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Math Help - Why's that these are normal vectors N1 and N2 from equation of 2 lines?

  1. #1
    Senior Member x3bnm's Avatar
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    Why's that these are normal vectors N1 and N2 from equation of 2 lines?

    There's an example on "Elementary Geometry of Differentiable Curves:An Undergraduate Introduction" by Gibson on page 9 which is:


    "Example 1.10 Let N = (a,b) be a non-zero vector, and let L be a line with equation  z \cdot N = c. By a direction for the line we mean any
    vector T orthogonal to N: it is an example of a 'tangent' vector.(Chapter 2) Thus we could take T = (-b,a). Alternatively, we could choose
    any two distinct points p, q on the line, and take T = q - p: the relation T \cdot N = 0 then follows from p \cdot N + c = 0 , q \cdot N + c = 0
    on subtraction. On this basis we say that two lines a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0 are orthogonal when the corresponding 'normal' vectors
    N_{1} =  (a_1, b_1), N_{2} =  (a_2, b_2) are orthogonal, or equivalently the corresponding 'tangent' vectors T_1 = (-b_1, a_1), T_2 = (-b_2, a_2) are orthogonal:
    either way, the condition is that a_1 a_2 + b_1 b_2 = 0."


    My problem is: How did the author picked (a_1, b_1) and (a_2, b_2) to be normal vectors from the two equations?

    Can anyone kindly tell me how and why are these normal vectors? And also how did the author(I know he's is 100% right) find out that (a_1 , b_1) and (a_2, b_2)
    are orthogonal? I can't seem to see the big picture.
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    Senior Member x3bnm's Avatar
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    Re: Why's that these are normal vectors N1 and N2 from equation of 2 lines?

    Quote Originally Posted by x3bnm View Post
    There's an example on "Elementary Geometry of Differentiable Curves:An Undergraduate Introduction" by Gibson on page 9 which is:


    "Example 1.10 Let N = (a,b) be a non-zero vector, and let L be a line with equation  z \cdot N = c. By a direction for the line we mean any
    vector T orthogonal to N: it is an example of a 'tangent' vector.(Chapter 2) Thus we could take T = (-b,a). Alternatively, we could choose
    any two distinct points p, q on the line, and take T = q - p: the relation T \cdot N = 0 then follows from p \cdot N + c = 0 , q \cdot N + c = 0
    on subtraction. On this basis we say that two lines a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0 are orthogonal when the corresponding 'normal' vectors
    N_{1} =  (a_1, b_1), N_{2} =  (a_2, b_2) are orthogonal, or equivalently the corresponding 'tangent' vectors T_1 = (-b_1, a_1), T_2 = (-b_2, a_2) are orthogonal:
    either way, the condition is that a_1 a_2 + b_1 b_2 = 0."


    My problem is: How did the author picked (a_1, b_1) and (a_2, b_2) to be normal vectors from the two equations?

    Can anyone kindly tell me how and why are these normal vectors? And also how did the author(I know he's is 100% right) find out that (a_1 , b_1) and (a_2, b_2)
    are orthogonal? I can't seem to see the big picture.


    I found the solution. Suppose N = (a, b) is a normal to a line that goes through a given point (x_0, y_0)

    N \cdot (x - x_0, y - y_0) = 0

    (a,b) \cdot (x - x_0, y - y_0) = 0

     a x + b y - (a x_0 + b y_0) = 0

    Let c = -( a x_0 + b y_0 )

    so the equation becomes:
    a x + b y + c = 0 and So (a,b) is normal. Because two lines are perpendicular their normals also has to be orthogonal to each other.
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    Re: Why's that these are normal vectors N1 and N2 from equation of 2 lines?

    It's in analogy to normals to planes. If you look at a plane

    ax+by+cz+d = 0

    a normal vector to this plane is N = (a,b,c). It's 'normal' to every line through the plane. Now if you set c = 0 you obtain the intersection of the plane with the xy-plane. A normal vector is N = (a,b,0), which can be seen as lying in the xy-plane. All you do now is disregard the third dimension and you get your situation.
    Thanks from x3bnm
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    Senior Member x3bnm's Avatar
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    Re: Why's that these are normal vectors N1 and N2 from equation of 2 lines?

    Quote Originally Posted by Vlasev View Post
    It's in analogy to normals to planes. If you look at a plane

    ax+by+cz+d = 0

    a normal vector to this plane is N = (a,b,c). It's 'normal' to every line through the plane. Now if you set c = 0 you obtain the intersection of the plane with the xy-plane. A normal vector is N = (a,b,0), which can be seen as lying in the xy-plane. All you do now is disregard the third dimension and you get your situation.
    Thanks.
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    Re: Why's that these are normal vectors N1 and N2 from equation of 2 lines?

    i think it's important to note that the equation:

    z \cdot N = c

    only is a line in a space of two dimensions. in more dimensions, say n, what you get is an affine hyperplane (with n-1 dimensions).

    let's look at this from a "high-school perspective".

    say we have a line given by:

    ax + by = c. because of the peculiarities of x as a "preferred independent variable", we have two cases to consider:

    case 1) b = 0.

    here, our line is: ax + c = 0. again, we have two cases to consider:

    case 1a) a = 0.

    in this case, c must be 0, as well, and any (x,y) will suffice (0x + 0y + 0 = 0 is true for any point in the plane). so our "line" is actually the entire plane! however, this case is actually ruled out by taking N to be non-zero (if N = 0 were allowed, then every vector in the plane is indeed orthogonal to the origin. strange, but true).

    case 1b) a ≠ 0 (which must be true if b = 0).

    here, x = c/a, which is a vertical line. so a tangent vector is (x,0) (for any x) and a normal vector is (0,y) for any y. it is perfectly acceptable to choose x = y = a.

    case 2a) a = 0, b ≠ 0. this is similar to case 1b) and you can work out the details yourself. (0,b) is a tangent vector, and (-b,0) is a normal vector (you could use (b,0) as well...at every point on a line, there are "two" normals. usually we pick an orientation of the plane, to choose a "positive" one (the one at 90 degrees, instead of -90 degrees)).

    case 2b) a ≠ 0, b ≠ 0. this is the "interesting one".

    typically, we define two lines:

    y = m1x + b1
    y = m2x + b2

    to be perpendicular if m1m2 = -1.

    so suppose we have our lines as:

    a1x + b1y = c1
    a2x + b2y = c2, what does perpendicularity mean in terms of the a's,b's and c's?

    putting the two lines in slope-intercept form we have:

    y = (-a1/b1)x + c1/b1 and
    y = (-a2/b2)x + c2/b2

    hmm...note that the slopes don't involve "c's" at all, just a's and b's.

    thus m1m2 = -1 becomes:

    (-a1/b1)(-a2/b2) = -1 that is:

    a1a2 = -b1b2 or:

    a1a2 + b1b2 = 0.

    one often identifies the direction of a line with its slope: so a "parallel" vector to the first line going through the origin lies on the line:

    y = m1x = (-a1/b1)x. we can choose any x we please, so let's choose x = -b1.

    this gives us a parallel vector (x,y) = (-b1,a1).

    now all we need to do is verify that N = (a1,b1) lies on "the perpendicular line":

    y = m2x. so we plug in the value x = a1 into this formula:

    y = m2a1 = (-a2/b2)a1 = (-a1/b1)(-a2/b2)(-b1)

    = (-1)(-b1) = b1.

    one caveat: these vectors, T and N actually lie in the "tangent space at the point P on the line L" (which is where the "c" comes in). so technically, we should write:

    N = (a,b)P, T = (-b,a)P to indicate the "tails" of the vectors (arrows) start at the point P, rather than at the origin. geometrically, this means translating the origin to the point P on the line.
    Last edited by Deveno; August 29th 2012 at 05:48 PM.
    Thanks from x3bnm
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  6. #6
    Senior Member x3bnm's Avatar
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    Re: Why's that these are normal vectors N1 and N2 from equation of 2 lines?

    Quote Originally Posted by Deveno View Post
    i think it's important to note that the equation:

    z \cdot N = c

    only is a line in a space of two dimensions. in more dimensions, say n, what you get is an affine hyperplane (with n-1 dimensions).

    let's look at this from a "high-school perspective".

    say we have a line given by:

    ax + by = c. because of the peculiarities of x as a "preferred independent variable", we have two cases to consider:

    case 1) b = 0.

    here, our line is: ax + c = 0. again, we have two cases to consider:

    case 1a) a = 0.

    in this case, c must be 0, as well, and any (x,y) will suffice (0x + 0y + 0 = 0 is true for any point in the plane). so our "line" is actually the entire plane! however, this case is actually ruled out by taking N to be non-zero (if N = 0 were allowed, then every vector in the plane is indeed orthogonal to the origin. strange, but true).

    case 1b) a ≠ 0 (which must be true if b = 0).

    here, x = c/a, which is a vertical line. so a tangent vector is (x,0) (for any x) and a normal vector is (0,y) for any y. it is perfectly acceptable to choose x = y = a.

    case 2a) a = 0, b ≠ 0. this is similar to case 1b) and you can work out the details yourself. (0,b) is a tangent vector, and (-b,0) is a normal vector (you could use (b,0) as well...at every point on a line, there are "two" normals. usually we pick an orientation of the plane, to choose a "positive" one (the one at 90 degrees, instead of -90 degrees)).

    case 2b) a ≠ 0, b ≠ 0. this is the "interesting one".

    typically, we define two lines:

    y = m1x + b1
    y = m2x + b2

    to be perpendicular if m1m2 = -1.

    so suppose we have our lines as:

    a1x + b1y = c1
    a2x + b2y = c2, what does perpendicularity mean in terms of the a's,b's and c's?

    putting the two lines in slope-intercept form we have:

    y = (-a1/b1)x + c1/b1 and
    y = (-a2/b2)x + c2/b2

    hmm...note that the slopes don't involve "c's" at all, just a's and b's.

    thus m1m2 = -1 becomes:

    (-a1/b1)(-a2/b2) = -1 that is:

    a1a2 = -b1b2 or:

    a1a2 + b1b2 = 0.

    one often identifies the direction of a line with its slope: so a "parallel" vector to the first line going through the origin lies on the line:

    y = m1x = (-a1/b1)x. we can choose any x we please, so let's choose x = -b1.

    this gives us a parallel vector (x,y) = (-b1,a1).

    now all we need to do is verify that N = (a1,b1) lies on "the perpendicular line":

    y = m2x. so we plug in the value x = a1 into this formula:

    y = m2a1 = (-a2/b2)a1 = (-a1/b1)(-a2/b2)(-b1)

    = (-1)(-b1) = b1.

    one caveat: these vectors, T and N actually lie in the "tangent space at the point P on the line L" (which is where the "c" comes in). so technically, we should write:

    N = (a,b)P, T = (-b,a)P to indicate the "tails" of the vectors (arrows) start at the point P, rather than at the origin. geometrically, this means translating the origin to the point P on the line.

    Thanks Deveno for a comprehensive and well-thought-out proof. It will help me a lot in understanding the underlying properties of tangent and normal. Again thanks.
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