I am preparing for my exams and facing a problem with the projection in linear algebra.
Problem: Find the direction of the line $$ p \ \epsilon \ \mathbb{R}^3 $$, which intersects the origin, and we know that all three points:
$$ u = (1,2,3)^T, v = (0,3,1)^T, w = (8,1,2)^T $$
are projected to the same point when projecting on the line $p$.
As I understand, we're projecting from the space of higher dimension to the subspace $ U = span\{p\} $. The orthonormal base of the space $U$ is $ x_U = <x,z>z = \frac{1}{||p||^2}<x,p>p = \frac{x^Tp}{p^Tp}p$, where $x_U$ is the projected vector. I am not sure how to work with the information that all vectors are projected to the same point on $p$.
Thank you in advance for any clue.
Propose $p = (a,b,c)^{T}$. Note that the orthogonal projection of a vector $x$ on $Span\{p\}$ is $\frac{<x,p>}{<p,p>}p$. Therefore, we know that $\frac{<u,p>}{<p,p>}p = \frac{<v,p>}{<p,p>}p = \frac{<w,p>}{<p,p>}p$, So $<u,p> = <v,p> = <w,p>$. From this, we get $<u - v,p> = 0$ and $<u - w,p> = 0$, which are equivalent to $-a + b -2c = 0$ and $7a - b - c = 0$ respectively. Now we just need to solve for these 2 equations to find $p$.
(Note that there are infinitely many solutions, but every 2 solutions are linearly dependant, so we do end up with 1 direction.)