# Thread: Distance to the origin

1. ## Distance to the origin

The distance from a plane (a^t)x = c (in m-dimensional space) to the origin is
|c| / ||a||.
How far is the plane x1+x2-x3-x4=8 from the origin, and what point on it is nearest?

2. Originally Posted by veronicak5678
The distance from a plane (a^t)x = c (in m-dimensional space) to the origin is
|c| / ||a||.
How far is the plane x1+x2-x3-x4=8 from the origin, and what point on it is nearest?
$x1+ x2- x3- x4= \begin{bmatrix}1 & 1 & -1 & -1\end{bmatrix}\begin{bmatrix}x1 \\ x2 \\ x3\\ x4\end{bmatrix}= 8$.
What are a and c? Apply your formula to them.

Another way to do this is to note that the vector <1, 1, -1, -1> is perpendicular to the plane and so the point on the plane closest to the origin lies on the line through the origin having that direction vector. Write parametric equations for the line and determine where it crosses the plane.

3. So the distance is 8 / root 4 = 4. I'm not sure about the point, though. How do I write parametric equations? What it the plane were not perpendicular?

4. The 4 dimensional line through $(x_{10}$, $x_{20}$, $x_{30}$, $x_{40})$ in the same direction as vector <A, B, C, D> can be written $x_1= At+ x_{10}$, $x_2= Bt+ x_{20}$, $x3= Ct+ x_{30}$, $x_4= Dt+ x_{40}$. The line through (0, 0, 0) in the same direction as vector <1, 1, -1, -1> is $x_1= t, x_2= t, x_3= -t, x_4= -t$.