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

In summary: R to P d = sqrt((-1--2)^2+(1-4)^2+(2-0)^2)= 4.69. I used the distance formula it worked out to be 4.69.Luff, you are incorrect. The vertices are P(-2,4,0), Q(1,2,-1), and R(-1,1,2). You just forgot to include the negative sign for the z-coordinate of point Q in your calculation for the distance from R to Q. The correct distance is 4.69. Sorry, you are right. I have made a careless mistake. I am sorry.
  • #1
mr_coffee
1,629
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
//http://img217.imageshack.us/img217/919/13192yq.jpg
//http://img217.imageshack.us/img217/1842/22403rs.jpg

#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 each other 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:
 
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  • #2
mr_coffee said:
#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?
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.
 
  • #3
mr_coffee said:
#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 each other as shown in the drawing. SO doesn't that right there prove it isn't an equialteral triangle? or did i screw up?

#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?
#7 you did screw up
recall
d^2=(x1-x2)^2+(y1-y2)^2+(z1-z2)^2

#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
(x-x0)^2+(y-y0)^2+(z-z0)^2=R^2
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.
 
  • #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
 
  • #5
mr_coffee said:
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
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.
say
#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/
http://home.xnet.com/~fidler/triton/math/review/mat135/vector/L_Psp/Distpl1.htm
http://mathworld.wolfram.com/Point-PlaneDistance.html
http://omega.albany.edu:8008/calc3/distance-to-point-dir/ptoplane-maple.html
http://softsurfer.com/Archive/algorithm_0104/algorithm_0104.htm

http://mathforum.org/library/drmath/view/51956.html
 
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What is Calc III?

Calc III, short for Calculus III, is a college-level math course that builds upon the concepts of Calculus I and II. It covers topics such as multivariable calculus, vector calculus, and partial derivatives.

What are 3D drawings?

3D drawings are graphical representations of objects or spaces that have three dimensions: length, width, and height. They are created using mathematical equations and computer software to give a realistic and detailed view of objects in a three-dimensional space.

What are sphere problems?

Sphere problems are mathematical problems that involve finding the properties or relationships of spheres, which are three-dimensional objects with a curved surface and all points equidistant from a central point. These problems often involve concepts from geometry, trigonometry, and calculus.

What are some real-life applications of Calc III and 3D drawings?

Calc III and 3D drawings have many practical applications in fields such as engineering, physics, computer graphics, and architecture. For example, they can be used to design and analyze complex structures, simulate physical phenomena, and create realistic animations and video games.

What are some tips for solving Calc III and 3D drawing problems?

Some tips for solving Calc III and 3D drawing problems include understanding the concepts and equations involved, practicing with different types of problems, visualizing the objects and spaces in question, and using appropriate mathematical tools and software. It is also helpful to break down complex problems into smaller, more manageable parts and to check your work for accuracy and logic.

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