Question

Prove the property\((ca) \times b = c(a \times b) = a \times (cb)\).

Short answer

The property \((ca) \times b = c(a \times b) = a \times (cb)\) is proved.

Step-by-step solution

Formula used

Consider,\({\rm{a}} = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \)and\({\rm{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \).

Constant vector multiplication rule states that\(c{\rm{a}} = \left\langle {c{a_1},c{a_2},c{a_3}} \right\rangle \)

Calculate the cross product \(c{\bf{a}}\)and \({\bf{b}}\)

Let,

\(\begin{array}{l}{\bf{a}} = {a_1}{\bf{i}} + {a_2}{\bf{j}} + {a_3}{\bf{k}}\\{\bf{b}} = {b_1}{\bf{i}} + {b_2}{\bf{j}} + {b_3}{\bf{k}}\end{array}\)

Then,

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

Using property of determinants, we can write

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

Using property of determinants, we can write

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

Therefore, we have

\((c{\bf{a}}) \times {\bf{b}} = c({\bf{a}} \times {\bf{b}}) = {\bf{a}} \times (c{\bf{b}})\)

Hence, the property\((ca) \times b = c(a \times b) = a \times (cb)\)is proved.