Finding an equation describing the plane containing t1&t2?

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In summary, the conversation discusses finding the tangent of line T1 and T2 to the curve t1/t2 at the point (1,3,1). The person has found the vectors T1 and T2 and is now trying to find an equation describing the plane containing them. They mention two methods, one involving the cross product of the vectors and the other using tangent planes and linear approximation. The latter method is confusing and they are seeking advice on how to approach it. The conversation ends with a suggestion to define the curve in terms of a parameter and then work out the partial derivatives to find the tangent plane at the given point.
  • #1
mr_coffee
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Hello everyone, this problem is very long so if i screw up one part the whole thing is wrong. But I'm pretty sure I'm doing it right...
The last 5 questions asked me to find the tangent of line T1 and T2 to the Curve t1/t2, at the point (1,3,1). So i found everything, and now i have a vector T1 and T2;
T1 = <1+t,3,1+8t>
T2 = <1,3+3t,1-4t>
Now he wants me to find an equation descrbing the plane containing T1 and T2, one way i could do this is to take the cross product of the 2 vectors because they both intersect at the same point, (1,3,1)...but that's not how he wants us to do it, he wants us to use a method with tangent planes and linear approximation but I'm confused on how to do that@! the chapter doesn't talk about that at all...all he says is
We know that any plan passing through the point P(xo,yo,zo) has an eqwuation of the form
A(x-xo) + B(y-yo) + C(z-zo) = 0;
then it says we can rewrite it as z - zo = a(x-xo) + b(y-yo);
Then it says
Suppose f has continuous partial derivatives. An equation of the tangent plane to the surface z = f(x,y) at the point P(xo,yo,zo) is
z-zo = fx(xo,yo)(x-xo) + fy(xo,yo)(y-yo);
note: fx means partial derivative with respect of x;

Any ideas on what i could do ? Thanks@
 
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  • #2
You seem to have your curve descibed in terms of a parameter, t.

You will have to define the curve as,

z = f(x,y)

If you can do that, then P(xo,yo,zo) is the tangent point. And the tangent plane at this point will contain any tangent vectors at this point.
Work out the partial derivatives, e.g. fx(xo,yo) as numerical values, where fx(xo,yo) is the partial derivative of f wrt x where x = xo and y = yo.

Then fill in that eqn,

z-zo = fx(xo,yo)(x-xo) + fy(xo,yo)(y-yo);
 
  • #3
thank u! worked great
 

1. What is the process for finding an equation describing the plane containing two points?

The process for finding an equation describing the plane containing two points, t1 and t2, involves three steps: finding the normal vector, finding the point on the plane, and plugging these values into the general equation of a plane.

2. How do you find the normal vector of the plane?

The normal vector of a plane is a vector that is perpendicular to the plane. To find it, you can use the cross product of the two vectors formed by the two points, t1 and t2.

3. Can you use any two points on the plane to find the equation?

Yes, any two distinct points on the plane can be used to find the equation. However, if the points are collinear, the equation will not be unique.

4. What is the general equation of a plane?

The general equation of a plane is Ax + By + Cz + D = 0, where A, B, and C are the components of the normal vector, and D is a constant term. This equation represents all the points that lie on the plane.

5. Can you find the equation of a plane if you only have one point and the normal vector?

Yes, you can find the equation of a plane if you have one point and the normal vector. You can use the point to find the constant term, D, in the general equation of a plane.

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