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

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To find the scalar equation of a line given a normal vector [3,1] and a point (2,4), the initial attempt used the parametric form R=[2,4] + T[3,1]. The equations derived were X=2+3T and Y=4+T, leading to the relation (X-2)/3=Y-4. However, the user struggled with finding the correct parallel vector, leading to confusion over the dot product and the resulting equations. Ultimately, the user noted that their final answer was also incorrect, indicating a need for further clarification on 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.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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