Question

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

Short answer

The angle between vectors \(a\) and \(b\) is \({52^\circ }\).

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\(v\).

Calculation for the dot product \({\rm{a}}{\rm{.b}}\)

The given vectors are,

\(\begin{aligned}{l}a &= 4i - 3j + k\\b &= 2i - k\end{aligned}\)

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

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

In above equation, substitute \(4\) for \({a_1}, - 3\) for \({a_2},1\) for \({a_3},2\) for \({b_1},0\) for \({b_2}\) and \(1\) for \({b_3}\).

\(\begin{aligned}{l}a \cdot b &= (4)(2) + ( - 3)(0) + (1)( - 1)\\a \cdot b &= 8 - 0 - 1\\a \cdot b &= 7\end{aligned}\)

Calculation of the magnitude of three dimensional vector \({\rm{a}}\)and \({\rm{b}}\)

Consider a general expression to find the magnitude of a three-dimensional vector that is\(a = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \).

\(|{\rm{a}}| = \sqrt {a_1^2 + a_2^2 + a_3^2} \)

In the above equation, substitute \(4\) for \({a_1}, - 3\) for \({a_2}\) and \(1\) for \({a_3}\).

\(\begin{aligned}{l}|a| &= \sqrt {{{(4)}^2} + {{( - 3)}^2} + {{(1)}^2}} \\|a| &= \sqrt {16 + 9 + 1} \\|a| &= \sqrt {26} \end{aligned}\)

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

\({\rm{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \)

\(|{\rm{b}}| = \sqrt {b_1^2 + b_2^2 + b_3^2} \)

In the above equation, substitute \(2\) for \({b_1},0\) for \({b_2}\) and \( - 1\) for \({b_3}\).

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