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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 \( - 15\)

See the step by step solution

Step by Step Solution

Step 1: Concept of Dot Product

Formula used:

The dot product of two vectors \({\rm{a}}\) and \({\rm{b}}\) in terms of \(\theta \) is \({\rm{a}} \cdot {\rm{b}} = |{\rm{a}}||{\rm{b}}|\cos \theta \).

Step 2: Calculation of the dot product

The given magnitude of two vectors are \(|{\rm{a}}| = 6\) and \(|{\rm{b}}| = 5\) with angle \(\frac{{2\pi }}{3}\) between them.

Substitute \(|{\rm{a}}| = 6,|\;{\rm{b}}| = 5\), and \(\theta = \frac{{2\pi }}{3}\) in above formula and obtain the dot product as follows.

\(\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= (6)(5)\cos \left( {\frac{{2\pi }}{3}} \right)\\{\rm{a}} \cdot {\rm{b}} &= 30( - 0.5)\\{\rm{a}} \cdot {\rm{b}} &= - 15\end{aligned}\)

Thus, the dot product between two vector \({\rm{a}}\) and \({\rm{b}}\) is \(a \cdot b = - 15\).

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