Find equation of plane given conditions

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To find the equation of a plane that passes through the point (1,2,-2) and contains the line defined by the parametric equations x=2t, y=3-t, z=1+3t, one must first determine if the point lies on the line. If it does, a unique plane cannot be established. If not, two points on the line can be derived from its parametric equations, allowing the formation of two vectors in the plane. The cross product of these vectors will yield the normal vector, which, along with the given point, can be used to formulate the plane's equation. Understanding these relationships is crucial for solving the problem effectively.
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Homework Statement


Find an equation of the plane that passes through the point (1,2,-2) and that contains the line
x=2t,y=3-t,z=1+3t.

Homework Equations

The Attempt at a Solution



I know that a plane is determined by a base vector and a normal vector, and the equation of the plane is ##\vec{n} \cdot (\vec{r} - \vec{r_0})##, where n is the normal vector, and r0 is the base vector.

So we know the base vector, and now must determine what the normal vector is to the plane. We need to find two vectors in the plane, and take the cross product of those vectors. I am not sure where to find those two vectors... Would any two non-parallel vectors whose endpoints are on the line suffice?
 
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Hi Mr Davis:
I suggest thinking about the problem a bit differently.

I think it would help if you write down, in section 2, the general form of the "equation of the plane". The identity of three points in the plane should lead you to three simultaneous equations to solve.

Good luck.

Regards,
Buzz
 
Is the given point on the line? If so, there is no unique plane.

If the given point is not on the line, use the parametric equations of the line to find two points. then you'll have three points, from which you can get two vectors in the plane. From these vectors, you can get a normal to the plane, and from it and the given point, getting the equation of the plane is straightforward.
 
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