Why do we divide by $ sqrt{A^2 + B^2} $ to convert a function from the standard form to the normal form?
$begingroup$
To convert an equation in the standard form, $ Ax + By + C = 0 $ to the normal form $ cos omega x + sin omega y = p $, we first divide by $ sqrt{A^2 + B^2} $. Why?
For example, to convert the equation $ sqrt 3 x + y - 8 = 0 $, we divide by $ sqrt{A^2 + B^2} = 2 $ to get $ frac{sqrt 3}{2} x + frac 12 y = 4 $ and then solve for $ cos omega = frac{sqrt 3}{2} $ and $ sin omega = frac 12 $to find the value of $ omega $.
I understand that we can't possibly convert this directly, since there is no $ omega $ for which $ cos omega = sqrt 3 $--but why do we specifically divide by $ sqrt{A^2 + B^2}?$ Why not any other number to bring the values of $ cos omega $ and $ sin omega $ between 0 and 1 and then solve it?
linear-algebra algebra-precalculus
edited Jan 13 at 17:43
J. W. Tanner
3,4601320
asked Jan 13 at 16:57
WorldGovWorldGov
324211
$endgroup$
add a comment |
$begingroup$
To convert an equation in the standard form, $ Ax + By + C = 0 $ to the normal form $ cos omega x + sin omega y = p $, we first divide by $ sqrt{A^2 + B^2} $. Why?
For example, to convert the equation $ sqrt 3 x + y - 8 = 0 $, we divide by $ sqrt{A^2 + B^2} = 2 $ to get $ frac{sqrt 3}{2} x + frac 12 y = 4 $ and then solve for $ cos omega = frac{sqrt 3}{2} $ and $ sin omega = frac 12 $to find the value of $ omega $.
I understand that we can't possibly convert this directly, since there is no $ omega $ for which $ cos omega = sqrt 3 $--but why do we specifically divide by $ sqrt{A^2 + B^2}?$ Why not any other number to bring the values of $ cos omega $ and $ sin omega $ between 0 and 1 and then solve it?
linear-algebra algebra-precalculus
edited Jan 13 at 17:43
J. W. Tanner
3,4601320
asked Jan 13 at 16:57
WorldGovWorldGov
324211
$endgroup$
3
$begingroup$
Because $cos^2 omega + sin^2 omega = 1$
$endgroup$
– Martin R
Jan 13 at 16:59
add a comment |
$begingroup$
To convert an equation in the standard form, $ Ax + By + C = 0 $ to the normal form $ cos omega x + sin omega y = p $, we first divide by $ sqrt{A^2 + B^2} $. Why?
For example, to convert the equation $ sqrt 3 x + y - 8 = 0 $, we divide by $ sqrt{A^2 + B^2} = 2 $ to get $ frac{sqrt 3}{2} x + frac 12 y = 4 $ and then solve for $ cos omega = frac{sqrt 3}{2} $ and $ sin omega = frac 12 $to find the value of $ omega $.
I understand that we can't possibly convert this directly, since there is no $ omega $ for which $ cos omega = sqrt 3 $--but why do we specifically divide by $ sqrt{A^2 + B^2}?$ Why not any other number to bring the values of $ cos omega $ and $ sin omega $ between 0 and 1 and then solve it?
linear-algebra algebra-precalculus
edited Jan 13 at 17:43
J. W. Tanner
3,4601320
asked Jan 13 at 16:57
WorldGovWorldGov
324211
$endgroup$
To convert an equation in the standard form, $ Ax + By + C = 0 $ to the normal form $ cos omega x + sin omega y = p $, we first divide by $ sqrt{A^2 + B^2} $. Why?
For example, to convert the equation $ sqrt 3 x + y - 8 = 0 $, we divide by $ sqrt{A^2 + B^2} = 2 $ to get $ frac{sqrt 3}{2} x + frac 12 y = 4 $ and then solve for $ cos omega = frac{sqrt 3}{2} $ and $ sin omega = frac 12 $to find the value of $ omega $.
I understand that we can't possibly convert this directly, since there is no $ omega $ for which $ cos omega = sqrt 3 $--but why do we specifically divide by $ sqrt{A^2 + B^2}?$ Why not any other number to bring the values of $ cos omega $ and $ sin omega $ between 0 and 1 and then solve it?
linear-algebra algebra-precalculus
linear-algebra algebra-precalculus
edited Jan 13 at 17:43
J. W. Tanner
3,4601320
asked Jan 13 at 16:57
WorldGovWorldGov
324211
edited Jan 13 at 17:43
J. W. Tanner
3,4601320
asked Jan 13 at 16:57
WorldGovWorldGov
324211
edited Jan 13 at 17:43
J. W. Tanner
3,4601320
edited Jan 13 at 17:43
J. W. Tanner
3,4601320
edited Jan 13 at 17:43
J. W. Tanner
3,4601320
3,4601320
asked Jan 13 at 16:57
WorldGovWorldGov
324211
asked Jan 13 at 16:57
WorldGovWorldGov
324211
asked Jan 13 at 16:57
WorldGovWorldGov
324211
324211
3
$begingroup$
Because $cos^2 omega + sin^2 omega = 1$
$endgroup$
– Martin R
Jan 13 at 16:59
add a comment |
3
$begingroup$
Because $cos^2 omega + sin^2 omega = 1$
$endgroup$
– Martin R
Jan 13 at 16:59
3
3
$begingroup$
Because $cos^2 omega + sin^2 omega = 1$
$endgroup$
– Martin R
Jan 13 at 16:59
$begingroup$
Because $cos^2 omega + sin^2 omega = 1$
$endgroup$
– Martin R
Jan 13 at 16:59
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Given any two numbers $A$ and $B$ (not both equal zero), the numbers $a=A/sqrt{A^2+B^2}$ and $b=B/sqrt{A^2+B^2}$ satisfy $a^2+b^2=1$, so $a$ and $b$ are, respectively, the cosine and the sine of some angle. You are just finding a unit normal vector out of the given normal vector $(A,B)$ in the equation of the line. The same can be done in any number of dimensions: you can transform a given non-zero normal vector into a unit normal vector.
answered Jan 13 at 17:09
GReyesGReyes
2,22815
$endgroup$
add a comment |
$begingroup$
We can solve $(cosomega,sinomega)=(a,b)$ if and only if $(a,b)$ is a unit vector (i.e. the point $(a,b)$ is on the unit circle). The original vector $(A,B)$ is not unit, so we need to normalize it, that is, to find another vector of the length one that has the same direction. It is done by dividing the vector by its length
$$
(a,b)=frac{(A,B)}{|(A,B)|}=frac{(A,B)}{sqrt{A^2+B^2}}.
$$
answered Jan 13 at 17:07
A.Γ.A.Γ.
22.9k32656
$endgroup$
add a comment |
$begingroup$
Because the most important property of $sin$ and $cos$ must be preserved ($sin^2x+cos^2 x=1$).
answered Jan 13 at 17:12
Mostafa AyazMostafa Ayaz
16.7k3939
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given any two numbers $A$ and $B$ (not both equal zero), the numbers $a=A/sqrt{A^2+B^2}$ and $b=B/sqrt{A^2+B^2}$ satisfy $a^2+b^2=1$, so $a$ and $b$ are, respectively, the cosine and the sine of some angle. You are just finding a unit normal vector out of the given normal vector $(A,B)$ in the equation of the line. The same can be done in any number of dimensions: you can transform a given non-zero normal vector into a unit normal vector.
answered Jan 13 at 17:09
GReyesGReyes
2,22815
$endgroup$
add a comment |
$begingroup$
Given any two numbers $A$ and $B$ (not both equal zero), the numbers $a=A/sqrt{A^2+B^2}$ and $b=B/sqrt{A^2+B^2}$ satisfy $a^2+b^2=1$, so $a$ and $b$ are, respectively, the cosine and the sine of some angle. You are just finding a unit normal vector out of the given normal vector $(A,B)$ in the equation of the line. The same can be done in any number of dimensions: you can transform a given non-zero normal vector into a unit normal vector.
answered Jan 13 at 17:09
GReyesGReyes
2,22815
$endgroup$
add a comment |
$begingroup$
Given any two numbers $A$ and $B$ (not both equal zero), the numbers $a=A/sqrt{A^2+B^2}$ and $b=B/sqrt{A^2+B^2}$ satisfy $a^2+b^2=1$, so $a$ and $b$ are, respectively, the cosine and the sine of some angle. You are just finding a unit normal vector out of the given normal vector $(A,B)$ in the equation of the line. The same can be done in any number of dimensions: you can transform a given non-zero normal vector into a unit normal vector.
answered Jan 13 at 17:09
GReyesGReyes
2,22815
$endgroup$
Given any two numbers $A$ and $B$ (not both equal zero), the numbers $a=A/sqrt{A^2+B^2}$ and $b=B/sqrt{A^2+B^2}$ satisfy $a^2+b^2=1$, so $a$ and $b$ are, respectively, the cosine and the sine of some angle. You are just finding a unit normal vector out of the given normal vector $(A,B)$ in the equation of the line. The same can be done in any number of dimensions: you can transform a given non-zero normal vector into a unit normal vector.
answered Jan 13 at 17:09
GReyesGReyes
2,22815
edited Jan 13 at 18:59
answered Jan 13 at 17:09
GReyesGReyes
2,22815
answered Jan 13 at 17:09
GReyesGReyes
2,22815
answered Jan 13 at 17:09
GReyesGReyes
2,22815
2,22815
add a comment |
add a comment |
$begingroup$
We can solve $(cosomega,sinomega)=(a,b)$ if and only if $(a,b)$ is a unit vector (i.e. the point $(a,b)$ is on the unit circle). The original vector $(A,B)$ is not unit, so we need to normalize it, that is, to find another vector of the length one that has the same direction. It is done by dividing the vector by its length
$$
(a,b)=frac{(A,B)}{|(A,B)|}=frac{(A,B)}{sqrt{A^2+B^2}}.
$$
answered Jan 13 at 17:07
A.Γ.A.Γ.
22.9k32656
$endgroup$
add a comment |
$begingroup$
We can solve $(cosomega,sinomega)=(a,b)$ if and only if $(a,b)$ is a unit vector (i.e. the point $(a,b)$ is on the unit circle). The original vector $(A,B)$ is not unit, so we need to normalize it, that is, to find another vector of the length one that has the same direction. It is done by dividing the vector by its length
$$
(a,b)=frac{(A,B)}{|(A,B)|}=frac{(A,B)}{sqrt{A^2+B^2}}.
$$
answered Jan 13 at 17:07
A.Γ.A.Γ.
22.9k32656
$endgroup$
add a comment |
$begingroup$
We can solve $(cosomega,sinomega)=(a,b)$ if and only if $(a,b)$ is a unit vector (i.e. the point $(a,b)$ is on the unit circle). The original vector $(A,B)$ is not unit, so we need to normalize it, that is, to find another vector of the length one that has the same direction. It is done by dividing the vector by its length
$$
(a,b)=frac{(A,B)}{|(A,B)|}=frac{(A,B)}{sqrt{A^2+B^2}}.
$$
answered Jan 13 at 17:07
A.Γ.A.Γ.
22.9k32656
$endgroup$
We can solve $(cosomega,sinomega)=(a,b)$ if and only if $(a,b)$ is a unit vector (i.e. the point $(a,b)$ is on the unit circle). The original vector $(A,B)$ is not unit, so we need to normalize it, that is, to find another vector of the length one that has the same direction. It is done by dividing the vector by its length
$$
(a,b)=frac{(A,B)}{|(A,B)|}=frac{(A,B)}{sqrt{A^2+B^2}}.
$$
answered Jan 13 at 17:07
A.Γ.A.Γ.
22.9k32656
answered Jan 13 at 17:07
A.Γ.A.Γ.
22.9k32656
answered Jan 13 at 17:07
A.Γ.A.Γ.
22.9k32656
answered Jan 13 at 17:07
A.Γ.A.Γ.
22.9k32656
22.9k32656
add a comment |
add a comment |
$begingroup$
Because the most important property of $sin$ and $cos$ must be preserved ($sin^2x+cos^2 x=1$).
answered Jan 13 at 17:12
Mostafa AyazMostafa Ayaz
16.7k3939
$endgroup$
add a comment |
$begingroup$
Because the most important property of $sin$ and $cos$ must be preserved ($sin^2x+cos^2 x=1$).
answered Jan 13 at 17:12
Mostafa AyazMostafa Ayaz
16.7k3939
$endgroup$
add a comment |
$begingroup$
Because the most important property of $sin$ and $cos$ must be preserved ($sin^2x+cos^2 x=1$).
answered Jan 13 at 17:12
Mostafa AyazMostafa Ayaz
16.7k3939
$endgroup$
Because the most important property of $sin$ and $cos$ must be preserved ($sin^2x+cos^2 x=1$).
answered Jan 13 at 17:12
Mostafa AyazMostafa Ayaz
16.7k3939
answered Jan 13 at 17:12
Mostafa AyazMostafa Ayaz
16.7k3939
answered Jan 13 at 17:12
Mostafa AyazMostafa Ayaz
16.7k3939
answered Jan 13 at 17:12
Mostafa AyazMostafa Ayaz
16.7k3939
16.7k3939
add a comment |
add a comment |
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3
$begingroup$
Because $cos^2 omega + sin^2 omega = 1$
$endgroup$
– Martin R
Jan 13 at 16:59