Finding scalar equation of a line with a vector and a point given

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Homework Help Overview

The discussion revolves around finding the scalar equation of a line given a normal vector and a point. The subject area includes vector mathematics and line equations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the normal vector and the line's equation, with attempts to derive the scalar equation from given parameters. Some question the validity of their approaches and whether the chosen points and vectors are appropriate.

Discussion Status

There is ongoing exploration of different methods to derive the equation, with some participants expressing uncertainty about their calculations and results. Multiple interpretations of the problem are being discussed, but no consensus has been reached regarding the correct approach or solution.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is also mention of incorrect answers, indicating potential misunderstandings or miscalculations in the process.

BlazeKH
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I would really appreciate the help, I've been trying to figure this out for the last three hours no joke.

Homework Statement



Write the scalar equation the line given the normal vector [3,1] and point (2,4)

Homework Equations



R=[X0,Y0]+ T[M1,M2]


The Attempt at a Solution



R=[2,4] + T[3,1]

X=2+3T Y=4+T

(X-2)/3=T Y-4=T

(X-2)/3=Y-4

X-2=3Y-12

X-3Y+10=0

ANSWER:3X+2Y-10
 
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You need to find the vector parallel to the line you want i.e. the vector normal to the normal.

The dot product of the parallel line and the normal is zero. Use that to get the required parallel vector.
 
I tried that but it didn't work, am I doing something wrong?

[X,Y].[3,1]=0

3X+Y=0

Y=-3X (I subbed in a point, to my understanding it shouldn't matter which)

Y=-3(1)

Y=-3

P=[1,-3]

R=[2,4]+T[1,-3]

X=2+T Y=4-3T

X-2=T (Y-4)/-3=T

-3X+6=Y-4

-3X-Y+10=0
 
Okay I figured it out the answer is also incorrect btw.
 

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