Question

Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement

For any vectors \({\rm{u}}\)and\({\rm{v}}\)in \({{\rm{V}}_{\rm{3}}}\) and any scalar \({\rm{k,k(u \times v) = (ku) \times v}}{\rm{.}}\)

Short answer

Therefore, the answer for the given statement isTRUE.

Step-by-step solution

Assumption of Scalar.

Assume\({\rm{u = }}\left\langle {{{\rm{u}}_{\rm{1}}}{\rm{,}}{{\rm{u}}_{\rm{2}}}{\rm{,}}{{\rm{u}}_{\rm{3}}}} \right\rangle {\rm{,v = }}\left\langle {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\rangle {\rm{,}}\)then\({\rm{ku = }}\left\langle {{\rm{k}}{{\rm{u}}_{\rm{1}}}{\rm{,k}}{{\rm{u}}_{\rm{2}}}{\rm{,k}}{{\rm{u}}_{\rm{3}}}} \right\rangle \)and

\({\rm{(ku) \times v = }}\left| {\begin{aligned}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{\rm{k}}{{\rm{u}}_{\rm{1}}}}&{{\rm{k}}{{\rm{u}}_{\rm{2}}}}&{{\rm{k}}{{\rm{u}}_{\rm{3}}}}\\{{{\rm{v}}_{\rm{1}}}}&{{{\rm{v}}_{\rm{2}}}}&{{{\rm{v}}_{\rm{3}}}}\end{aligned}} \right|\)

Getting Left-hand-Side.

If one row in a determinant has a common factor, we are able to divide that row via that factor and locate it outside the determinant.

\(\begin{aligned}{c}{\rm{ = k}}\left| {\begin{aligned}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{{\rm{u}}_{\rm{1}}}}&{{{\rm{u}}_{\rm{2}}}}&{{{\rm{u}}_{\rm{3}}}}\\{{{\rm{v}}_{\rm{1}}}}&{{{\rm{v}}_{\rm{2}}}}&{{{\rm{v}}_{\rm{3}}}}\end{aligned}} \right|\\{\rm{ = k(u \times v)}}\end{aligned}\)

Thus, the given statement is TRUE.