Question

Show that\(c\) bisects the angle between\(a\) and\(b.\)\({\rm{c = |a|b + |b|a}}\).

Short answer

The \(c\) bisects the angle between\(a\) and\(b.\)is shown below.

Step-by-step solution

Formula used.

Write the expression to find\(a \cdot b\)in terms of\(\theta .\)

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

Here,

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

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

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

Given\({\rm{c = |a| b + |b| a}}\)

Rearrange the equation and use it for calculation.

Rearrange equation.

\(\cos \theta = \frac{{{\rm{a}} \cdot {\rm{b}}}}{{|{\rm{a}}||{\rm{b}}|}}\,\,\,\,\,\,\,\,\,\,......\left( 1 \right)\)

Consider\(\alpha \)be the angle between\(a\)and\(c\)vectors and\(\beta \) be the angle between\(c\)and\(b.\)vectors.

Rearrange equation (1) for angle\(\alpha \).

\(\cos \alpha = \frac{{{\rm{a}} \cdot {\rm{c}}}}{{|{\rm{a}}||{\rm{c}}|}}\)

Substitute\(|{\rm{a}}|{\rm{b}} + |{\rm{b}}|{\rm{a for c}}\).

\(\begin{aligned}{c}\cos \alpha &= \frac{{a \cdot (|a|b + |b|a)}}{{|a||c|}}\\ = \frac{{a \cdot |a|b + a \cdot |b|a}}{{|a||c|}}\\ &= \frac{{a|a| \cdot b + {{\left. {|a|} \right|}^2}|b|}}{{|a||c|}}\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 2 \right)\\ &= \frac{{|a|(a \cdot b + |a||b|)}}{{|a||c|}}\\\cos \alpha = \frac{{a \cdot b + |a||b|}}{{|c|}}\end{aligned}\)

Rearrange equation (1) for angle\(\beta \).

\(\cos \beta = \frac{{{\rm{b}} - {\rm{c}}}}{{|{\rm{b}}||{\rm{c}}|}}\)

Substitute\(|{\rm{a}}|{\rm{b}} + |{\rm{b}}|{\rm{a for c}}\).

\(\begin{aligned}{c}\cos \beta &= \frac{{{\rm{b}} \cdot (|{\rm{a}}|{\rm{b}} + |{\rm{b}}|{\rm{a}})}}{{|{\rm{b}}||{\rm{c}}|}}\\ &= \frac{{{\rm{b}} \cdot |{\rm{a}}|{\rm{b}} + {\rm{b}} \cdot |{\rm{b}}|{\rm{a}}}}{{|{\rm{b}}||{\rm{c}}|}}\\ &= \frac{{|{\rm{b}}{|^2}|{\rm{a}}| + {\rm{b}}|{\rm{b}}| \cdot {\rm{a}}}}{{|{\rm{b}}||{\rm{c}}|}}\,\,\,\,\,\,\,\,\,\,.......\left( 3 \right)\\ &= \frac{{|{\rm{b}}|(|{\rm{a}}||{\rm{b}}| + {\rm{b}} \cdot {\rm{a}})}}{{|{\rm{b}}||{\rm{c}}|}}\\\cos \beta &= \frac{{|{\rm{a}}||{\rm{b}}| + {\rm{b}} - {\rm{a}}}}{{|{\rm{c}}|}}\end{aligned}\)

Equate equation (1) and equation (2).

\(\begin{aligned}{l}\cos \alpha &= \cos \beta \\\alpha &= \beta \end{aligned}\)

However the angle\(\alpha \) and\(\beta \) lies from\({0^\circ } \le \alpha \le {180^\circ }\) and\({0^\circ } \le \beta \le {180^\circ },\) so the\(\alpha = \beta \) and the\(c\)bisects the angle between\(a\) and\(b.\)

Thus, the\(c\)bisects the angle between\(a\) and\(b.\)is shown.