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Q16E

Expert-verifiedFound in: Page 556

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 \(\theta \approx {37^\circ }\).

**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 **\({\rm{a}} \cdot {\rm{b}} = |{\rm{a}}||{\rm{b}}|\cos \theta \)**.**

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.

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

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\(\left( 1 \right).\)

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

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