91影视

The vector \(\widehat {\mathop{\rm y}\nolimits} \) is known as theorthogonal projection of y onto uand, the vector z is known as thecomponent of y orthogonal to u. Typically, \(\widehat {\mathop{\rm y}\nolimits} \) is represented by \({{\mathop{\rm proj}\nolimits} _L}{\mathop{\rm y}\nolimits} \) and is called the orthogonal projection of y onto L.

\(\widehat {\mathop{\rm y}\nolimits} = {{\mathop{\rm proj}\nolimits} _L}{\mathop{\rm y}\nolimits} = \frac{{{\mathop{\rm y}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} \)

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 \(\widehat {\mathop{\rm y}\nolimits} \), that is \(\left\| {{\mathop{\rm y}\nolimits} - \widehat {\mathop{\rm y}\nolimits} } \right\|\).

Compute \({\mathop{\rm y}\nolimits} \cdot {\mathop{\rm u}\nolimits} \) and \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} \) as shown below:

\(\begin{array}{c}{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm y}\nolimits} = \left( {\begin{array}{*{20}{c}}8\\6\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}3\\1\end{array}} \right)\\ = 30\\{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} = \left( {\begin{array}{*{20}{c}}8\\6\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}8\\6\end{array}} \right)\\ = 100\end{array}\)

Compute \({\mathop{\rm y}\nolimits} - \widehat {\mathop{\rm y}\nolimits} \) as shown below:

\(\begin{array}{c}{\mathop{\rm y}\nolimits} - \widehat {\mathop{\rm y}\nolimits} = {\mathop{\rm y}\nolimits} - \frac{{{\mathop{\rm y}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} \\ = \left( {\begin{array}{*{20}{c}}3\\1\end{array}} \right) - \frac{{30}}{{100}}\left( {\begin{array}{*{20}{c}}8\\6\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}3\\1\end{array}} \right) - \frac{3}{{10}}\left( {\begin{array}{*{20}{c}}8\\6\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}3\\1\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{\frac{{24}}{{10}}}\\{\frac{{18}}{{10}}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{\frac{6}{{10}}}\\{\frac{{ - 8}}{{10}}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{\frac{3}{5}}\\{ - \frac{4}{5}}\end{array}} \right)\end{array}\)

Therefore, the component of \({\mathop{\rm y}\nolimits} \) orthogonal to \({\mathop{\rm u}\nolimits} \) is \({\mathop{\rm y}\nolimits} - \widehat {\mathop{\rm y}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{3}{5}}\\{ - \frac{4}{5}}\end{array}} \right)\).

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

\(\begin{array}{c}\left\| {{\mathop{\rm y}\nolimits} - \widehat {\mathop{\rm y}\nolimits} } \right\| = \sqrt {{{\left( {\frac{3}{5}} \right)}^2} + {{\left( { - \frac{4}{5}} \right)}^2}} \\ = \sqrt {\left( {\frac{9}{{25}}} \right) + \left( {\frac{{16}}{{25}}} \right)} \\ = \sqrt {\frac{{25}}{{25}}} \\ = 1\end{array}\)

Thus, the distance from \({\mathop{\rm y}\nolimits} \) to the line through \({\mathop{\rm u}\nolimits} \) and the origin is 1.