# 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?
$\displaystyle x1+ x2- x3- x4= \begin{bmatrix}1 & 1 & -1 & -1\end{bmatrix}\begin{bmatrix}x1 \\ x2 \\ x3\\ x4\end{bmatrix}= 8$.
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 $\displaystyle (x_{10}$, $\displaystyle x_{20}$, $\displaystyle x_{30}$, $\displaystyle x_{40})$ in the same direction as vector <A, B, C, D> can be written $\displaystyle x_1= At+ x_{10}$, $\displaystyle x_2= Bt+ x_{20}$, $\displaystyle x3= Ct+ x_{30}$, $\displaystyle x_4= Dt+ x_{40}$. The line through (0, 0, 0) in the same direction as vector <1, 1, -1, -1> is $\displaystyle x_1= t, x_2= t, x_3= -t, x_4= -t$.

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### the distance if -6,8 from origin is 10 how

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