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Q9E

Expert-verifiedFound in: Page 556

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

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

The dot product \(a \cdot b\) is \( - 15\)

**Formula used:**

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

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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