Equation of a Plane - Find Your Solution

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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 line, represented parametrically. The point in question is (-2, 8, 10), and the line is defined by the equations x = 4 + t, y = 3t, z = 4 - 4t.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the normal vector of the plane and the direction vector of the line. There is an exploration of how to derive the normal vector from the line's direction and the implications of the point provided.

Discussion Status

Some participants have offered insights into identifying the direction vector of the line and its relationship to the normal vector of the plane. Others have reiterated the need to establish the normal vector to proceed with the equation of the plane.

Contextual Notes

There is mention of a study group that engaged in the problem for an extended period, indicating the complexity and collaborative nature of the discussion. The original poster expresses uncertainty about deriving the normal vector, which remains a focal point of the conversation.

fball558
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equation of a plane??

Homework Statement



Find an equation of the plane through the point and perpendicular to the given line.
(-2, 8, 10)
x = 4 + t, y = 3t, z = 4 - 4t


The Attempt at a Solution



i know the point (-2,8,10) is the starting positon (call it R0)
i know that N (normal) is perpendicular to R - R0 where R is ending position
R = (x,y,z) so i said
R= (4+t,3t,4-4t)
the N (normal) = (a,b,c) and i don't know how to get these numbers. if i can find out these i can plug it into the scalor equation of a plane formula and be set.

any help would be great!
thanks
 
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You need to find the direction of x = 4 + t, y = 3t, z = 4 - 4t, this direction is parallel to the normal of your plane

eg. x=2t,y=3t,z=t

[x] [0+2t] [0] [2t] [0] [2]
[y] = [0+3t] = [0] + [3t] = [0] =t[3]
[z] [0+t] [0] [t] [0] [1]

so <2,3,1> is the direction of that line

Do the same for your line.
 


fball558 said:

Homework Statement



Find an equation of the plane through the point and perpendicular to the given line.
(-2, 8, 10)
x = 4 + t, y = 3t, z = 4 - 4t


The Attempt at a Solution



i know the point (-2,8,10) is the starting positon (call it R0)
i know that N (normal) is perpendicular to R - R0 where R is ending position
R = (x,y,z) so i said
R= (4+t,3t,4-4t)
the N (normal) = (a,b,c) and i don't know how to get these numbers. if i can find out these i can plug it into the scalor equation of a plane formula and be set.

any help would be great!
thanks
In general, if vector <A, B, C> is perpendicular the plane and (p,q,r) is a point in the plane, then, for a general point (x,y,z) in the plane, <x- p, y- q, z- r> is a vector in the plane and so <A, B, C>.<x- p, y- q, z- r>= A(x- p)+ B(y- q)+ C(z- r)= 0.

In your case, a vector in the direction of the line x = 4 + t, y = 3t, z = 4 - 4t is <1, 3, -4>, the coefficients of t in each component.
 


thanks a lot everyone. i got in a little study group with a few of my friends and we strugled for a while but finally found it out.
(5 hours for 30 problems... not too bad) haha
 

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