91Ó°ÊÓ

The equation \({\bf{y}} = \widehat {\bf{y}} + {\bf{z}}\) is satisfied with \({\bf{z}}\) orthogonal to \({\bf{u}}\) such that if \(\alpha = \frac{{{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}\), and \(\widehat {\bf{y}} = \frac{{{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\). Then, the vector \(\widehat {\mathop{\rm y}\nolimits} \) is known as theorthogonal projection of y onto uand, the vector \({\bf{z}}\) is known as thecomponentof \({\mathop{\rm y}\nolimits} \) orthogonal to \({\bf{u}}\).

Typically, \(\widehat {\bf{y}}\) is represented by \({{\mathop{\rm proj}\nolimits} _L}{\bf{y}}\) and is called the orthogonal projection of \({\mathop{\rm y}\nolimits} \) onto L. Then, \(\widehat {\bf{y}} = {{\mathop{\rm proj}\nolimits} _L}{\bf{y}} = \frac{{{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\).

Compute \({\bf{y}} \cdot {\bf{u}}\) and \({\bf{u}} \cdot {\bf{u}}\) as shown below:

\(\begin{array}{c}{\bf{y}} \cdot {\bf{u}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\9\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right)\\ = - 3 + 18\\ = 15\end{array}\)

And,

\(\begin{array}{c}{\bf{u}} \cdot {\bf{u}} = \left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right)\\ = 1 + 4\\ = 5\end{array}\)

Compute \({\bf{y}} - \widehat {\bf{y}}\) as shown below:

\(\begin{array}{c}{\bf{y}} - \widehat {\bf{y}} = \frac{{{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\\ = \left( {\begin{array}{*{20}{c}}{ - 3}\\9\end{array}} \right) - \frac{{15}}{5}\left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 3}\\9\end{array}} \right) - 3\left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 3}\\9\end{array}} \right) - \left( {\begin{array}{*{20}{c}}3\\6\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 6}\\3\end{array}} \right)\end{array}\)

Therefore, the component of \({\bf{y}}\) orthogonal to \({\bf{u}}\) is \({\bf{y}} - \widehat {\bf{y}} = \left( {\begin{array}{*{20}{c}}{ - 6}\\3\end{array}} \right)\).

The distance from y to the line through u and the origin is the length of the perpendicular line segment from y to the orthogonal projection \(\widehat {\bf{y}}\), that is \(\left\| {{\bf{y}} - \widehat {\bf{y}}} \right\|\).

Compute \(\left\| {{\bf{y}} - \widehat {\bf{y}}} \right\|\) as shown below:

\(\begin{array}{c}\left\| {{\bf{y}} - \widehat {\bf{y}}} \right\| = \sqrt {{{\left( { - 6} \right)}^2} + {{\left( 3 \right)}^2}} \\ = \sqrt {\left( {36} \right) + \left( 9 \right)} \\ = \sqrt {45} \\ = 3\sqrt 5 \end{array}\)

Thus, the distance from \({\bf{y}}\) to the line through u and the origin is \(3\sqrt 5 \) units.