# Intersecon between a line and hypershere in R4 to RN

• MHB
• LMHmedchem
In summary: The equations for a plane in R7 would be:$$x= x_0 + at + bs\\ y= y_0 + bt + cs\\ z= z_0 + ct + ds$$The equation for a hyperplane in R7 would be:$$x= x_0 + at + bs + et\\ y= y_0 + bt + cs + ft\\z= z_0 + ct + ds + gt\\ u= u_0 + dt + es + ft\\v= v_0 + et + fs + gt\\ w= w_0 + LMHmedchem Hello Again, If I have point B in an orthogonal n-space and point C at the origin of the same space, Code: point_B = 0.03299720 0.00585822 -0.36979000 -0.43413200 -0.60787700 0.61335300 0.76003400 point_C = 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 this would make the vector and from point_B to point_C, Code:  vBC = -0.03299720 -0.00585822 0.36979000 0.43413200 0.60787700 -0.61335300 -0.76003400 |vBC| = 1.28440897 If I were to place a hypersphere around point_B at a radius, say 10% of |vBC| = 0.128, how would I define the equation of that hypersphere and the coordinates of the intersection between the hypersphere and vBC? My understanding is the the equation for a hypersphere in R6 would be,$${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$$I assume that we would find the equation of the line BC and then the simultaneous solution to both equations. I am not really sure how to generate the equation of the line BC in R6. I am never quite sure if this is a line, plane, or hyperplane. As far as the intersecting point, will there actually be a point that is both on the line and in the hypersphere, or can we only find the point on BC that is closest to the hypersphere? Thanks again, LMHmedchem LMHmedchem said: If I have point B in an orthogonal n-space and point C at the origin of the same space, If I were to place a hypersphere around point_B at a radius, say 10% of |vBC| = 0.128, how would I define the equation of that hypersphere and the coordinates of the intersection between the hypersphere and vBC? My understanding is the the equation for a hypersphere in R6 would be,$${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\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}$$LMHmedchem said: Hello Again, If I have point B in an orthogonal n-space and point C at the origin of the same space, Code: point_B = 0.03299720 0.00585822 -0.36979000 -0.43413200 -0.60787700 0.61335300 0.76003400 point_C = 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 This is odd. You titled this "R4 to RN" but this is clearly in R7. this would make the vector and from point_B to point_C, Code:  vBC = -0.03299720 -0.00585822 0.36979000 0.43413200 0.60787700 -0.61335300 -0.76003400 |vBC| = 1.28440897 If I were to place a hypersphere around point_B at a radius, say 10% of |vBC| = 0.128, how would I define the equation of that hypersphere and the coordinates of the intersection between the hypersphere and vBC? My understanding is the the equation for a hypersphere in R6 would be,$${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, ….

## 1. What is the formula for finding the intersection between a line and hypersphere in R4 to RN?

The formula for finding the intersection between a line and hypersphere in R4 to RN is given by setting the equation of the line equal to the equation of the hypersphere and solving for the point(s) of intersection. This can be done by using the distance formula to calculate the distance between the center of the hypersphere and the line, and then determining if this distance is equal to the radius of the hypersphere.

## 2. How many points of intersection can there be between a line and hypersphere in R4 to RN?

The number of points of intersection between a line and hypersphere in R4 to RN can vary. It can be either 0, 1, or 2 points of intersection. If the line is tangent to the hypersphere, there will be only 1 point of intersection. If the line is completely contained within the hypersphere, there will be no points of intersection. And if the line intersects the hypersphere at two distinct points, there will be 2 points of intersection.

## 3. Can the intersection between a line and hypersphere in R4 to RN be a line?

Yes, it is possible for the intersection between a line and hypersphere in R4 to RN to be a line. This can happen when the line is contained within the hypersphere and is parallel to the tangent plane of the hypersphere at its center. In this case, the line will intersect the hypersphere at every point along its length, resulting in a line of intersection.

## 4. What does the intersection between a line and hypersphere in R4 to RN represent?

The intersection between a line and hypersphere in R4 to RN represents the point or points where the line and the hypersphere intersect in n-dimensional space. This can have various applications in geometry, physics, and other fields of science. For example, in physics, the intersection can represent the location of an object in space that is moving along the line and is affected by the gravitational pull of the hypersphere.

## 5. How does the dimension of R4 to RN affect the intersection between a line and hypersphere?

The dimension of R4 to RN has a direct impact on the intersection between a line and hypersphere. As the dimension increases, the complexity of the intersection also increases. In higher dimensions, there can be multiple points of intersection, and the intersection can even be a higher-dimensional object, such as a plane or hyperplane. Moreover, the equations and methods used to find the intersection also become more complex as the dimension increases.

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