Range of the parameter of sphere intersecting with a plane

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SUMMARY

The discussion focuses on determining the range of the parameter d for which the intersection of the sphere defined by the equation x² + y² + z² = 1 and the plane given by x + y + z = d is non-empty. The solution involves substituting Cartesian coordinates into both equations and simplifying to find that the condition for intersection is based on the distance from the origin. The conclusion is that the plane intersects the sphere if the distance from the origin is less than or equal to 1, leading to the range of d being established as -√3 ≤ d ≤ √3.

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  • Understanding of Cartesian coordinates and their relation to spheres and planes.
  • Familiarity with trigonometric identities, specifically sin²θ + cos²θ = 1.
  • Knowledge of vector geometry and distance calculations in three-dimensional space.
  • Basic algebraic manipulation skills to solve equations involving multiple variables.
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  • Study the geometric interpretation of spheres and planes in three-dimensional space.
  • Learn about vector equations and their applications in determining distances from points to planes.
  • Explore the implications of spherical coordinates in solving three-dimensional problems.
  • Investigate methods for visualizing intersections of geometric shapes in 3D space.
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Students and educators in mathematics, particularly those studying geometry and algebra, as well as anyone interested in understanding the intersection of geometric shapes in three-dimensional space.

Jadehaan
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Homework Statement


Find the range of the parameter d for which the intersection of the sphere x2+y2+z2=1 and the plane x+y+z=d is non-empty.


Homework Equations


Cartesian coordinates of a sphere:
x=rcos[tex]\theta[/tex]sin[tex]\phi[/tex]
y=rsin[tex]\theta[/tex]sin[tex]\phi[/tex]
z=rcos[tex]\phi[/tex]


r=1

The Attempt at a Solution


I substitute x,y,z in both equations
d=sin[tex]\theta[/tex]sin[tex]\phi[/tex]+cos[tex]\phi[/tex]+cos[tex]\theta[/tex]sin[tex]\phi[/tex]
cos2[tex]\theta[/tex]sin2[tex]\phi[/tex]+sin2[tex]\theta[/tex]sin2[tex]\phi[/tex]+cos2[tex]\phi[/tex]=1

Since sin2[tex]\theta[/tex]+cos2[tex]\theta[/tex]=1
I get 1+cos2[tex]\phi[/tex]=1
This implies that [tex]\phi[/tex]=90
Which solves the first equation for d=sin[tex]\theta[/tex]+cos[tex]\theta[/tex]
Is this right?
Thanks for any help.
 
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I would think your answer would need to be in the form a ≤ d ≤ b. A simpler approach might be to observe the plane will intersect the sphere if the distance of the plane from the origin is ≤ 1. This can be easily done with vectors and no need for spherical coordinates.
 

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