Tangent plane, why is it orthongonal and not parallel?

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Homework Help Overview

The discussion revolves around finding the equation of a plane that passes through a specific point and is perpendicular to a given vector. The original poster questions the relationship between the normal vector of the plane and its perpendicularity to the specified vector.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definition of a normal vector and its relationship to the plane's orientation. There are questions about the implications of perpendicularity and whether the normal vector can be considered parallel to the plane.

Discussion Status

Participants are actively questioning the definitions and relationships between vectors and planes. Some guidance has been offered regarding the interpretation of normal vectors, but there is no explicit consensus on the underlying concepts yet.

Contextual Notes

There is a potential misunderstanding regarding the terminology used to describe the relationship between the plane and the vectors involved. The discussion includes references to scalar multiples and the nature of direction in relation to the plane.

flyingpig
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Homework Statement




Find equation of the plane

The plane though P(6,3,2) and is perpendicular to vector <-2,1,5>

Why would it be -2(x - 6) + (y - 3) + 5(z - 2) = 0?

If it is perpendicular to <-2,1,5>, shouldn't be the cross product of some other vector with this? Using <-2,1,5> wouldn't mean it is parallel to the vector instead?



The Attempt at a Solution

 
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Are you saying the CROSS product of two perpendicular vectors is zero? If not what are you saying?
 
I am saying that

[PLAIN]http://img34.imageshack.us/img34/9647/unledseh.jpg

A plane's direction is determined by it's normal vector right? So if it is perpendicular to the vector <2,-1,5>, why would its normal vector be the same as <2,-1,5>?
 
Last edited by a moderator:
flyingpig said:
I am saying that

[PLAIN]http://img34.imageshack.us/img34/9647/unledseh.jpg

A plane's direction is determined by it's normal vector right? So if it is perpendicular to the vector <2,-1,5>, why would its normal vector be the same as <2,-1,5>?

Perpendicular means the same thing as normal, as far as I know. If the plane is perpendicular to <2,-1,5> then it's also normal to <2,-1,5>. I'm really not sure what your question is. The normal vector is not parallel to the plane if that's what confusing you. Describing it as the 'direction of the plane' doesn't mean it's parallel to the plane.
 
Last edited by a moderator:
But <2,-1,5> is just a scalar multiple of <2,-1,5> (scalar = 1), that is parallel?
 
flyingpig said:
But <2,-1,5> is just a scalar multiple of <2,-1,5> (scalar = 1), that is parallel?

I may have figured out what is confusing you. A plane doesn't have a unique parallel direction. So the phrase 'direction of the plane' means its normal. Not its parallel. See my previous post.
 
Is it because of this?

[PLAIN]http://img841.imageshack.us/img841/6863/unleditl.jpg
 
Last edited by a moderator:
I have no clue what that is supposed to represent. However, I do note that you asked about a plane and a vector but your pictures show only two vectors, no plane.
 
Oh, that is supposed to be a plane and the two vectors lie on the plane
 

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