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A couple of calc III, 3d drawings & sphere problems! need help!

  1. Sep 4, 2005 #1
    Hello everyone, I'm stuck on some problems, This is the first homeowrk assignment and i'm pretty lost! i'm going to list the directions and show you what I have done so far, any help would be great!

    Here is my work to the problems:
    //http://img175.imageshack.us/img175/4090/4108rz.jpg [Broken]
    //http://img217.imageshack.us/img217/919/13192yq.jpg [Broken]
    //http://img217.imageshack.us/img217/1842/22403rs.jpg [Broken]

    #4. What are the projections of the point (2,3,5) on the xy-,yz-, and xz-planes? Draw a rectangular box with the origin and (2,3,5) as opposite verticies and with its face parallel to the coordinate planes. Label all vertices of the box. Find the length of the diagnoal of the box.

    #7. Show that the triangle with vertices P(-2,4,0), Q(1,2-1), and R(-1,1,2) is an equialteral triangle. I used the distance formula from RP and to PQ, and they didn't equal eachother as shown in the drawing. SO doesn't that right there prove it isn't an equialteral triangle? or did i screw up?

    #10. Find the distance from (3,7,-5) to each of the following.
    (a) THe xy-plane
    (b) The yz-plane
    (c) The xz-plane
    (d) the x-axis
    (e) the y-axis
    (f) the z-axis

    I know this is probably easy but I'm not sure on how you would start this problem. Of course you would use the distance formula, but what would u choose for the 2nd point?

    #19. Prove that the midpoint of the line segment from P1(x1,y1,z1) to P2(x2,y2,z2) is ([x1+x2]/2, [y1+y2]/2, [z1+z2]/2); (b) find the lengths of the medians of the triangle with vertices A(1,2,3),B(-2,0,5); C(4,1,5). The answer to B is: 5/2, 1/2 *sqrt(94), 1/2*sqrt(85);

    #25. Describe in words the region of R^3 representee by the equation or inequality. x > 3; answer: A half-space consisting of all points in front of plane x= 3. What is a half space? and what would it look like?

    #28. y = z; I said a plane parrallel to the xz-plane. I just treated z as being a constant like k. But i'm sure that isn't right. or is it?

    #34. xyz = 0; I have no idea what this would be in words or visually.

    #40. Consider the points P such that the distance from P to A(-1,5,3) is twice the distance from P to B(6,2,-2). Show that the set of all such points is a sphere, and find its center and radius. I'm fine on finding the center and radius of problems but i'm stuck on the first part, any hints?

    Thanks! :biggrin:
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Sep 4, 2005 #2


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    The point on the point or line closest to the given point. Also the point whose normal line contains the given point. If someone asked how far from the floor a point was how would you measure? Most would measure on a line perpendicular to the floor. These are quite easy as it is obvious how to minimize the distance, take the free parameters equal to the coresponding coordinate of the given point. For practice do a few with less special planes, line x+2y-z=11.
  4. Sep 4, 2005 #3


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    #7 you did screw up

    #25 a half space is just that half the space. Imagine the plane x=3 the stuff on each side of it is a half space. The analogy to R^2 might help consider x=3 the stuff to each side is a halfspace.

    #28 y=z is a plane where x can be anything, but y=z. In parametric form
    (s,t,t). To imagine it take a slice z=a then the line x=y,z=a all such lines form the plane you want. You could also think of a plane containing the two lines x=y,z=0 and the line x=0=y=

    #34 the set of all point were xyz=0
    The zero product property states if a product of real numbers is zero at least one number is zero. Thus you are looking for all points in which at least one of the x,y, or z coordinates is zero.

    #40 a sphere can be put into standard form
    where the point (x0,y0,z0) is called the center and R is called the radius
    use the distance formula to write an equation that represent the set of points described. Manipulate that equation into standard form to show that it represents a sphere.
  5. Sep 4, 2005 #4
    LuffLuff, how did i screw up 7? I used ur equation. vertices P(-2,4,0), Q(1,2-1), and R(-1,1,2) d = sqrt((-2-1)^2+(4-1)^2+(0-2)^2) = 4.69; d = sqrt((1+2)^2+(2-4)^2+(-1-0)^2) = 3.74. You also said... For practice do a few with less special planes, line x+2y-z=11. How did you figure out that line would work? THanks
  6. Sep 5, 2005 #5


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    d = sqrt((-2-1)^2+(4-1)^2+(0-2)^2) = 4.69
    should be
    d = sqrt((-2-(-1))^2+(4-1)^2+(0-2)^2) = 3.74

    For the point-plane distance you want the minimun distance from the point to any point on the plane.
    #10. Find the distance from (3,7,-5) to each of the following.
    (a) THe xy-plane
    we want
    d^2=min d((3,7,-5),(s,t,0))=min (3-s)^2+(7-t)^2+5^2=25
    since each term in the sum is >0 making the first two terms 0 is the min
    This was easy because of the coordinates, it is trickier is we want
    Find the distance from (3,7,-5) to each of the following.
    The plane x+2y-z=11
    There are several ways to work this out. You could minimize the distance for example
    d^2=min d((3,7,s+2t-11),(s,t,0))=min (3-s)^2+(7-t)^2+(6-s-2t)^2
    It is easier if one realizes that the minimum distance is alone a line normal to the plane and containing the point.
    also you could look for a sphere centered at the given point that intersects the given line at only one point, its radius is the desired distance.
    here are some links on the subject
    http://astronomy.swin.edu.au/~pbourke/geometry/pointplane/ [Broken]
    http://home.xnet.com/~fidler/triton/math/review/mat135/vector/L_Psp/Distpl1.htm [Broken]

    Last edited by a moderator: May 2, 2017
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