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

  • Thread starter yopy
  • Start date
  • #1
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Hi, Im currently doing a practice test for my final exam coming up, im 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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Hi, Im currently doing a practice test for my final exam coming up, im 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.

Homework Statement





Homework Equations





The Attempt at a Solution

 
  • #3
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Do you agree with the answers i have?
 
  • #4
33,154
4,838
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
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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
33,154
4,838
Looks fine.
 
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