Matrix, that projects, any point of the xy-plane, on the line y = 4 x The solution shoul

Blericker74

Blericker74

Answered question

2022-07-06

Matrix, that projects, any point of the xy-plane, on the line
y = 4 x
The solution should be:
T = ( 0.06 0.235 0.235 0.94 )
But somehow i dont know how to get this solution?

Answer & Explanation

vrtuljakc6

vrtuljakc6

Beginner2022-07-07Added 16 answers

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Suppose < math xmlns="http://www.w3.org/1998/Math/MathML"> A = ( a , b ) is an arbitrary point on the xy-plane. Denote the line joining the point A and perpendicular to the line y = m x as l 2 . Then l 2 has gradient 1 m .
Note that vector equation of l 2 is
( a , b ) + t ( 1 , 1 m ) ,
where t R. The intersection point, say A between l 1 and l 2 can be found using the equation t ( 1 , 1 m ) = m ( a , b )
Solving the equation gives us
t = m ( b m a ) m 2 + 1
Hence, A can be denoted as
( a , b ) + m ( b m a ) m 2 + 1 ( 1 , 1 m ) = ( a + m ( b m a ) m 2 + 1 , b b m a m 2 + 1 ) = ( a + m b m 2 + 1 , m ( a + m b ) m 2 + 1 )
Hence, M can be found using the equation
M ( a b ) = ( a + m b m 2 + 1 m ( a + m b ) m 2 + 1 ) = ( 1 m 2 + 1 m m 2 + 1 ) a + ( m m 2 + 1 m 2 m 2 + 1 ) b
Hence, the matrix M is
( 1 m 2 + 1 m m 2 + 1 m m 2 + 1 m 2 m 2 + 1 )

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