Finding the Normal Vector for a Plane in 3D Space - Is My Approach Correct?

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Homework Help Overview

The discussion revolves around finding the normal vector for a plane in 3D space, specifically from the scalar equation 2x − 3y + z = 5. The original poster attempts to derive a vector equation for the plane by identifying points and vectors on the plane, and then using the cross product to find the normal vector.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method of finding points on the plane and using them to create vectors. There is a focus on the use of the cross product to determine the normal vector, with questions about the correctness of the original poster's approach and calculations.

Discussion Status

Some participants have provided encouragement and asked for further calculations to identify potential mistakes. The conversation has shifted towards additional questions related to vectors in higher dimensions, indicating a broadening of the discussion beyond the initial problem.

Contextual Notes

There are mentions of challenges in posting matrices and the need for clarity in vector relationships, particularly regarding scalar multiples and their implications for the problems being discussed.

hkus10
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1) Find a vector equation for the plane in R3(3D) with scalar equation 2x − 3y + z = 5 .

First,I find three points on the plane and then I used one point as a fixed point in order to find two vectors on the plane by using two other points. Then, I tried to test whether the two vectors are perpendicular to the normal vector of this plane by using cross product. However, I do not get the right normal vector by using my two vectors on the plane?
My question is that whether my approach is wrong? If yes, what should I do?
 
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Your approach sounds fine. Show us your calculations so we can spot where the mistake lies.
 
hi hkus10! :wink:
hkus10 said:
First,I find three points on the plane and then I used one point as a fixed point in order to find two vectors on the plane by using two other points. Then, I tried to test whether the two vectors are perpendicular to the normal vector of this plane by using cross product.

do you mean that you used the cross product as the normal?

that should work :confused:

show us your full calculations, and then we'll see what went wrong! :smile:
 
is there any way to post a matrix in this forum. If yes, how?
 
easiest way is to use the CODE tags (because they preserve the spacing) :wink:

Code:
a b c
d e f
g h i
 
Oops! When I do it over, I found the right answer...this is math! However, I do have another question.

1) Let P(is a 4 by 1 vector) = (refer to quote) and v(is a 4 by 1 vector) = (refer to quote). Let Λ be the line in R4 through P and parallel to v.
P = 2 v = 1
1 -2
0 3
5 0
2,1,0,5 for p and the rest of them will be the pattern
The real question is which of the following vectors are in Λ?
a = 2 ; b = 4 ; c = -1
4 -3 -3
6 6 3
0 5 -5

For this question, do I look at whether each of a, b, or c is the scalar multiple of v? If yes, none of them will be right? If no, how to look at it?

2) Let P(is a 4 by 1 vector) = (refer to quote), u( is a 4 by 1 vector) = refer to quote) and v(is a 4 by 1 vector) = (refer to quote). Let Π be the line in R4 through P and parallel to v.
P = 0 u= 1 v = 1
2 2 2
3 -1 2
1 2 1

The real question is which of the following vectors are in Λ?
a = 2 ; b = 0 ; c = 5
4 2 12
4 0 4
4 2 9

For this question do I have to look at whether a, b, or c is not scalar multiple of u or v, or a, b, or c is not scalar multiple of each other(a,b, c)? If yes, none of them will be right?
If no, how to approach it?
 
push!
 

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