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

• August 28th 2012, 04:34 PM
x3bnm
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.
• August 28th 2012, 04:59 PM
x3bnm
Re: Why's that these are normal vectors N1 and N2 from equation of 2 lines?
Quote:

Originally Posted by x3bnm
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.
• August 28th 2012, 09:01 PM
Vlasev
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.
• August 29th 2012, 04:37 PM
x3bnm
Re: Why's that these are normal vectors N1 and N2 from equation of 2 lines?
Quote:

Originally Posted by Vlasev
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.
• August 29th 2012, 05:44 PM
Deveno
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.
• August 31st 2012, 08:36 AM
x3bnm
Re: Why's that these are normal vectors N1 and N2 from equation of 2 lines?
Quote:

Originally Posted by Deveno
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.