Plane Equations Homework: Parallel, Coincident, Distance

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Homework Help Overview

The discussion revolves around determining relationships between planes in three-dimensional space, specifically addressing whether two given planes are parallel, orthogonal, or coincident. Additionally, participants explore finding the equation of a plane parallel to a given plane and calculating the distance from a point to a plane.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using dot and cross products of normal vectors to assess relationships between planes. Questions arise about the conditions for planes being parallel or coincident. There are attempts to derive the equation of a plane given a point and a normal vector, along with inquiries about the validity of calculations and assumptions regarding distance.

Discussion Status

The conversation includes various attempts to clarify concepts and calculations. Some participants seek verification of their reasoning and results, while others pose questions about the underlying principles of plane equations and distances. There is no explicit consensus, but the discussion is active with multiple interpretations being explored.

Contextual Notes

Participants are encouraged to share their attempts and clarify where they are stuck, indicating a collaborative approach to problem-solving. There is a focus on ensuring understanding of the mathematical principles involved, particularly regarding the properties of planes and distances in three-dimensional geometry.

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Homework Statement



a)Determine whether the following two planes 3x-y+z=2 and 2x+2y-4z=5 are parallel, orthogonal,coincident (same) or none.

b)Find the equation of the plane that contains the point (2,3,-1) and parallel to the plane 5x-3y+2z=10

c)Find the distance from the point (3,5,-8) to the plane 6x-3y+2z=28


Homework Equations





The Attempt at a Solution

 
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Hi fazal! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:
 
for A)
take the dot product of the two normal vectors. if its zero then they're perpendicular ?
take the cross product of the two normal vectors. if its zero then they're parallel ?

For them to be the same they'd have to be linear multiples of each other? how?

B)Parallel to the plane 5x-3y+2z=10 means a vector normal to the plane is (5, -3, 2). So the equation has the form 5x - 3y + 2z = d. To get d, substitute the point (2,3,-1).

therefore is the equation 5x-3y+2z=-1 after sub the above??
plse check for me...

c)For c, use the formula:

so the answer is:- 41/7
 
plse assist to check
 
fazal said:
for A)
take the dot product of the two normal vectors. if its zero then they're perpendicular ?
take the cross product of the two normal vectors. if its zero then they're parallel ?
What are the normal vectors for these planes? Are they parallel? Are they perpendicular?

For them to be the same they'd have to be linear multiples of each other? how?
For two vectors to be parallel, not "the same", one has to be a multiple of the other: <3, 2, 1> and <6, 4, 2> are parallel because <6, 4, 2>= 2<3, 2, 1>.

B)Parallel to the plane 5x-3y+2z=10 means a vector normal to the plane is (5, -3, 2). So the equation has the form 5x - 3y + 2z = d. To get d, substitute the point (2,3,-1).

therefore is the equation 5x-3y+2z=-1 after sub the above??
plse check for me...
Why would you need someone else to check for you? Is 5(2)- 3(3)+ 2(-1)= 1?

c)For c, use the formula:

so the answer is:- 41/7
Is that supposed to be negative? A distance is never negative.
 

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