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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 $$- pq$$

See the step by step solution

## Step 1: Concept of the Dot Product

The minimum of two vectors are required to perform a dot product. The resultant dot product of two vectors is scalar. hence, the dot product is also known as a scalar product.

## Step 2: Calculation of the dot product

The given two vectors are$${\rm{a}} = \langle p, - p,2p\rangle$$ and $${\rm{b}} = \langle 2q,q, - q\rangle$$

Consider 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}

Substitute $$p$$ for $${a_1}, - p$$ for $${a_2},2p$$ for $${a_3},2q$$ for $${b_1},q$$ for $${b_2}$$ and $$- q$$ for $${b_3}$$.

\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= (p)(2q) + ( - p)(q) + (2p)( - q)\\{\rm{a}} \cdot {\rm{b}} &= 2pq - pq - 2pq\\{\rm{a}} \cdot {\rm{b}} &= - pq\end{aligned}

Thus, $$a \cdot b$$ is $$- pq$$.