Question

To find the angle between vectors \(a\) and \(b\) vectors.

Short answer

The angle between vectors \({\bf{a}}\) and \({\bf{b}}\) is \({63^\circ }\).

Step-by-step solution

Concept of Dot Product

Formula Used:

Write the expression to find\({\rm{u}} \cdot {\rm{v}}\)in terms of\(\theta \).

\({\rm{u}} \cdot {\rm{v}} = |{\rm{u}}||{\rm{v}}|\cos \theta \)

Here,

\(|u|\)is the magnitude of vector\(u\),

\(|{\rm{v}}|\)is the magnitude of vector\(v\), and

\(\theta \)is the angle between vectors \(u\) and \(v\).a

Calculation for the dot product\(a \cdot b\)

Write a general expression to find the dot product between two three-dimensional vectors.

\(\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= \left\langle {{a_1},{a_2},{a_3}} \right\rangle \cdot \left\langle {{b_1},{b_2},{b_3}} \right\rangle \\{\rm{a}} \cdot {\rm{b}} &= {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}\end{aligned}\)

Rearrange the formula,

\(\begin{aligned}{l}\cos \theta &= \frac{{{\rm{a}} \cdot {\rm{b}}}}{{|{\rm{a}}||{\rm{b}}|}}\\\theta &= {\cos ^{ - 1}}\left( {\frac{{{\rm{a}} \cdot {\rm{b}}}}{{|{\rm{a}}||{\rm{b}}|}}} \right)\end{aligned}\)

Consider a general expression to find the magnitude of a two-dimensional vector that is \(a = \left\langle {{a_1},{a_2}} \right\rangle \).

\(|{\rm{a}}| = \sqrt {a_1^2 + a_2^2} \)

Similarly, consider a general expression to find the magnitude of a two- dimensional vector that is

\({\rm{b}} = \left\langle {{b_1},{b_2}} \right\rangle .\)

\(|{\rm{b}}| = \sqrt {b_1^2 + b_2^2} \)

In formula, substitute\(4\)for\({a_1},3\)for \({a_2},2\) for \({b_1}\) and \( - 1\) for \({b_2}\).

\(\begin{aligned}{l}a \cdot b &= (4)(2) + (3)( - 1)\\a \cdot b &= 8 - 3\\a \cdot b &= 5\end{aligned}\)

Calculation for the angle between vectors \(a\)and \(b\)

In equation of\(|{\rm{a}}|\), substitute \(4\) for \({a_1}\) and \(3\) for \({a_2}\).

\(\begin{aligned}{l}|{\rm{a}}| &= \sqrt {{4^2} + {3^2}} \\|{\rm{a}}| &= \sqrt {16 + 9} \\|{\rm{a}}| &= \sqrt {25} \\|{\rm{a}}| &= 5\end{aligned}\)

In equation of\(|b|\), substitute\(2\)\({b_1}\)and \( - 1\) for \({b_2}\).

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

In equation of\(\theta \),substitute\(5\)for\({\rm{a}} \cdot {\rm{b}},5\)for \(|{\rm{a}}|\), and \(\sqrt 5 \) for \(|{\rm{b}}|\).

\(\begin{aligned}{l}\theta &= {\cos ^{ - 1}}\left( {\frac{5}{{5\sqrt 5 }}} \right)\\\theta &= {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 5 }}} \right)\\\theta &= {63^\circ }\end{aligned}\)

Thus, the angle between vectors \(a\)and \(b\) is \({63^\circ }\).