MHB Plane that is tangent to two curves at an intersection

brunette15
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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.
 
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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.
 
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