How do you find a normal of a direction vector in 3 space?

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Homework Help Overview

The discussion revolves around finding a normal vector to a direction vector in three-dimensional space, as well as determining the scalar equation of a plane that contains the x-axis and a specific point.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the concept of normals in three-dimensional space, questioning the existence of a single normal vector and discussing the implications of a plane containing the x-axis. There are inquiries about the steps needed to find a normal vector and the scalar equation of the plane.

Discussion Status

The discussion is ongoing, with participants expressing uncertainty about the steps involved in finding a normal vector and the corresponding plane equation. Some guidance has been offered regarding the cross product and the relationship between vectors and planes, but clarity is still sought.

Contextual Notes

Participants are navigating the complexities of vector mathematics and plane equations, with some confusion about terminology and the process involved in deriving the normal vector and plane equation.

thomasrules
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Question #1. How do you find a normal of a direction vector in 3 space?
Question #2.:

What is the scalar equation of the plane that contains the x-axis and the point (4,-2,1)?
 
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There is no single normal to a vector in 3-space. There exist an entire plane normal to any vector and any vector in that plane is normal to the original plane.

If a plane includes the entire x-axis, then it also includes the vector from (0,0,0) to (1, 0, 0): i. Since this plane includes the point (4, -2, 1) as well as (0,0,0) it includes the vector 4i- 2j+ k and so the cross product of those two vectors is perpendicular to the plane: -j- 2k is perpendicular to the entire plane. If (x, y, z) is any point in the plane, then the vector from (0,0,0) to (x,y,z), xi+ yj+ zk, is perpendicular to that vector: (xi+ yj+ zk).(-j- 2k)= 0 or -y- z= 0. The desired plane is z= -y.
 
hallsofivy...that is not the answer...I think you are wrong...can you please make it a bit more clear on the steps made to get the answer.

for my first question ok sorry maybe I worded it wrong...how do i find a normal plane of a direction vector in 3 space?
 
or maybe ur right ...thats the plane...but not the answer

I'm not sure how to find the normal...(1,0,0)cross what? to get a normal...then after that put the (x,y,z) values in the equation Ax+By+Cz+D=0
Then find D by inserting the point (4,-2,1) and you've got the equation?

Is this process correct
 
Last edited:

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