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Q16E

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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 the angle between vectors $$a$$ and $$b$$ vectors.

The exact angle between two vectors a and $${\rm{b}}$$ is $$\theta = {\cos ^{ - 1}}\left( {\frac{4}{5}} \right)$$.

The approximate angle between two vectors a and $${\rm{b}}$$ is $$\theta \approx {37^\circ }$$.

See the step by step solution

## Step 1: Concept of Dot Product of two vectors

Formula used:

1. The magnitude of a vector $${\bf{a}} = \left\langle {{a_1},{a_2},{a_3}} \right\rangle$$ is $$|{\rm{a}}| = \sqrt {a_1^2 + a_2^2 + + a_3^2}$$.

2. 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 of two vector $$a \cdot b$$

The given two vectors are $${\rm{a}} = \langle 4,0,2\rangle$$ and $${\rm{b}} = \langle 2, - 1,0\rangle$$.

Consider a general expression to find the dot product between 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 the value of components $${a_1} = 4,{a_2} = 0,{a_3} = 2,{b_1} = 2,{b_2} = - 1$$, and $${b_3} = 0$$ in the above result.

\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= (4)(2) + (0)( - 1) + (2)(0)\\{\rm{a}} \cdot {\rm{b}} &= 8 + 0 + 0\\{\rm{a}} \cdot {\rm{b}} &= 8\end{aligned}

## Step 3: Calculation of the magnitude of vector $$a$$and $$b$$

Calculate the magnitude of the vector a by Substituting $${a_1} = 4,{a_2} = 0$$, and $${a_3} = 2$$ in the formula $$\left( 1 \right).$$

\begin{aligned}{l}|{\rm{a}}| &= \sqrt {{{(4)}^2} + {0^2} + {2^2}} \\|{\rm{a}}| &= \sqrt {16 + 4} \\|{\rm{a}}| &= \sqrt {20} \\|{\rm{a}}| &= 2\sqrt 5 \end{aligned}

Calculate the magnitude of the vector b by Substituting $${b_1} = 2,{b_2} = - 1$$, and $${b_3} = 0$$ in the formula$$\left( 1 \right).$$

\begin{aligned}{l}|{\rm{b}}| &= \sqrt {{2^2} + {{( - 1)}^2} + {0^2}} \\|{\rm{b}}| &= \sqrt {4 + 1} \\|{\rm{b}}| &= \sqrt 5 \end{aligned}