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Found in: Page 556

### Essential Calculus: Early Transcendentals

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 $$- 15$$

See the 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 $$a \cdot b = - 15$$.