Question

If \({\bf{a}} \cdot {\bf{b}} = \sqrt 3 \) and \({\bf{a}} \times {\bf{b}} = \langle 1,2,2\rangle \), find the angle between \({\bf{a}}\) and \({\bf{b}}\).

Short answer

The angle between \({\bf{a}}\) and \({\bf{b}}\) is \(\underline {{{60}^^\circ }} .\)

Step-by-step solution

Formula and concept used to find the angle between a diagonal of a cube and one of its edges

The expression to find\({\rm{a}} \cdot {\rm{b}}\)in terms of\(\theta \).

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

Here,

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

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

\(\theta \)is the angle between vectors\({\bf{a}}\)and\({\bf{b}}\).

Rearrange the equation.

\(|{\rm{a}}||{\rm{b}}| = \frac{{{\rm{a}} \cdot {\rm{b}}}}{{\cos \theta }}\) …… (1)

Write the expression to find\(|{\rm{a}} \times {\rm{b}}|\)in terms of\(\theta \).

\(|{\rm{a}} \times {\rm{b}}| = |{\rm{a}}||{\rm{b}}|\sin \theta \) …… (2)

Calculation to find the angle between \({\bf{a}}\) and \(b\)

Substitute equation (1) in equation (2).

\(\begin{array}{l}|{\rm{a}} \times {\rm{b}}| = \frac{{{\rm{a}} \cdot {\rm{b}}}}{{\cos \theta }}\sin \theta \\ = {\rm{a}} \cdot {\rm{b}}\tan \theta \end{array}\)

Rearrange the equation to find angle between \({\bf{a}}\) and \({\bf{b}}\).

\(\theta = {\tan ^{ - 1}}\left( {\frac{{|a \times b|}}{{a \cdot b}}} \right)\)

Substitute \(\sqrt 3 \) for \(a \cdot b\) and \(\langle 1,2,2\rangle \) for \(a \times b\),

\(\begin{array}{l}\theta = {\tan ^{ - 1}}\left( {\frac{{|(1,2,2)|}}{{\sqrt 3 }}} \right)\\ = {\tan ^{ - 1}}\left( {\frac{{\sqrt {{{(1)}^2} + {{(2)}^2} + {{(2)}^2}} }}{{\sqrt 3 }}} \right)\\ = {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + 4 + 4} }}{{\sqrt 3 }}} \right)\\ = {\tan ^{ - 1}}\left( {\frac{{\sqrt 9 }}{{\sqrt 3 }}} \right)\end{array}\)

Simplify the equation.

\(\begin{array}{l}\theta = {\tan ^{ - 1}}\left( {\frac{3}{{\sqrt 3 }}} \right)\\ = {\tan ^{ - 1}}(\sqrt 3 )\\ = {60^^\circ }\end{array}\)

Thus, the angle between \({\bf{a}}\) and \({\bf{b}}\) is \({60^^\circ }\).