Post History
#2: Post edited
by
Bennyshinichi
·
2024-05-19T13:43:38Z (6 months ago)
- Given two lines, $l: 3x-y=9$ and $m: x-3y=3$, and an angle bisector $x-y=3$. Point $A\left(3,0\right)$ is where these three lines intersect. The problem asks us to find a point on the angle bisector that is at a perpendicular distance of $\sqrt{10}$ from line $m$. I know that we can use the distance formula from point to line. But I'm curious if it's possible to use vectors to solve this problem.
I have written the angle bisector and line $m$ in vector equations as $\begin{pmatrix} 3\\0 \end{pmatrix} + t \begin{pmatrix} 1\\1\end{pmatrix}$ and $\begin{pmatrix} 3\\0 \end{pmatrix} + t \begin{pmatrix} 3\\1\end{pmatrix}$ respectively.- I was considering a vector $\begin{pmatrix} x\\y \end{pmatrix}$ denoted as $s$ that has a magnitude of $\sqrt{10}$ and is perpendicular to $m$. By using the dot product we get $3x+y=0$ (with x and y that are the components from the vector $s$, 3 and 1 from the direction vector of $m$).
- But this doesn't seem to help me anywhere. It feels like I'm missing an important piece of the puzzle that could solve this problem. So is there something that I have overlooked? Or is there another formula or property that should be applied to solve this problem?
- Given two lines, $l: 3x-y=9$ and $m: x-3y=3$, and an angle bisector $x-y=3$. Point $A\left(3,0\right)$ is where these three lines intersect. The problem asks us to find a point on the angle bisector that is at a perpendicular distance of $\sqrt{10}$ from line $m$. I know that we can use the distance formula from point to line. But I'm curious if it's possible to use vectors to solve this problem.
- I have written the angle bisector and line $m$ in vector equations as $\vec{u}=\begin{pmatrix} 3\\0 \end{pmatrix} + t \begin{pmatrix} 1\\1\end{pmatrix}$ and $\vec{v}=\begin{pmatrix} 3\\0 \end{pmatrix} + t \begin{pmatrix} 3\\1\end{pmatrix}$ respectively.
- I was considering a vector $\begin{pmatrix} x\\y \end{pmatrix}$ denoted as $s$ that has a magnitude of $\sqrt{10}$ and is perpendicular to $m$. By using the dot product we get $3x+y=0$ (with x and y that are the components from the vector $s$, 3 and 1 from the direction vector of $m$).
- But this doesn't seem to help me anywhere. It feels like I'm missing an important piece of the puzzle that could solve this problem. So is there something that I have overlooked? Or is there another formula or property that should be applied to solve this problem?
#1: Initial revision
by
Bennyshinichi
·
2024-05-18T21:39:37Z (6 months ago)
How to find a point on a vector equation with another vector equation and perpendicular distance?
Given two lines, $l: 3x-y=9$ and $m: x-3y=3$, and an angle bisector $x-y=3$. Point $A\left(3,0\right)$ is where these three lines intersect. The problem asks us to find a point on the angle bisector that is at a perpendicular distance of $\sqrt{10}$ from line $m$. I know that we can use the distance formula from point to line. But I'm curious if it's possible to use vectors to solve this problem.
I have written the angle bisector and line $m$ in vector equations as $\begin{pmatrix} 3\\0 \end{pmatrix} + t \begin{pmatrix} 1\\1\end{pmatrix}$ and $\begin{pmatrix} 3\\0 \end{pmatrix} + t \begin{pmatrix} 3\\1\end{pmatrix}$ respectively.
I was considering a vector $\begin{pmatrix} x\\y \end{pmatrix}$ denoted as $s$ that has a magnitude of $\sqrt{10}$ and is perpendicular to $m$. By using the dot product we get $3x+y=0$ (with x and y that are the components from the vector $s$, 3 and 1 from the direction vector of $m$).
But this doesn't seem to help me anywhere. It feels like I'm missing an important piece of the puzzle that could solve this problem. So is there something that I have overlooked? Or is there another formula or property that should be applied to solve this problem?