Transformation Matrix project

Question

My task is to find the Transformation Matrix, that projects, any point of the xy-plane, on the line $$ y = 4x$$

The solution should be:

$$T=\pmatrix{0.06&0.235\\0.235&0.94}$$

But somehow i dont know how to get this solution?

I tried to take $A$ as a point of the xy-plane $A= \pmatrix{1\\1}$

Next i got: $g:x = \pmatrix{x\\y}+ t* \pmatrix{1\\1}$

This did i put into $y-4x =0$

So i got as $t = -4/3x -1/3y$

As last step i calculated $g:x$ with the value of $t$

$$g:x = \pmatrix{x\\y}+ (-4/3x -1/3y)* \pmatrix{1\\1}$$

What i got was:

$$T= \pmatrix{-1/3&-1/3\\-4/3&2/3}$$

But that is not the right solution! What did i wrong? How to get the solution of above? Thanks

In R3 i suppose the matrix would look like this: Am i correct? Thanks

$$\begin{pmatrix} \frac{1}{m^2+1} & \frac{m}{m^2+1} & 0 \\ \frac{m}{m^2+1} & \frac{m^2}{m^2+1} & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

Answer 1

This question can be generalized as follow:

Aim: Find the transformation matrix $M$ such that it projects every point on xy-plane into the line $l_1$ where $l_1$ has equation $y=mx$.

Suppose $A=(a,b)$ is an arbitrary point on the xy-plane. Denote the line joining the point $A$ and perpendicular to the line $y=mx$ as $l_2$. Then $l_2$ has gradient $-\frac{1}{m}$.

Note that vector equation of $l_2$ is $$(a,b)+t(1,-\frac{1}{m}),$$ where $t \in \mathbb{R}$. The intersection point, say $A^{'}$ between $l_1$ and $l_2$ can be found using the equation $t(1,-\frac{1}{m})=m(a,b)$.

Solving the equation gives us $$t=\frac{m(b-ma)}{m^2+1}$$. Hence, $A^{'}$ can be denoted as $$(a,b)+\frac{m(b-ma)}{m^2+1}(1,-\frac{1}{m})=\left(a+\frac{m(b-ma)}{m^2+1}, b-\frac{b-ma}{m^2+1} \right) = \left( \frac{a+mb}{m^2+1}, \frac{m(a+mb)}{m^2+1} \right)$$

Hence, $M$ can be found using the equation $$M\left( \begin{array}{c} a\\ b \end{array} \right)= \left( \begin{array}{c} \frac{a+mb}{m^2+1}\\ \frac{m(a+mb)}{m^2+1} \end{array} \right)= \left( \begin{array}{c} \frac{1}{m^2+1}\\ \frac{m}{m^2+1} \end{array} \right)a + \left( \begin{array}{c} \frac{m}{m^2+1}\\ \frac{m^2}{m^2+1} \end{array} \right)b$$

Hence, the matrix $M$ is $$\begin{pmatrix} \frac{1}{m^2+1} & \frac{m}{m^2+1}\\ \frac{m}{m^2+1} & \frac{m^2}{m^2+1} \end{pmatrix}$$

Answer 2

suppose $A=(a,b)$, because you want the project of $A$ on the line $y=4x$, $A$ should be on a line which is perpendiculaire to the known line. So the tangent of the new line is $-1/4$. Then the points on the new line could be expressed as $A+t(1,-1/4)=(a+t,b-t/4)$. The project of $A$ on the line is denoted by $A'$, then there should be some $t$ satisfying $A'=(a+t,b-t/4)$. $A'$ is on the line $y=4x$, so $b-t/4=4a+4t$, thus $t=4(b-4a)/17$. then $$A'=(a+t,b-\frac{t}{4})=(\frac{4b+a}{17},\frac{4a+16b}{17})$$ We would find a matrix that satisfies $A'=MA$, that is $$\begin{pmatrix}\frac{4b+a}{17}\\\frac{4a+16b}{17}\end{pmatrix}=M\begin{pmatrix}a\\b\end{pmatrix}$$ So$$M=\begin{pmatrix}\frac{1}{17} & \frac{4}{17}\\\frac{4}{17} & \frac{16}{17}\end{pmatrix}\approx\begin{pmatrix}0.0588 & 0.2353\\0.2353 & 0.9412\end{pmatrix}$$ This is the solution.