Question

Determine the cross-product between\(a\)and\(b\)and verify\(a \times b\)is orthogonal to both\(a\)and\(b\).

Short answer

\(a \times b\) is \((1 - t){\rm{i}} + \left( {{t^3} - {t^2}} \right){\rm{k}}\)And it is orthogonal to both\(a\) and\(b.\)

Step-by-step solution

The cross-product formula.

Consider the given expression to find the cross-product of\(a\)and\(b.\)

\({\rm{a}} \times {\rm{b}} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|\) …… (1)

Use the cross-product formula for calculation.

Consider to verify\(a \times b\) is orthogonal to\(a.\)

\((a \times b) \cdot a = 0\)

Consider to verify\(a \times b\)is orthogonal to\(b.\)
\((a \times b) \cdot b = 0\)

In equation (1), substitute\(t{\rm{ for }}{a_1},\cos t{\rm{ for }}{a_2},\sin t{\rm{ for }}{a_3},1{\rm{ for }}{b_1}, - \sin t{\rm{ for }}{b_2}\)and\(\cos t{\rm{ for }}{b_3}{\rm{. }}\)

\(\begin{array}{l}{\rm{a}} \times {\rm{b}} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\t&1&{\frac{1}{t}}\\{{t^2}}&{{t^2}}&1\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}1&{\frac{1}{t}}\\{{t^2}}&1\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{c}}t&{\frac{1}{t}}\\{{t^2}}&1\end{array}} \right|{\rm{j}} + \left| {\begin{array}{*{20}{c}}t&1\\{{t^2}}&{{t^2}}\end{array}} \right|{\rm{k}}\\ = (1 - t){\rm{i}} - (t - t){\rm{j}} + \left( {{t^3} - {t^2}} \right){\rm{k}}\\ = (1 - t){\rm{i}} + \left( {{t^3} - {t^2}} \right){\rm{k}}\end{array}\)

Thus, the\(a \times b\) is\((1 - t){\rm{i}} + \left( {{t^3} - {t^2}} \right){\rm{k}}\)

For \((a \times b) \cdot a\).

Substitute\((1 - t){\rm{i}} + \left( {{t^3} - {t^2}} \right){\rm{k for a}} \times {\rm{b}}\) and\(\left\langle {t,1,\frac{1}{t}} \right\rangle {\rm{ for a}}\)

\(\begin{aligned}{l}({\rm{a}} \times {\rm{b}}) \cdot {\rm{b}} = \left( {(1 - t){\rm{i}} + \left( {{t^3} - {t^2}} \right){\rm{k}}} \right) \cdot \left\langle {{t^2},{t^2},1} \right\rangle \\\begin{array}{*{20}{c}}{ = \left( {(1 - t){\rm{i}} + 0{\rm{j}} + \left( {{t^3} - {t^2}} \right){\rm{k}}} \right) \cdot \left( {{t^2}{\rm{i}} + {t^2}{\rm{j}} + {\rm{k}}} \right)}\\{ = {t^2} - {t^3} + 0 + {t^3} - {t^2}}\\{ = 0}\end{array}\\\end{aligned}\)

For\((a \times b) \cdot b\)

Substitute\((1 - t){\rm{i}} + \left( {{t^3} - {t^2}} \right){\rm{k for a}} \times {\rm{b}}\) and\(\left\langle {{t^2},{t^2},1} \right\rangle {\rm{ for b}}{\rm{. }}\).

\(\begin{array}{l}({\rm{a}} \times {\rm{b}}) \cdot {\rm{b}} = \left( {(1 - t){\rm{i}} + \left( {{t^3} - {t^2}} \right){\rm{k}}} \right) \cdot \left\langle {{t^2},{t^2},1} \right\rangle \\ = \left( {(1 - t){\rm{i}} + 0{\rm{j}} + \left( {{t^3} - {t^2}} \right){\rm{k}}} \right) \cdot \left( {{t^2}{\rm{i}} + {t^2}{\rm{j}} + {\rm{k}}} \right)\\ = {t^2} - {t^3} + 0 + {t^3} - {t^2}\\ = 0\end{array}\)

\((a \times b) \cdot b = 0\)

As\((a \times b) \cdot a\) and\((a \times b) \cdot b\) is equal to\(0\), it satisfies the condition of orthogonal.

Thus, the\(a \times b\)is orthogonal to both\(a\) and\(b.\)