Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 21/Solution 1

Give the formula in the hint, first we write the lines in the standard form $y=mx$.

$L_{1}:y=-{\frac {1}{2}}x\ \Longrightarrow m=-{\frac {1}{2}}$,

$L_{2}:y=2x\ \Longrightarrow m=2$.

The matrix of projection onto $L_{1}$ is $P_{1}={\begin{pmatrix}{\frac {4}{5}}&-{\frac {2}{5}}\\-{\frac {2}{5}}&{\frac {1}{5}}\end{pmatrix}}$,

and the matrix of projection onto $L_{2}$ is $P_{2}={\begin{pmatrix}{\frac {1}{5}}&{\frac {2}{5}}\\\ \\{\frac {2}{5}}&{\frac {4}{5}}\end{pmatrix}}$

Thus projecting any point in 2D plane first on $L_{1}$ and then on $L_{2}$ can be represented by

$\mathbb {R} ^{2}\ni {\begin{pmatrix}x\\y\end{pmatrix}}\rightarrow P_{2}{\Big (}P_{1}{\begin{pmatrix}x\\y\end{pmatrix}}{\Big )}$ so the transformation matrix is $T=P_{2}\circ P_{1}$

$P_{2}\circ P_{1}={\begin{pmatrix}{\frac {1}{5}}&{\frac {2}{5}}\\\ \\{\frac {2}{5}}&{\frac {4}{5}}\end{pmatrix}}{\begin{pmatrix}{\frac {4}{5}}&-{\frac {2}{5}}\\-{\frac {2}{5}}&{\frac {1}{5}}\end{pmatrix}}=0$

In fact, since these two lines are orthogonal, projecting a vector (point) first on $L_{1}$ and then orthogonally projecting the new point on $L_{2}$ will return us to $0$.