Question: Show that orthogonal projection of a vector y onto a line L through the origin in \({\mathbb{R}^{\bf{2}}}\) does not depend on the choice of the nonzero u in L used in the formula for \({\bf{\hat y}}\). To do this, suppose y and u are given and \({\bf{\hat y}}\) has been computed by formula (2) in this section. Replace u in that formula by \(c{\bf{u}}\), where c is an unspecified nonzero scalar. Show that the new formula gives the same \({\bf{\hat y}}\).

The orthogonal projection of yonto the line L can be expressed as:

\(\widehat {\bf{y}} = \frac{{{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\)

The formula \(\widehat {\bf{y}} = \frac{{{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\) becomes:

\(\begin{array}{c}\widehat {\bf{y}} = \frac{{{\bf{y}} \cdot \left( {c{\bf{u}}} \right)}}{{\left( {c{\bf{u}}} \right) \cdot \left( {c{\bf{u}}} \right)}}\left( {c{\bf{u}}} \right)\\ = \frac{{c\left( {{\bf{y}} \cdot {\bf{u}}} \right)}}{{{c^2}\left( {{\bf{u}} \cdot {\bf{u}}} \right)}}\left( c \right){\bf{u}}\\ = \widehat {\bf{y}}\end{array}\)

Therefore, y does not depend on the choice of nonzero u in L.

Hence, it is proved that the new formula gives the same \(\widehat {\bf{y}}\).