How to Find the Normal Vector for a Plane Perpendicular to Another?

Click For Summary

Homework Help Overview

The discussion revolves around finding the scalar equation of a plane that is perpendicular to another plane defined by the normal vector [3,1,2]. The problem involves points A(2,-6,-1) and B(1,2,-4) and seeks to determine the normal vector for the new plane.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to understand how to derive the normal vector for the new plane, questioning the relationship between direction vectors and normal vectors. Some participants clarify that the normal vector of the original plane lies within the context of the new plane, while others introduce the concept of the cross product as a method to find a perpendicular vector.

Discussion Status

The discussion is ongoing, with participants exploring the concept of perpendicularity in relation to vectors. Some guidance has been offered regarding the use of the cross product to find a vector that is perpendicular to both the given normal vector and the vector between points A and B.

Contextual Notes

There is a mention of the original poster's uncertainty regarding the definition of a direction vector in the context of planes, as well as the need for clarification on how to compute a vector that is perpendicular to two given vectors.

eme_girl
Messages
5
Reaction score
0
Find the scalar eq'n of a plane that is perpendicular to the plane with normal vector [3,1,2] and passes through points A(2,-6,-1) and B(1,2,-4).

I think that the normal vector can be the direction vector of this new plane. But then, in order to find the scalar eq'n I need a normal vector of this new plane. How do I find this?
 
Physics news on Phys.org
Strictly speaking a plane doesn't have a "direction vector". What is true is that the vector [3,1,2] is in the plane you want. You also know that the vector from A(2,6,-1) to B(1,2,-4) (which is, of course, [1-2,2-6,-4-(-1)]= [-1, -4, -3] is in the plane. Do you know how to find a vector that is perpendicular to both [3,1,2] and [1,4,-3]?
 
I understand what you just found. But no, I do not know how to find a vector's that perpendicular to both those vectors.
 
eme_girl,
do u know the direction of a vector that is a cross product of two vectors?

-- AI
 
TenaliRaman's point: the cross product of two vectors is always perpendicular to both.

The cross product of [a1,a2,a3] and [b1,b2,b3] is the vector [a2b3-a3b2,a3b1-a1b3,a1b2-a2b1].
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
1K
Replies
16
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
4K
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
16
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K