关于过原点与点Q(1,1,1)且垂直于地面的平面方程及交线求解的技术问询

Hey there! Let's unpack your questions one by one to get everything sorted out.

First: Clarifying the Plane Equation & Your Confusion About Point Q #

First off, let's clear up the wording confusion: A point can't be "perpendicular to the ground"—that phrase doesn't make sense geometrically. It's far more likely the problem is asking for a plane that passes through point Q(1,1,1) (and possibly the origin, as you assumed) while being perpendicular to the ground. Let's verify your current answer and adjust if needed.

Your core approach using the plane formula n · (P - P₀) = 0 is totally correct! But let's check the assumptions you made:

• You chose n = <1,1,1> as the normal vector and P₀ = (0,0,0) as the plane's reference point, leading to the equation x + y + z = 0. Mathematically, this is a valid plane passing through the origin with that normal. However, this plane isn't perpendicular to the ground—let's confirm why.

In standard coordinate systems, the ground is the xy-plane, with the equation z = 0. Its normal vector is <0,0,1>. For two planes to be perpendicular, their normal vectors must have a dot product of 0 (meaning they're perpendicular to each other). Your plane's normal <1,1,1> dotted with the ground's normal <0,0,1> gives 1*0 + 1*0 + 1*1 = 1 ≠ 0—so your current plane doesn't meet the "perpendicular to ground" requirement, if that's part of the problem.

What's the Correct Plane Equation (If Perpendicular to Ground is Required)? #

If the plane needs to:

1. Pass through the origin and point Q(1,1,1)
2. Be perpendicular to the ground (xy-plane)

We need a normal vector that's perpendicular to both the vector OQ (<1,1,1>) and the ground's normal (<0,0,1>). Calculate this cross product:

<1,1,1> × <0,0,1> = <(1*1 - 1*0), (1*0 - 1*1), (1*0 - 1*0)> = <1, -1, 0>

Using this normal and the origin as the reference point, the plane equation becomes:

1*(x-0) - 1*(y-0) + 0*(z-0) = 0 → x - y = 0

This plane passes through Q(1,1,1) (1-1=0 checks out) and is perpendicular to the ground (its normal <1,-1,0> dotted with <0,0,1> equals 0).

If the problem doesn't require perpendicularity to the ground, your original equation x + y + z = 0 is perfectly correct.

Second: Finding the Line of Intersection with the Ground #

First, confirm the ground plane equation: In standard settings, ground is the xy-plane, so z = 0. Your approach to find the intersection line is spot-on—here's how to execute it for both possible planes:

Case 1: Your Original Plane x + y + z = 0 & Ground z = 0 #

1. Direction vector: Cross product of the two planes' normals

   <1,1,1> × <0,0,1> = <1, -1, 0>
   

2. Common point: Solve both plane equations. Set z=0, so x + y = 0. We can use the origin (0,0,0) (it lies on both planes)
3. Parametric line equation:

   x = t
   y = -t
   z = 0
   

Case 2: Perpendicular-to-Ground Plane x - y = 0 & Ground z = 0 #

1. Direction vector: Cross product of normals

   <1,-1,0> × <0,0,1> = <-1, -1, 0> (or simplify to <1,1,0> by multiplying by -1)
   

2. Common point: Use (0,0,0) (satisfies both x=y and z=0)
3. Parametric line equation:

   x = t
   y = t
   z = 0
   

Key Takeaways #

• If perpendicularity to the ground isn't required, your original plane equation x + y + z = 0 is correct.
• If the plane must be perpendicular to the ground (and pass through origin + Q), use x - y = 0.
• For intersection lines, always start with the ground plane z=0, compute the cross product of normals for direction, find a common point, and write the parametric equations.

备注：内容来源于stack exchange，提问作者Haider H.H