Mixing
why do we use cosine as the expression of vector dot product?
$begingroup$
When we do vector products, we use two different methods.
One is the vector dot product, another is vector cross product.
The equation of the vector dot product is
$$textbf A cdot textbf B =|textbf A| | textbf B| costheta,$$
where $theta$ is the angle between the vectors $textbf A$ and $textbf B$.
Why do we use cosine as the expression?
vectors
$endgroup$
migrated from physics.stackexchange.com Feb 15 '15 at 15:54
This question came from our site for active researchers, academics and students of physics.
add a comment |
$begingroup$
When we do vector products, we use two different methods.
One is the vector dot product, another is vector cross product.
The equation of the vector dot product is
$$textbf A cdot textbf B =|textbf A| | textbf B| costheta,$$
where $theta$ is the angle between the vectors $textbf A$ and $textbf B$.
Why do we use cosine as the expression?
vectors
$endgroup$
migrated from physics.stackexchange.com Feb 15 '15 at 15:54
This question came from our site for active researchers, academics and students of physics.
1
$begingroup$
It really doesn't matter, use $cos x$ or $sin(x+pi /2)$ if you like, they are equivalent.
$endgroup$
– JamalS
Feb 15 '15 at 9:01
$begingroup$
possible duplicate of What is the physical meaning of a product of vectors?
$endgroup$
– Ali
Feb 15 '15 at 12:45
add a comment |
$begingroup$
When we do vector products, we use two different methods.
One is the vector dot product, another is vector cross product.
The equation of the vector dot product is
$$textbf A cdot textbf B =|textbf A| | textbf B| costheta,$$
where $theta$ is the angle between the vectors $textbf A$ and $textbf B$.
Why do we use cosine as the expression?
vectors
$endgroup$
When we do vector products, we use two different methods.
One is the vector dot product, another is vector cross product.
The equation of the vector dot product is
$$textbf A cdot textbf B =|textbf A| | textbf B| costheta,$$
where $theta$ is the angle between the vectors $textbf A$ and $textbf B$.
Why do we use cosine as the expression?
vectors
vectors
asked Feb 15 '15 at 6:24
Shams Tarek
migrated from physics.stackexchange.com Feb 15 '15 at 15:54
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com Feb 15 '15 at 15:54
This question came from our site for active researchers, academics and students of physics.
1
$begingroup$
It really doesn't matter, use $cos x$ or $sin(x+pi /2)$ if you like, they are equivalent.
$endgroup$
– JamalS
Feb 15 '15 at 9:01
$begingroup$
possible duplicate of What is the physical meaning of a product of vectors?
$endgroup$
– Ali
Feb 15 '15 at 12:45
add a comment |
1
$begingroup$
It really doesn't matter, use $cos x$ or $sin(x+pi /2)$ if you like, they are equivalent.
$endgroup$
– JamalS
Feb 15 '15 at 9:01
$begingroup$
possible duplicate of What is the physical meaning of a product of vectors?
$endgroup$
– Ali
Feb 15 '15 at 12:45
1
1
$begingroup$
It really doesn't matter, use $cos x$ or $sin(x+pi /2)$ if you like, they are equivalent.
$endgroup$
– JamalS
Feb 15 '15 at 9:01
$begingroup$
It really doesn't matter, use $cos x$ or $sin(x+pi /2)$ if you like, they are equivalent.
$endgroup$
– JamalS
Feb 15 '15 at 9:01
$begingroup$
possible duplicate of What is the physical meaning of a product of vectors?
$endgroup$
– Ali
Feb 15 '15 at 12:45
$begingroup$
possible duplicate of What is the physical meaning of a product of vectors?
$endgroup$
– Ali
Feb 15 '15 at 12:45
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
The dot product of two vectors $A$ and $B$ is just the product of the magnitude of one vector with the scalar projection of the other one on itself. Hence the $cos$ term. Also, note that the $cos$ function is greater for smaller angles, and lesser for larger ones, just like the length of the projection. (both are the same thing, actually.)
answered Feb 15 '15 at 6:52
Hritik NarayanHritik Narayan
372312
$endgroup$
add a comment |
$begingroup$
The dot product is defined in that way.
Note that $costheta$ is a suitable function; since by the Schwarz inequality:
$$|mathbf{A} cdot mathbf{B}| leq |mathbf{A}| |mathbf{B}|$$
and thus the dot product ranges continuosly between -1 and 1, as $cos theta$ for $theta in [0,pi] $.
edited Jul 29 '16 at 6:28
rubik
6,76132661
answered Feb 15 '15 at 8:46
NNecNNec
31717
$endgroup$
add a comment |
$begingroup$
Let $vec{mathbf a} = (x_1, y_1) = (a cos alpha, a sin alpha)$
Let $vec{mathbf b} = (x_2, y_2) = (b cos beta, b sin beta)$
Then $theta = beta - alpha$
By definition,
begin{align}
vec{mathbf a} circ vec{mathbf b}
&= x_1x_2 + y_1y_2 \
&= ab(cos alpha cos beta + sin alpha sin beta) \
&= ab cos(beta - alpha) \
&= ab cos theta
end{align}
answered Jan 15 at 23:00
steven gregorysteven gregory
18.2k32258
$endgroup$
add a comment |
$begingroup$
This is entirely determined by what we consider a rotation in the plane.
Let $u, v$ be two unit vectors. Let the angle between them be $theta$, and we can naturally write $v$ as
$$v = u cos theta + u_perp sin theta$$
where $u_perp$ is a unit vector perpendicular to $u$. Then clearly, the dot product is $u cdot v = cos theta$.
But, if you're not in a Euclidean plane anymore, this relationship no longer holds. For example, in a Lorentzian space, instead of cosine and sine, we get hyperbolic functions instead:
$$v = u cosh theta + u_perp sinh theta$$
And the dot product is $u cdot v = cosh theta$. A physicist should recognize that this $theta$ is the "rapidity", and that the form of $v$ given here is exactly that of any Lorentz boost.
So the reason we use sine and cosine in Euclidean space is because they are dictated by the use of sine and cosine in rotations. In other spaces, with different rotation operators, you use the functions that are associated with those rotations instead.
answered Feb 15 '15 at 16:06
MuphridMuphrid
15.6k11542
$endgroup$
$begingroup$
What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
$endgroup$
– got it--thanks
Feb 22 '15 at 20:22
1
$begingroup$
Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
$endgroup$
– Muphrid
Feb 22 '15 at 21:17
add a comment |
$begingroup$
In dot product we use cos theta because in this type of product
1.) One vector is the projection over the other.
2.) The distance is covered along one axis or in the direction of force and there is no need of perpendicular axis or sin theta.
In cross product the angle between must be greater than 0 and less than 180 degree it is max at 90 degree. let take the example of torque if the angle between applied force and moment arm is 90 degree than torque will be max.
That's why we use cos theta for dot product and sin theta for cross product.
answered Jul 29 '16 at 5:56
muhammad mubeenmuhammad mubeen
1
$endgroup$
add a comment |
$begingroup$
Because here cosine is responsible for doing any type of work e.g work done cosine is use . Here one axis is use for work done I.e x -axis and for x -axis we are using cosine
answered May 25 '18 at 8:52
Bibi AlamBibi Alam
1
$endgroup$
add a comment |
$begingroup$
It is actually the definition of the dot product of two vectors.
answered Feb 15 '15 at 8:38
omehoqueomehoque
199212
$endgroup$
7
$begingroup$
The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
$endgroup$
– Danu
Feb 15 '15 at 11:31
3
$begingroup$
Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
$endgroup$
– ACuriousMind
Feb 15 '15 at 13:57
add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The dot product of two vectors $A$ and $B$ is just the product of the magnitude of one vector with the scalar projection of the other one on itself. Hence the $cos$ term. Also, note that the $cos$ function is greater for smaller angles, and lesser for larger ones, just like the length of the projection. (both are the same thing, actually.)
answered Feb 15 '15 at 6:52
Hritik NarayanHritik Narayan
372312
$endgroup$
add a comment |
$begingroup$
The dot product of two vectors $A$ and $B$ is just the product of the magnitude of one vector with the scalar projection of the other one on itself. Hence the $cos$ term. Also, note that the $cos$ function is greater for smaller angles, and lesser for larger ones, just like the length of the projection. (both are the same thing, actually.)
answered Feb 15 '15 at 6:52
Hritik NarayanHritik Narayan
372312
$endgroup$
add a comment |
$begingroup$
The dot product of two vectors $A$ and $B$ is just the product of the magnitude of one vector with the scalar projection of the other one on itself. Hence the $cos$ term. Also, note that the $cos$ function is greater for smaller angles, and lesser for larger ones, just like the length of the projection. (both are the same thing, actually.)
answered Feb 15 '15 at 6:52
Hritik NarayanHritik Narayan
372312
$endgroup$
The dot product of two vectors $A$ and $B$ is just the product of the magnitude of one vector with the scalar projection of the other one on itself. Hence the $cos$ term. Also, note that the $cos$ function is greater for smaller angles, and lesser for larger ones, just like the length of the projection. (both are the same thing, actually.)
answered Feb 15 '15 at 6:52
Hritik NarayanHritik Narayan
372312
answered Feb 15 '15 at 6:52
Hritik NarayanHritik Narayan
372312
answered Feb 15 '15 at 6:52
Hritik NarayanHritik Narayan
372312
answered Feb 15 '15 at 6:52
Hritik NarayanHritik Narayan
372312
372312
add a comment |
add a comment |
$begingroup$
The dot product is defined in that way.
Note that $costheta$ is a suitable function; since by the Schwarz inequality:
$$|mathbf{A} cdot mathbf{B}| leq |mathbf{A}| |mathbf{B}|$$
and thus the dot product ranges continuosly between -1 and 1, as $cos theta$ for $theta in [0,pi] $.
edited Jul 29 '16 at 6:28
rubik
6,76132661
answered Feb 15 '15 at 8:46
NNecNNec
31717
$endgroup$
add a comment |
$begingroup$
The dot product is defined in that way.
Note that $costheta$ is a suitable function; since by the Schwarz inequality:
$$|mathbf{A} cdot mathbf{B}| leq |mathbf{A}| |mathbf{B}|$$
and thus the dot product ranges continuosly between -1 and 1, as $cos theta$ for $theta in [0,pi] $.
edited Jul 29 '16 at 6:28
rubik
6,76132661
answered Feb 15 '15 at 8:46
NNecNNec
31717
$endgroup$
add a comment |
$begingroup$
The dot product is defined in that way.
Note that $costheta$ is a suitable function; since by the Schwarz inequality:
$$|mathbf{A} cdot mathbf{B}| leq |mathbf{A}| |mathbf{B}|$$
and thus the dot product ranges continuosly between -1 and 1, as $cos theta$ for $theta in [0,pi] $.
edited Jul 29 '16 at 6:28
rubik
6,76132661
answered Feb 15 '15 at 8:46
NNecNNec
31717
$endgroup$
The dot product is defined in that way.
Note that $costheta$ is a suitable function; since by the Schwarz inequality:
$$|mathbf{A} cdot mathbf{B}| leq |mathbf{A}| |mathbf{B}|$$
and thus the dot product ranges continuosly between -1 and 1, as $cos theta$ for $theta in [0,pi] $.
edited Jul 29 '16 at 6:28
rubik
6,76132661
answered Feb 15 '15 at 8:46
NNecNNec
31717
edited Jul 29 '16 at 6:28
rubik
6,76132661
edited Jul 29 '16 at 6:28
rubik
6,76132661
edited Jul 29 '16 at 6:28
rubik
6,76132661
6,76132661
answered Feb 15 '15 at 8:46
NNecNNec
31717
answered Feb 15 '15 at 8:46
NNecNNec
31717
answered Feb 15 '15 at 8:46
NNecNNec
31717
31717
add a comment |
add a comment |
$begingroup$
Let $vec{mathbf a} = (x_1, y_1) = (a cos alpha, a sin alpha)$
Let $vec{mathbf b} = (x_2, y_2) = (b cos beta, b sin beta)$
Then $theta = beta - alpha$
By definition,
begin{align}
vec{mathbf a} circ vec{mathbf b}
&= x_1x_2 + y_1y_2 \
&= ab(cos alpha cos beta + sin alpha sin beta) \
&= ab cos(beta - alpha) \
&= ab cos theta
end{align}
answered Jan 15 at 23:00
steven gregorysteven gregory
18.2k32258
$endgroup$
add a comment |
$begingroup$
Let $vec{mathbf a} = (x_1, y_1) = (a cos alpha, a sin alpha)$
Let $vec{mathbf b} = (x_2, y_2) = (b cos beta, b sin beta)$
Then $theta = beta - alpha$
By definition,
begin{align}
vec{mathbf a} circ vec{mathbf b}
&= x_1x_2 + y_1y_2 \
&= ab(cos alpha cos beta + sin alpha sin beta) \
&= ab cos(beta - alpha) \
&= ab cos theta
end{align}
answered Jan 15 at 23:00
steven gregorysteven gregory
18.2k32258
$endgroup$
add a comment |
$begingroup$
Let $vec{mathbf a} = (x_1, y_1) = (a cos alpha, a sin alpha)$
Let $vec{mathbf b} = (x_2, y_2) = (b cos beta, b sin beta)$
Then $theta = beta - alpha$
By definition,
begin{align}
vec{mathbf a} circ vec{mathbf b}
&= x_1x_2 + y_1y_2 \
&= ab(cos alpha cos beta + sin alpha sin beta) \
&= ab cos(beta - alpha) \
&= ab cos theta
end{align}
answered Jan 15 at 23:00
steven gregorysteven gregory
18.2k32258
$endgroup$
Let $vec{mathbf a} = (x_1, y_1) = (a cos alpha, a sin alpha)$
Let $vec{mathbf b} = (x_2, y_2) = (b cos beta, b sin beta)$
Then $theta = beta - alpha$
By definition,
begin{align}
vec{mathbf a} circ vec{mathbf b}
&= x_1x_2 + y_1y_2 \
&= ab(cos alpha cos beta + sin alpha sin beta) \
&= ab cos(beta - alpha) \
&= ab cos theta
end{align}
answered Jan 15 at 23:00
steven gregorysteven gregory
18.2k32258
edited Jan 16 at 6:15
answered Jan 15 at 23:00
steven gregorysteven gregory
18.2k32258
answered Jan 15 at 23:00
steven gregorysteven gregory
18.2k32258
answered Jan 15 at 23:00
steven gregorysteven gregory
18.2k32258
18.2k32258
add a comment |
add a comment |
$begingroup$
This is entirely determined by what we consider a rotation in the plane.
Let $u, v$ be two unit vectors. Let the angle between them be $theta$, and we can naturally write $v$ as
$$v = u cos theta + u_perp sin theta$$
where $u_perp$ is a unit vector perpendicular to $u$. Then clearly, the dot product is $u cdot v = cos theta$.
But, if you're not in a Euclidean plane anymore, this relationship no longer holds. For example, in a Lorentzian space, instead of cosine and sine, we get hyperbolic functions instead:
$$v = u cosh theta + u_perp sinh theta$$
And the dot product is $u cdot v = cosh theta$. A physicist should recognize that this $theta$ is the "rapidity", and that the form of $v$ given here is exactly that of any Lorentz boost.
So the reason we use sine and cosine in Euclidean space is because they are dictated by the use of sine and cosine in rotations. In other spaces, with different rotation operators, you use the functions that are associated with those rotations instead.
answered Feb 15 '15 at 16:06
MuphridMuphrid
15.6k11542
$endgroup$
$begingroup$
What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
$endgroup$
– got it--thanks
Feb 22 '15 at 20:22
1
$begingroup$
Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
$endgroup$
– Muphrid
Feb 22 '15 at 21:17
add a comment |
$begingroup$
This is entirely determined by what we consider a rotation in the plane.
Let $u, v$ be two unit vectors. Let the angle between them be $theta$, and we can naturally write $v$ as
$$v = u cos theta + u_perp sin theta$$
where $u_perp$ is a unit vector perpendicular to $u$. Then clearly, the dot product is $u cdot v = cos theta$.
But, if you're not in a Euclidean plane anymore, this relationship no longer holds. For example, in a Lorentzian space, instead of cosine and sine, we get hyperbolic functions instead:
$$v = u cosh theta + u_perp sinh theta$$
And the dot product is $u cdot v = cosh theta$. A physicist should recognize that this $theta$ is the "rapidity", and that the form of $v$ given here is exactly that of any Lorentz boost.
So the reason we use sine and cosine in Euclidean space is because they are dictated by the use of sine and cosine in rotations. In other spaces, with different rotation operators, you use the functions that are associated with those rotations instead.
answered Feb 15 '15 at 16:06
MuphridMuphrid
15.6k11542
$endgroup$
$begingroup$
What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
$endgroup$
– got it--thanks
Feb 22 '15 at 20:22
1
$begingroup$
Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
$endgroup$
– Muphrid
Feb 22 '15 at 21:17
add a comment |
$begingroup$
This is entirely determined by what we consider a rotation in the plane.
Let $u, v$ be two unit vectors. Let the angle between them be $theta$, and we can naturally write $v$ as
$$v = u cos theta + u_perp sin theta$$
where $u_perp$ is a unit vector perpendicular to $u$. Then clearly, the dot product is $u cdot v = cos theta$.
But, if you're not in a Euclidean plane anymore, this relationship no longer holds. For example, in a Lorentzian space, instead of cosine and sine, we get hyperbolic functions instead:
$$v = u cosh theta + u_perp sinh theta$$
And the dot product is $u cdot v = cosh theta$. A physicist should recognize that this $theta$ is the "rapidity", and that the form of $v$ given here is exactly that of any Lorentz boost.
So the reason we use sine and cosine in Euclidean space is because they are dictated by the use of sine and cosine in rotations. In other spaces, with different rotation operators, you use the functions that are associated with those rotations instead.
answered Feb 15 '15 at 16:06
MuphridMuphrid
15.6k11542
$endgroup$
This is entirely determined by what we consider a rotation in the plane.
Let $u, v$ be two unit vectors. Let the angle between them be $theta$, and we can naturally write $v$ as
$$v = u cos theta + u_perp sin theta$$
where $u_perp$ is a unit vector perpendicular to $u$. Then clearly, the dot product is $u cdot v = cos theta$.
But, if you're not in a Euclidean plane anymore, this relationship no longer holds. For example, in a Lorentzian space, instead of cosine and sine, we get hyperbolic functions instead:
$$v = u cosh theta + u_perp sinh theta$$
And the dot product is $u cdot v = cosh theta$. A physicist should recognize that this $theta$ is the "rapidity", and that the form of $v$ given here is exactly that of any Lorentz boost.
So the reason we use sine and cosine in Euclidean space is because they are dictated by the use of sine and cosine in rotations. In other spaces, with different rotation operators, you use the functions that are associated with those rotations instead.
answered Feb 15 '15 at 16:06
MuphridMuphrid
15.6k11542
answered Feb 15 '15 at 16:06
MuphridMuphrid
15.6k11542
answered Feb 15 '15 at 16:06
MuphridMuphrid
15.6k11542
answered Feb 15 '15 at 16:06
MuphridMuphrid
15.6k11542
15.6k11542
$begingroup$
What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
$endgroup$
– got it--thanks
Feb 22 '15 at 20:22
1
$begingroup$
Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
$endgroup$
– Muphrid
Feb 22 '15 at 21:17
add a comment |
$begingroup$
What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
$endgroup$
– got it--thanks
Feb 22 '15 at 20:22
1
$begingroup$
Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
$endgroup$
– Muphrid
Feb 22 '15 at 21:17
$begingroup$
What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
$endgroup$
– got it--thanks
Feb 22 '15 at 20:22
$begingroup$
What does this have to do with rotations? This just looks like decomposition of $v$ into $v_{|}$ and $v_{bot}$ parts.
$endgroup$
– got it--thanks
Feb 22 '15 at 20:22
1
1
$begingroup$
Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
$endgroup$
– Muphrid
Feb 22 '15 at 21:17
$begingroup$
Whether you use cosine or hyperbolic cosine depends on what the rotation operations are, on which set of vectors is the set of unit vectors.
$endgroup$
– Muphrid
Feb 22 '15 at 21:17
add a comment |
$begingroup$
In dot product we use cos theta because in this type of product
1.) One vector is the projection over the other.
2.) The distance is covered along one axis or in the direction of force and there is no need of perpendicular axis or sin theta.
In cross product the angle between must be greater than 0 and less than 180 degree it is max at 90 degree. let take the example of torque if the angle between applied force and moment arm is 90 degree than torque will be max.
That's why we use cos theta for dot product and sin theta for cross product.
answered Jul 29 '16 at 5:56
muhammad mubeenmuhammad mubeen
1
$endgroup$
add a comment |
$begingroup$
In dot product we use cos theta because in this type of product
1.) One vector is the projection over the other.
2.) The distance is covered along one axis or in the direction of force and there is no need of perpendicular axis or sin theta.
In cross product the angle between must be greater than 0 and less than 180 degree it is max at 90 degree. let take the example of torque if the angle between applied force and moment arm is 90 degree than torque will be max.
That's why we use cos theta for dot product and sin theta for cross product.
answered Jul 29 '16 at 5:56
muhammad mubeenmuhammad mubeen
1
$endgroup$
add a comment |
$begingroup$
In dot product we use cos theta because in this type of product
1.) One vector is the projection over the other.
2.) The distance is covered along one axis or in the direction of force and there is no need of perpendicular axis or sin theta.
In cross product the angle between must be greater than 0 and less than 180 degree it is max at 90 degree. let take the example of torque if the angle between applied force and moment arm is 90 degree than torque will be max.
That's why we use cos theta for dot product and sin theta for cross product.
answered Jul 29 '16 at 5:56
muhammad mubeenmuhammad mubeen
1
$endgroup$
In dot product we use cos theta because in this type of product
1.) One vector is the projection over the other.
2.) The distance is covered along one axis or in the direction of force and there is no need of perpendicular axis or sin theta.
In cross product the angle between must be greater than 0 and less than 180 degree it is max at 90 degree. let take the example of torque if the angle between applied force and moment arm is 90 degree than torque will be max.
That's why we use cos theta for dot product and sin theta for cross product.
answered Jul 29 '16 at 5:56
muhammad mubeenmuhammad mubeen
1
answered Jul 29 '16 at 5:56
muhammad mubeenmuhammad mubeen
1
answered Jul 29 '16 at 5:56
muhammad mubeenmuhammad mubeen
1
answered Jul 29 '16 at 5:56
muhammad mubeenmuhammad mubeen
1
1
add a comment |
add a comment |
$begingroup$
Because here cosine is responsible for doing any type of work e.g work done cosine is use . Here one axis is use for work done I.e x -axis and for x -axis we are using cosine
answered May 25 '18 at 8:52
Bibi AlamBibi Alam
1
$endgroup$
add a comment |
$begingroup$
Because here cosine is responsible for doing any type of work e.g work done cosine is use . Here one axis is use for work done I.e x -axis and for x -axis we are using cosine
answered May 25 '18 at 8:52
Bibi AlamBibi Alam
1
$endgroup$
add a comment |
$begingroup$
Because here cosine is responsible for doing any type of work e.g work done cosine is use . Here one axis is use for work done I.e x -axis and for x -axis we are using cosine
answered May 25 '18 at 8:52
Bibi AlamBibi Alam
1
$endgroup$
Because here cosine is responsible for doing any type of work e.g work done cosine is use . Here one axis is use for work done I.e x -axis and for x -axis we are using cosine
answered May 25 '18 at 8:52
Bibi AlamBibi Alam
1
answered May 25 '18 at 8:52
Bibi AlamBibi Alam
1
answered May 25 '18 at 8:52
Bibi AlamBibi Alam
1
answered May 25 '18 at 8:52
Bibi AlamBibi Alam
1
1
add a comment |
add a comment |
$begingroup$
It is actually the definition of the dot product of two vectors.
answered Feb 15 '15 at 8:38
omehoqueomehoque
199212
$endgroup$
7
$begingroup$
The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
$endgroup$
– Danu
Feb 15 '15 at 11:31
3
$begingroup$
Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
$endgroup$
– ACuriousMind
Feb 15 '15 at 13:57
add a comment |
$begingroup$
It is actually the definition of the dot product of two vectors.
answered Feb 15 '15 at 8:38
omehoqueomehoque
199212
$endgroup$
7
$begingroup$
The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
$endgroup$
– Danu
Feb 15 '15 at 11:31
3
$begingroup$
Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
$endgroup$
– ACuriousMind
Feb 15 '15 at 13:57
add a comment |
$begingroup$
It is actually the definition of the dot product of two vectors.
answered Feb 15 '15 at 8:38
omehoqueomehoque
199212
$endgroup$
It is actually the definition of the dot product of two vectors.
answered Feb 15 '15 at 8:38
omehoqueomehoque
199212
answered Feb 15 '15 at 8:38
omehoqueomehoque
199212
answered Feb 15 '15 at 8:38
omehoqueomehoque
199212
answered Feb 15 '15 at 8:38
omehoqueomehoque
199212
199212
7
$begingroup$
The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
$endgroup$
– Danu
Feb 15 '15 at 11:31
3
$begingroup$
Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
$endgroup$
– ACuriousMind
Feb 15 '15 at 13:57
add a comment |
7
$begingroup$
The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
$endgroup$
– Danu
Feb 15 '15 at 11:31
3
$begingroup$
Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
$endgroup$
– ACuriousMind
Feb 15 '15 at 13:57
7
7
$begingroup$
The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
$endgroup$
– Danu
Feb 15 '15 at 11:31
$begingroup$
The point of this question is to ask why one would want to express the dot product as something involving the cosine, not what the definition is.
$endgroup$
– Danu
Feb 15 '15 at 11:31
3
3
$begingroup$
Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
$endgroup$
– ACuriousMind
Feb 15 '15 at 13:57
$begingroup$
Besides, it is more commonly seen the other way around - the angle $theta$ is defined by this formula, while the dot product is defined by a symmetric matrix.
$endgroup$
– ACuriousMind
Feb 15 '15 at 13:57
add a comment |
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$begingroup$
It really doesn't matter, use $cos x$ or $sin(x+pi /2)$ if you like, they are equivalent.
$endgroup$
– JamalS
Feb 15 '15 at 9:01
$begingroup$
possible duplicate of What is the physical meaning of a product of vectors?
$endgroup$
– Ali
Feb 15 '15 at 12:45