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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 find a dot product between \({\rm{a}}\) and \({\rm{b}}\).

The dot product \(a \cdot b\) is \(7\)

See the step by step solution

Step by Step Solution

Step 1: Concept of the Dot Product between two three- dimensional vectors

The 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 of the dot product

The given two vectors are,

\(\begin{aligned}{l}a &= 3i + 2j - k\\b &= 4i + 5k\end{aligned}\)

Consider vectors \({\bf{a}}\) and \({\bf{b}}\) as follows.

\(\begin{aligned}{l}{\rm{a}} &= {a_1}{\rm{i}} + {a_2}{\rm{j}} + {a_3}{\rm{k}}\\{\rm{b}} &= {b_1}{\rm{i}} + {b_2}{\rm{j}} + {b_3}{\rm{k}}\end{aligned}\)

Compare the equations of\(a\)

\(\left\langle {{a_1},{a_2},{a_3}} \right\rangle = \langle 3,2, - 1\rangle \)

Compare the equations of\(b\)

\(\left\langle {{b_1},{b_2},{b_3}} \right\rangle = \langle 4,0,5\rangle \)

Substitute \(3\)for \({a_1},2\) for \({a_2}, - 1\) for \({a_3},4\) for \({b_1},0\) for \({b_2}\) and \(5\) for \({b_3}\)in equation of\(a \cdot b\)

\(\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= (3)(4) + (2)(0) + ( - 1)(5)\\{\rm{a}} \cdot {\rm{b}} &= 12 + 0 - 5\\{\rm{a}} \cdot {\rm{b}} &= 7\end{aligned}\)

Thus, \(a \cdot b\) is \(7.\)

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