How Do You Determine the Implicit Equation of a Plane from a Line and Origin?

[Loppyfoot]

Homework Statement

Find the implicit form for the plane that contains the origin and the line:

L(t) = <1+t,1-t,2t>

The Attempt at a Solution

'

So, the point P = (1,1,0) and vector v= <1,-1,2>.

To find the implicit equation for the plane using the form ax + by + cz = d , I will substitute the vector values in for a, b, and c. So I get

x -y + 2z = d.

My question is, should I use the origin points, (o,o,o) and substitute them in for x,y, and z to get d=0, or is there another process to this?

__________________________________________________________________________________________________________________#2. The function f(t) = (t^@,1/t) represents a curve in the plane parametrically. write an equation in parametric form for the tangent line to this curve at the point where t= 2.

So I solve the gradient: <2t, -1/(t^2)> and at t=2 the point is (4, 1/4).
and the gradient normal to t=2 is <4,-1/4>.

So would the parametric equation be (4,1/4) +t<4,-1/4> ?

Thanks a lot in advance!

 

[Mark44]

Homework Statement

Find the implicit form for the plane that contains the origin and the line:

L(t) = <1+t,1-t,2t>

The Attempt at a Solution

'

So, the point P = (1,1,0) and vector v= <1,-1,2>.

What about them? A clearer way of saying this is that P = (1, 1, 0) is on the line, and the vector v = <1, -1, 2> has the same direction as the line.

What you know about the plane is that it contains the point (0, 0, 0) and the line.

To find the equation of a plane, you need one point on the plane and the normal to the plane. If you have two vectors that lie in the plane, how can you get a third vector that is normal to the plane?

To find the implicit equation for the plane using the form ax + by + cz = d , I will substitute the vector values in for a, b, and c. So I get

x -y + 2z = d.

My question is, should I use the origin points, (o,o,o) and substitute them in for x,y, and z to get d=0, or is there another process to this?

__________________________________________________________________________________________________________________


#2. The function f(t) = (t^@,1/t) represents a curve in the plane parametrically. write an equation in parametric form for the tangent line to this curve at the point where t= 2.

So I solve the gradient: <2t, -1/(t^2)> and at t=2 the point is (4, 1/4).
and the gradient normal to t=2 is <4,-1/4>.

So would the parametric equation be (4,1/4) +t<4,-1/4> ?

Thanks a lot in advance!

 

[Loppyfoot]

Ok, so since the vector is pointing in the same direction as the line, I understand that I should use the cross product between two vectors. I understand that I have one vector, <1,-1,2> but how would I find the second vector? Should I use the vector <-1,1,-2> and take the cross product between those two?

 

[Mark44]

Ok, so since the vector is pointing in the same direction as the line, I understand that I should use the cross product between two vectors. I understand that I have one vector, <1,-1,2> but how would I find the second vector? Should I use the vector <-1,1,-2> and take the cross product between those two?

No. If you cross those two vectors you will only get the zero vector, since the 2nd vector is just -1 times the first vector.

Draw a sketch of the plane and identify three points that you know are in the plane. From these three points, form two vectors that aren't parallel (or antiparallel, as are your two vectors).

 

[Loppyfoot]

Is there a way to find that point computationally without sketching?

Well, could I use the points (0,0,0), (1,1,0) and (2,0,2)? I got (2,0,2) by f(1).

 

[Loppyfoot]

Is that the process that should be done?

 

[Mark44]

Is there a way to find that point computationally without sketching?

Maybe, but why would you want to handicap yourself by using only half of your brain on the problem?

Well, could I use the points (0,0,0), (1,1,0) and (2,0,2)? I got (2,0,2) by f(1).

Sure, those are all points in the plane. Now form two vectors and use them to get a third vector that is perpendicular to the plane.