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Expert-verified 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 $$7$$

See the 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 $$7.$$ ### Want to see more solutions like these? 