MHB Intersecon between a line and hypershere in R4 to RN

[LMHmedchem]

Hello Again,

If I have point B in an orthogonal n-space and point C at the origin of the same space,

point_B = 0.03299720  0.00585822  -0.36979000 -0.43413200  -0.60787700  0.61335300  0.76003400
point_C = 0.00000000  0.00000000   0.00000000  0.00000000   0.00000000  0.00000000  0.00000000

this would make the vector and from point_B to point_C,

 vBC  = -0.03299720   -0.00585822   0.36979000   0.43413200   0.60787700   -0.61335300   -0.76003400
|vBC| = 1.28440897

If I were to place a hypersphere around point_B at a radius, say 10% of |vBC| = 0.128, how would I define the equation of that hypersphere and the coordinates of the intersection between the hypersphere and vBC?

My understanding is the the equation for a hypersphere in R6 would be,
$${x}_{1}^{2} + {x}_{2}^{2}+{x}_{3}^{2}+{x}_{4}^{2}+{x}_{5}^{2}+{x}_{6}^{2}={R}^{2}$$

or in this case, the set of all points that satisfy,
$${x}_{1}^{2} + {x}_{2}^{2}+{x}_{3}^{2}+{x}_{4}^{2}+{x}_{5}^{2}+{x}_{6}^{2}=1.28$$

I assume that we would find the equation of the line BC and then the simultaneous solution to both equations.

I am not really sure how to generate the equation of the line BC in R6. I am never quite sure if this is a line, plane, or hyperplane. As far as the intersecting point, will there actually be a point that is both on the line and in the hypersphere, or can we only find the point on BC that is closest to the hypersphere?

Thanks again,

LMHmedchem

 

[I like Serena]

If I have point B in an orthogonal n-space and point C at the origin of the same space,

If I were to place a hypersphere around point_B at a radius, say 10% of |vBC| = 0.128, how would I define the equation of that hypersphere and the coordinates of the intersection between the hypersphere and vBC?

My understanding is the the equation for a hypersphere in R6 would be,
$${x}_{1}^{2} + {x}_{2}^{2}+{x}_{3}^{2}+{x}_{4}^{2}+{x}_{5}^{2}+{x}_{6}^{2}={R}^{2}$$

or in this case, the set of all points that satisfy,
$${x}_{1}^{2} + {x}_{2}^{2}+{x}_{3}^{2}+{x}_{4}^{2}+{x}_{5}^{2}+{x}_{6}^{2}=1.28$$

I assume that we would find the equation of the line BC and then the simultaneous solution to both equations.

I am not really sure how to generate the equation of the line BC in R6. I am never quite sure if this is a line, plane, or hyperplane. As far as the intersecting point, will there actually be a point that is both on the line and in the hypersphere, or can we only find the point on BC that is closest to the hypersphere?

Hi LMHmedchem,

A hypersphere consists of all points with the same distance to a point.
The equation of a hypersphere around B with radius R is therefore:
$$\| \overrightarrow x - \overrightarrow B\|= R \tag 1$$

The equation of a line through the origin and with direction $\overrightarrow{BC}$ is:
$$\overrightarrow x= \lambda \overrightarrow{BC} \tag 2$$

And since $C$ is actually the origin, we have:
$$\overrightarrow B = \overrightarrow{CB} = -\overrightarrow{BC} \tag 3$$

Substitute (2) and (3) in (1) to get:
$$\| \lambda \overrightarrow{BC} - (-\overrightarrow{BC})\| = R \quad\Rightarrow\quad
\| (\lambda+1) \overrightarrow{BC}\| = |\lambda+1| \cdot\|\overrightarrow{BC}\| =R \quad\Rightarrow\quad
|\lambda+1| = \frac{R}{BC}\quad\Rightarrow\quad
\lambda = -1\pm\frac R{BC} \tag 4
$$
That is, there are 2 points of intersection:
$$\overrightarrow x = \left(-1\pm\frac R{BC}\right)\overrightarrow{BC}$$

 

[HOI]

Hello Again,

If I have point B in an orthogonal n-space and point C at the origin of the same space,

point_B = 0.03299720  0.00585822  -0.36979000 -0.43413200  -0.60787700  0.61335300  0.76003400
point_C = 0.00000000  0.00000000   0.00000000  0.00000000   0.00000000  0.00000000  0.00000000

This is odd. You titled this "R4 to RN" but this is clearly in R7.

this would make the vector and from point_B to point_C,

 vBC  = -0.03299720   -0.00585822   0.36979000   0.43413200   0.60787700   -0.61335300   -0.76003400
|vBC| = 1.28440897

If I were to place a hypersphere around point_B at a radius, say 10% of |vBC| = 0.128, how would I define the equation of that hypersphere and the coordinates of the intersection between the hypersphere and vBC?

My understanding is the the equation for a hypersphere in R6 would be,
$${x}_{1}^{2} + {x}_{2}^{2}+{x}_{3}^{2}+{x}_{4}^{2}+{x}_{5}^{2}+{x}_{6}^{2}={R}^{2}$$

or in this case, the set of all points that satisfy,
$${x}_{1}^{2} + {x}_{2}^{2}+{x}_{3}^{2}+{x}_{4}^{2}+{x}_{5}^{2}+{x}_{6}^{2}=1.28$$

No, if the radius is 0.128, that would be equal to $$0.128^2= 0.016384$$

I assume that we would find the equation of the line BC and then the simultaneous solution to both equations.

I am not really sure how to generate the equation of the line BC in R6. I am never quite sure if this is a line, plane, or hyperplane. As far as the intersecting point, will there actually be a point that is both on the line and in the hypersphere, or can we only find the point on BC that is closest to the hypersphere?

Thanks again,

LMHmedchem

A line is one dimensional. In 7 dimensions, you would need to write 6 equations in the 7 variables to represent a line. In higher than two dimensions, the simplest way to write a line is in parametric equations with one parameter.

The line from (0.00000000, 0.00000000, 0.00000000, 0.00000000, 0.00000000, 0.00000000, 0.00000000) to (0.03299720, 0.00585822, -0.36979000, -0.43413200, -0.60787700, 0.61335300, 0.76003400) will be represented by
x1= 0.03299720t, x2= 0.00585822t, x3= -0.36979000t, x4= -0.43413200t, x5= -0.60787700t, x6= 0.61335300t, x7= 0.76003400t.
It should be easy to see that t= 0 gives the first point, t= 1 gives the second point.

In general, the line from (A1, A2, A3, …) to (B1, B2, B3, …) is given by
x1= (B1- A1)t+ A1, x2= (B2- A2)t+ A2, x3= (B3- A3)t+ A3, …

You see that when t= 0, x1= A1, x2= B2, x3= B3, … and when t= 1, x1= (B1- A1)+ A1= B1, x2= (B2- A2)+ A2= B2, x3= (B3- A3)+ A3= B3, etc.

In the case that A= (0, 0, 0, ….) as here, that reduces to
x1= B1t, x2= B2t x3= B3t, ….