Finding the Normal Vector and Equation of a Line Perpendicular to 2x + y = 4

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SUMMARY

The discussion focuses on finding the normal vector and the equation of a line perpendicular to the line represented by the equation 2x + y = 4. The normal vector can be derived from the coefficients of the line equation, yielding the vector n = <2, 1>. To find the unit normal vector from the origin, it is essential to normalize this vector. Additionally, the equation of a line passing through the point P(0,2) and normal to the given line can be established using the normal vector and point-slope form.

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the equation of a straight line is given by 2x + y = 4

(a) find the vector equation of a unit normal from the origin to the line and (b) the equation of a line passing through P(0,2) and normal to 2x + y = 4.


i know i need to use the dot product some how but i am utterly confused as to where to begin! please help.

i think first the vector equation for the given line is r = < 1 , -2 > then i need to dot r with rperp and set equal to 0? i am very confused. thanks!
 
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Welcome to PF, imb3cile! :smile:

The general equation of a line is [itex](\vec n, \vec x) = d[/itex]
In this equation [itex]\vec n[/itex] is a normal vector to the line, [itex]\vec x[/itex] is the vector with your coordinates x and y, and d is an arbitrary constant, that creates an offset to the origin.
The parenthesis around the 2 vectors indicate that the dot product is taken.

If the vector n would have components a and b, this would turn out as: ax + by = d

Can you see from your own equation what the normal vector n would be?

For part (a) you would need a multiple of this vector, such that if you fill it in for x and y, it matches the equation.
 

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