1. A cube located in R^3 has one face in the plane x=-4. The vertices for this f

Question

1. A cube located in R^3 has one face in the plane x=-4. The vertices for this face are: (-4,24), (-4,5,1), (-4,2,-2), (-4,-1,1). Determine all possible coordinates of the center of this cube. Explain your reasoning.

2. Find an equation for the set of all points P(x,y,z) satisfying the property that | PA | is two times the distance between P and the yz-plane where A is the point A(2,3,-1). Explain your reasoning.

3.Consider two tangent spheres with the following details. On one sphere the endpoint of the diameter having its other endpoint at the point of tangency (3,4,-5). On the other sphere, the endpoint of the diameter having its other endpoint at the point of tangency is (-2, 1, 6). The radius of one sphere is 3 times the radius of the other sphere. Determine all possible equations for these spheres. Explain your reasoning.

Explanation / Answer

1)given  The vertices for this face are: (-4,2,4), (-4,5,1), (-4,2,-2), (-4,-1,1)

distance between (-4,2,4), (-4,5,1)=32
distance between (-4,2,4), (-4,2,-2)=6
distance between (-4,2,4), (-4,-1,1)=32
distance between (-4,5,1),(-4,2,-2)=32
distance between (-4,5,1), (-4,-1,1)=6
distance between (-4,2,-2), (-4,-1,1)=32

center of face= midpoint of face diagonal joining   (-4,2,4), (-4,2,-2) =((-4-4)/2,(2+2)/2,(4-2)/2)=(-4,2,1)

length of edge of cube=32

distance of center from the side of cube=(32)/2

possible x coordinates of center =-4-((32)/2) , -4+((32)/2)
possible y coordinates of center =2-((32)/2) , 2+((32)/2)
possible z coordinates of center =1-((32)/2) , 1+((32)/2)

possible coordinates of the center of this cube are (-4-((32)/2)),(2-((32)/2)),(1-((32)/2)), (-4+((32)/2)),(2+((32)/2)),(1+((32)/2))

possible coordinates of the center of this cube are ((-8-32)/2,(4-32)/2,(2-32)/2), ((-8+32)/2,(4+32)/2,(2+32)/2)

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2)

distance between P(x,y,z) and the yz-plane is the absolute value of x coordinate. i.e, x

| PA | is two times the distance between P and the yz-plane

=>[(x-2)2+(y-3)2+(z+1)2] =2x

=>[(x-2)2+(y-3)2+(z+1)2] =(2x)2

=>[(x2-4x+4)+(y2-6y+9)+(z2+2z+1)] =4x2

=>-3x2-4x+y2-6y+z2+2z+14 =0

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