Question

If \({\rm{u = }}\left\langle {{{\rm{u}}_{\rm{1}}}{\rm{,}}{{\rm{u}}_{\rm{2}}}} \right\rangle \) and \({\rm{v = }}\left\langle {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}} \right\rangle \), then \({\rm{u}} \cdot {\rm{v = }}\left\langle {{{\rm{u}}_{\rm{1}}}{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{u}}_{\rm{2}}}{{\rm{v}}_{\rm{2}}}} \right\rangle \)

Short answer

The statement is false.

Step-by-step solution

Determine whether the dot product is a scalar or a vector.

The dot product (marked between two vectors by the "." sign) is defined as the sum of the products of the corresponding components, which results in a scalar.

\({\rm{u}} \cdot {\rm{v = }}{{\rm{u}}_{\rm{1}}}{{\rm{v}}_{\rm{1}}}{\rm{ + }}{{\rm{u}}_{\rm{2}}}{{\rm{v}}_{\rm{2}}}\)

\({\rm{ < }}{{\rm{u}}_{\rm{1}}}{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{u}}_{\rm{2}}}{{\rm{v}}_{\rm{2}}}{\rm{ > }}\) is a vector.

Result.

Therefore, the statement is false.