Calculus Question: Equations of Lines in 2-Space

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SUMMARY

The discussion focuses on determining the scalar equation of a line that passes through the point (1, -4) and is perpendicular to the line defined by the equation 3x + 2y - 6 = 0. The normal vector to the given line is identified as <3, 2>, which serves as the direction vector for the new line. By using the point (1, -4) and the direction vector, participants are guided to formulate the vector equation and subsequently eliminate the parameter to derive the scalar equation.

PREREQUISITES
  • Understanding of scalar and vector equations of lines
  • Knowledge of normal vectors in linear equations
  • Familiarity with the dot product in vector mathematics
  • Ability to manipulate linear equations in two-dimensional space
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  • Study the derivation of scalar equations from vector equations in 2D geometry
  • Learn about normal vectors and their applications in determining perpendicular lines
  • Explore the dot product and its significance in vector analysis
  • Practice solving problems involving lines and planes in higher dimensions
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Students studying calculus, particularly those focusing on geometry and linear algebra, as well as educators seeking to enhance their teaching methods in these subjects.

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Homework Statement



A line passes through the point (1, -4) and is perpendicular to the line 3x + 2y – 6 = 0. Determine a scalar equation for the line.

Was also given this: Find a vector which is normal to the line and then use the dot product of this vector and P0P.

Homework Equations



scalar equation: ax+by+c=0

The Attempt at a Solution



i'm not sure if this gets me anywhere but i turned the scalar equation into a vector equation:
Random point on line: (0,-3), therefore vector equation r=(0,-3)+t(2,-3)
Not sure where i can go from here, i just know i have to find a line perpendicular to 3x + 2y - 6 =0 that goes through the point (1, -4)

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
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Remember that a the coefficients of 3x+2y=6 give you a normal vector to the line. So use <3,2> for your direction vector and (1,-4) for your point. Write the vector equation using those and eliminate the parameter to get your scalar equation.
 
Okay thank you very much
 

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