Find a nonzero vector normal to a plane

Click For Summary
SUMMARY

The discussion focuses on finding a nonzero vector normal to the plane defined by the equation -5x - y + z + 9 = 0. The correct normal vector is identified as (-5, -1, 1), which is a scalar multiple of other valid normal vectors such as (5, 1, -1) or (1, 0.2, -0.2). The user expresses confusion regarding the final form of the vector equation, specifically whether to use the parameterized form r = ri + tv. Clarification is provided that there are infinitely many normal vectors, all scalar multiples of the original vector.

PREREQUISITES
  • Understanding of vector mathematics and normal vectors
  • Familiarity with the equation of a plane in three-dimensional space
  • Knowledge of parameterized equations of lines
  • Basic concepts of scalar multiplication in vector algebra
NEXT STEPS
  • Study the properties of normal vectors in three-dimensional geometry
  • Learn about parameterized equations of lines and their applications
  • Explore the concept of unit vectors and how to derive them from normal vectors
  • Investigate the relationship between scalar multiples of vectors and their geometric interpretations
USEFUL FOR

Students studying vector calculus, geometry enthusiasts, and anyone needing to understand the properties of planes and normal vectors in three-dimensional space.

PsychonautQQ
Messages
781
Reaction score
10

Homework Statement


Find a nonzero vector normal to the plane -5x -y +z +9 = 0



Homework Equations





The Attempt at a Solution


so the direction of the vector would be (a,b,c) = (-5,-1,1)
I'm not exactly sure what the final form should look like..
is r = ri + tv on the right track? where i plug (-5,-1,1) in for v and for ri i can plug in any numbers that make -5x - 6 + z +9 = 0?
If so I could theoretically plug in (1,4,0)
so
r= (1i + 4j + 0k) + t(-5i -1j +1k)
r = (1-5t)i + (4-t)j + (t)k

The internet program says I'm wrong.. halp?
 
Physics news on Phys.org
PsychonautQQ said:

Homework Statement


Find a nonzero vector normal to the plane -5x -y +z +9 = 0



Homework Equations





The Attempt at a Solution


so the direction of the vector would be (a,b,c) = (-5,-1,1)
I'm not exactly sure what the final form should look like..
is r = ri + tv on the right track? where i plug (-5,-1,1) in for v and for ri i can plug in any numbers that make -5x - 6 + z +9 = 0?
If so I could theoretically plug in (1,4,0)
so
r= (1i + 4j + 0k) + t(-5i -1j +1k)
r = (1-5t)i + (4-t)j + (t)k

The internet program says I'm wrong.. halp?

Your normal looks fine, but there are an infinite number of vectors that are perpendicular to (normal to) a given plane. However, all of the vectors are scalar multiples of one another. I suspect that the answer the program was looking for was <5, 1, -1>. Or maybe <1, .2, -.2>.

Did it mention anything about wanting a unit vector?
 

Similar threads

Replies
1
Views
1K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K