4. Application of vectors to metric problems: NORMAL VECTOR
Block:Geometry
 
Application of vectors to metric problems
4.1. NORMAL VECTOR TO A STRAIGHT LINE
 
A line perpendicular to a vector is called a normal vector.
Line in parameters The vector (d,-b) is normal to r, then it is perpendicular to its direction vector (b,d):  (d,-b).(b,d) = db-bd = 0
Implicit line Ax + By + C = 0 The vector (A,B) is normal to r

 
Justification that (A,B) is perpendicular to Ax + By + C = 0:
 

If P(x1,y1) and Q(x2,y2) belong to the line, its coordinates satisfy the equation:  Ax2 + By2 + C = 0 Ax1 + By1 + C = 0. Subtracting:    A(x2-x1) + B(y2-y1) = 0 

This last equality shows that  (A,B).(x2-x1,y2-y1) = 0

That is the vector of coordinates (A,B) is perpendicular to a direction vector of r, PQ and, therefore, is normal to r.

 
1.-If the line is  the direction vector is v(7,-1) and the normal vector is n(1,7) .

In this figure we have called them v(a,b) and n(c,d) 

Change a and b, so they become v(6,-3)  

You can drag the end point of the vector v, click on the buttons at the bottom, or type in the values and press enter

You will now see that the angle n formed with the line is not 90º. 
2.-Change c and d such that n(3,6) and then we will have the angle of 90º, or that now n is perpendicular to the line. Attempt to change a, b, c and d, yourself in such a way that the angle formed will be 90º.

3.-If the direction vector is v(5,-1) What could n be? Make a note in your exercise book and verify your answer in the previous figure. Write as well how that leaves the equation of the line.

4.- If the direction vector is v(-4,2) could the normal vector be n(4,8)? Write the answer in your book, justify it and verify it in the previous figure. Write also how that leaves the equation of the line.

5.- If v=(-3,1), what value should be given to d so that n(2,d) becomes a normal vector to the line. Write the answer in your book, justify it and verify it in the previous figure. Write also how that leaves the equation of the line.
4.1.1. normal vector to the line:  GENERAL EQUATION
In the figure we have the line 5x - 2y + 4 = 0. A possible normal vector is n=(c,d)=(5,-2) 

In this case the direction vector could be v=(a,b)=(2,5)

 1.-Change a and b, such that v(3,2). Now you will see that the equation of the line is 2x-3y+4=0.  

2.-If you do not change n, and continue with n(5,-2), you will see that the angle is NOT 90º. You will have to change c and d such that n(2,-3), or that c=A=2, d=B=-3, and effectively n will be the normal to the line.

Attention! there is not only one normal vector and one direction vector for the same line. They may take others which have the same direction. 
3.-Observe how you can take  v(3,2) and n(1,-1.5), and n will continue to be normal to the line. Or as well v(1.5,1) and n(-2,3) 

4.- If the direction vector is v(-5,1) What could be the value of n? Make a note of it in your workbook and verify your answer in the previous figure. Also write how this leaves the equation of the line Ax + By + C = 0.

5.- If the direction vector is v(-4,2) Could the normal vector be n(1,2) Write down your answer, justify it and verify it in the previous figure. Also write down how that leaves the equation of the line. Now it is not true that c=A and d=B. Could the the coordinates of n be correct then? Why?

6- If v=(-3,1), what value must be given to d so that n(2,d) becomes the normal vector to the line? Write down your answer, justify it and verify it in the previous figure. Also write down how that leaves the equation of the line. Now it is not true that c=A and d=B ¿Could the coordinates of n be correct then? Why?

       
           
  Ángela Núñez Castaín
 
Ministry of Education , Social Afairs and Sport. Year 2001
 
 

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