step solution

Step by Step Solution

Step1: Concept of Dot Product

Formula:

Write a general expression to find the dot product between two three-dimensional vectors.

\(\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= \left\langle {{a_1},{a_2},{a_3}} \right\rangle \cdot \left\langle {{b_1},{b_2},{b_3}} \right\rangle \\{\rm{a}} \cdot {\rm{b}} &= {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}\end{aligned}\)

Step 2: Calculation to represent vectors in three- dimensional form

Represent \(i,j,k\)vectors in three-dimensional form.

\(\begin{aligned}{l}{\bf{i}} &= 1{\rm{i}} + 0{\rm{j}} + 0{\rm{k}}\\{\bf{i}} &= \langle 1,0,0\rangle \\{\rm{j}} &= 0{\rm{i}} + 1{\rm{j}} + 0{\rm{k}}\\{\rm{j}} &= \langle 0,1,0\rangle \\{\rm{k}} &= 0{\rm{i}} + 0{\rm{j}} + 1{\rm{k}}\\{\rm{k}} &= \langle 0,0,1\rangle \end{aligned}\)

Step 3: Calculation for dot product of\({\rm{i}} \cdot {\rm{j}},{\rm{j}} \cdot {\rm{k}}\)and\({\rm{k}} \cdot {\rm{i}}\)

Find the dot product of \({\bf{i}}\) and \({\bf{j}}\) using formula.

\(\begin{aligned}{l}{\rm{i}} \cdot {\rm{j}} &= \langle 1,0,0\rangle \cdot \langle 0,1,0\rangle \\{\rm{i}} \cdot {\rm{j}} &= (1)(0) + (0)(1) + (0)(0)\\{\rm{i}} \cdot {\rm{j}} &= 0 + 0 + 0\\{\rm{i}} \cdot {\rm{j}} &= 0\end{aligned}\)

Find the dot product of \({\bf{j}}\) and \({\bf{k}}\) using formula.

\(\begin{aligned}{l}{\rm{j}} \cdot {\rm{k}} &= \langle 0,1,0\rangle \cdot \langle 0,0,1\rangle \\{\rm{j}} \cdot {\rm{k}} &= (0)(0) + (1)(0) + (0)(1)\\{\rm{j}} \cdot {\rm{k}} &= 0 + 0 + 0\\{\rm{j}} \cdot {\rm{k}} &= 0\end{aligned}\)

Find the dot product of \({\bf{k}}\) and \({\bf{i}}\) using formula.

\(\begin{aligned}{l}{\rm{k}} \cdot {\rm{i}} &= \langle 0,0,1\rangle \cdot \langle 1,0,0\rangle \\{\rm{k}} \cdot {\rm{i}} &= (0)(1) + (0)(0) + (1)(0)\\{\rm{k}} \cdot {\rm{i}} &= 0 + 0 + 0\\{\rm{k}} \cdot {\rm{i}} &= 0\end{aligned}\)

Thus, \({\rm{i}} \cdot {\rm{j}} = {\rm{j}} \cdot {\rm{k}} = {\rm{k}} \cdot {\rm{i}} = 0\) is shown.

Step 4: Calculation for dot product of\({\rm{i}} \cdot {\rm{i}},{\rm{j}} \cdot {\rm{j}}\)and\({\rm{k}} \cdot {\rm{k}}\)

Find the dot product of \({\bf{i}}\) and \({\bf{i}}\) using formula.

\(\begin{aligned}{l}{\rm{i}} \cdot {\rm{i}} &= \langle 1,0,0\rangle \cdot \langle 1,0,0\rangle \\{\rm{i}} \cdot {\rm{i}} &= (1)(1) + (0)(0) + (0)(0)\\{\rm{i}} \cdot {\rm{i}} &= 1 + 0 + 0\\{\rm{i}} \cdot {\rm{i}} &= 1\end{aligned}\)

Find the dot product of \({\bf{j}}\) and \({\bf{j}}\) using formula.

\(\begin{aligned}{l}{\rm{j}} \cdot {\rm{j}} &= \langle 0,1,0\rangle \cdot \langle 0,1,0\rangle \\{\rm{j}} \cdot {\rm{j}} &= (0)(0) + (1)(1) + (0)(0)\\{\rm{j}} \cdot {\rm{j}} &= 0 + 1 + 0\\{\rm{j}} \cdot {\rm{j}} &= 1\end{aligned}\)

Find the dot product of \({\bf{k}}\) and \({\bf{k}}\) using formula.

\(\begin{aligned}{l}{\rm{k}} \cdot {\rm{k}} &= \langle 0,0,1\rangle \cdot \langle 0,0,1\rangle \\{\rm{k}} \cdot {\rm{k}} &= (0)(0) + (0)(0) + (1)(1)\\{\rm{k}} \cdot {\rm{k}} &= 0 + 0 + 1\\{\rm{k}} \cdot {\rm{k}} &= 1\end{aligned}\)

Thus, \({\rm{i}} \cdot {\rm{i}} = {\rm{j}} \cdot {\rm{j}} = {\rm{k}} \cdot {\rm{k}} = 1\) is shown.

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