Calculus 3 problems (equations of planes and lines) 3 Space

In summary, the homework statement is that if x=2, y=5, and z=1, then the equation becomes (2-2)/(-3)= (4-4)/(-4)= (1-1)/2 or 0=0=0 which is true. If x=-1, y=0, and z=3, the equation becomes (-1-2)/(-3)= (0-4)/(-4)= (3-1)/2 or 1=1=1 which is also true.
  • #1
yopy
43
0
Hi, I am currently doing a practice test for my final exam coming up, I am wondering anyone can double check the questions to see if i did them write, below is a picture of the questions, the answers i got are listed at the bottom,

If you could, please post whether you agree with my answers to any of the following or if you disagree.

28k3d47.jpg

Homework Statement




for 1 A) i got, (X-2)/-3=(Y-4)/-4=(Z-1)/2

for 1 B) after simplifying i came up with -3x-4y+2z=-24


For 2) i got the following,

x= 3+5t
y=1+2t
z=2+5t

for 3) after simplifying i got, 3x-2y-5z=-42
 
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  • #2
yopy said:
Hi, I am currently doing a practice test for my final exam coming up, I am wondering anyone can double check the questions to see if i did them write, below is a picture of the questions, the answers i got are listed at the bottom,

If you could, please post whether you agree with my answers to any of the following or if you disagree.

28k3d47.jpg

Homework Statement




for 1 A) i got, (X-2)/-3=(Y-4)/-4=(Z-1)/2
If x= 2, y= 4, and z= 1, your equation becomes (2- 2)/(-3)= (4- 4)/(-4)= (1-1)/2 or 0= 0= 0 which is true. If x= -1, y= 0, and z= 3, your equation becomes (-1-2)/(-3)= (0-4)/(-4)= (3-1)/2 or 1= 1= 1 which is also true.

for 1 B) after simplifying i came up with -3x-4y+2z=-24
If x= 2, y= 5, and z= 1, this becomes -3(2)- 4(5)+ 2(1)= -6-20+ 2= -24. Yes, that point is in the plane. A vector perpendicular to this plane is <-2, -4, 2> which is the direction vector for the given line.

For 2) i got the following,

x= 3+5t
y=1+2t
z=2+5t
When t= 0 that gives (3,1,2) which is also the point given by t= 1 in the given parametric equations. Further, <5, 2, 5> is the direction vector and is the tangent vector to the given curve at t= 1.

for 3) after simplifying i got, 3x-2y-5z=-42
If x= -2, y= 3, and z= 6, your equation becomes 3(-2)- 2(3)- 5(6)= - 6- 6- 30= -42 so the point is on that plane. However, the normal vector to that plane is <3, -2, -5> while the the gradient of the given function is <3x^2, -3y^2, 2z> and at (-2, 3, 6), that is <12, -27, 12> which is NOT a multiple of <3, -2, -5> and so not in the same direction.

 
  • #3


Do you agree with the answers i have?
 
  • #4
HallsOfIvy disagrees with your answer to problem 3. Show us how you got the answer you've shown, and we'll figure it out.
 
  • #5
Mark44 said:
HallsOfIvy disagrees with your answer to problem 3. Show us how you got the answer you've shown, and we'll figure it out.


I figured the problem i did wrong, i was thinking of a tangent plane to a line instead of a tangent plane to a surface, to fix it i ended up taking the gradient of the surface function, plugging in the point, then dot producting it with (x-xi),(y-yi),(z-zi)

i ended up with12x-27y+12z = -33


pretty sure this is right now
 
  • #6
Looks fine.
 

1. What is Calculus 3 and why is it important?

Calculus 3 is a branch of mathematics that deals with functions of multiple variables in three-dimensional space. It is important because it helps us understand and model complex systems in the physical world, such as motion, fluid flow, and electric fields.

2. What are equations of planes and lines in three-dimensional space?

Equations of planes and lines in three-dimensional space are mathematical representations of these geometric objects. They describe the relationship between their coordinates and their orientation in space.

3. How do you find the equation of a plane given three points?

To find the equation of a plane given three points, you can use the formula (x-x1)(y-y2)-(x-x2)(y-y1)=(x-x1)(z-z2)-(x-x2)(z-z1).

4. How do you find the equation of a line given a point and a direction vector?

To find the equation of a line given a point and a direction vector, you can use the formula (x-x0)/a=(y-y0)/b=(z-z0)/c, where (x0, y0, z0) is the given point and a, b, c are the components of the direction vector.

5. How is Calculus 3 used in real life?

Calculus 3 is used in many fields, including physics, engineering, economics, and computer graphics. It is used to model and analyze complex systems and make predictions about their behavior. For example, it is used in designing airplanes, predicting stock market trends, and creating 3D animations in movies.

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