MHB Plane that is tangent to two curves at an intersection

Click For Summary
To construct a plane tangent to both curves at their intersection, first identify the point of intersection by solving for the values of u and v, which is found to be at u=0. The next step involves calculating the tangent vectors for both curves at this intersection point using their derivatives. The normal vector to the plane can then be determined by taking the cross product of these two tangent vectors. Finally, the equation of the plane can be formed using the identified point and the normal vector. Understanding these steps is crucial for successfully solving the problem.
brunette15
Messages
58
Reaction score
0
Can someone please help me with how to approach/solve this question? construct a plane that is tangent to both curves at the point of intersection.

1st curve:
x(v)=3
y(v)=4
z(v)=v
0<v<2

2nd curve:
x(u)=3+sin(u)
y(u)=4−u
z(u)=1−u
−1<u<1

My first approach was to find a point of intersection then form the equation of the plane but i am unsure how to make the plane tangent to both curves.
 
Physics news on Phys.org
Hi brunette15,

To find the equation of a plane, it suffices to find a point on the plane and a normal to that plane, that is, $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$. As you have surmised, we can find a point on the plane simply by noticing that $4=4-u$ (y-component), so $u=0$. Putting $u$ into one of the equations will give us $(x_0,y_0,z_0)$.

The normal vector is a vector that is orthogonal to both curves at that point. How can we find that?
 
Last edited:
Rido12 said:
Hi brunette15,

To find the equation of a plane, it suffices to find a point on the plane and a normal to that plane, that is, $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$. As you have surmised, we can find a point on the plane simply by noticing that $4=4-u$ (y-component), so $u=0$. Putting $u$ into one of the equations will give us $(x_0,y_0,z_0)$.

The normal vector is a vector that is orthogonal to both curves at that point. How can we find that?

I am still a bit confused as to how to find the normal vector. I am trying to use the cross product but i don't know which other point to use :/
 
brunette15 said:
I am still a bit confused as to how to find the normal vector. I am trying to use the cross product but i don't know which other point to use :/

Hi brunette15, (Wave)

Did you find the $u$ and $v$ for which the curves intersect? (Wondering)
As Rido already remarked, it is at $u=0$.

The tangent vector of a curve (x(v), y(v), z(v)) is given by (x'(v), y'(v), z'(v)).
Can you find the tangent vectors of both curves at the intersection point? (Wondering)

The normal vector is the cross product of those 2 vectors.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K