Mixing
Applying Law of Cosines to get Polarization Identity
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I am teaching myself some basic linear algebra using MATH51 textbook and I got to a problem which states following:
The parallelogram in R3 formed by two 3-vectors v and w. The vectors
on the diagonals of the parallelogram are v - w and v + w.
For the case n = 2, apply the Law of Cosines to each of the triangles
made by cutting theparallelogram in half along a diagonal (regard the
chosen diagonal as the side opposite the “angle” being used in the Law
of Cosines) to establish the polarization identity.
After some Googling I got some intution (although not a clear one) to use dot product of v and w, the thing is dot product operation has not been introduced yet. Is there any way to use Law of Cosines to get Polarization Identity without using dot product?
I am including a screenshot with the entire problem and a diagram for better illustration.
linear-algebra
asked Jan 22 at 18:06
MitoMito
31
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add a comment |
$begingroup$
I am teaching myself some basic linear algebra using MATH51 textbook and I got to a problem which states following:
The parallelogram in R3 formed by two 3-vectors v and w. The vectors
on the diagonals of the parallelogram are v - w and v + w.
For the case n = 2, apply the Law of Cosines to each of the triangles
made by cutting theparallelogram in half along a diagonal (regard the
chosen diagonal as the side opposite the “angle” being used in the Law
of Cosines) to establish the polarization identity.
After some Googling I got some intution (although not a clear one) to use dot product of v and w, the thing is dot product operation has not been introduced yet. Is there any way to use Law of Cosines to get Polarization Identity without using dot product?
I am including a screenshot with the entire problem and a diagram for better illustration.
linear-algebra
asked Jan 22 at 18:06
MitoMito
31
$endgroup$
add a comment |
$begingroup$
I am teaching myself some basic linear algebra using MATH51 textbook and I got to a problem which states following:
The parallelogram in R3 formed by two 3-vectors v and w. The vectors
on the diagonals of the parallelogram are v - w and v + w.
For the case n = 2, apply the Law of Cosines to each of the triangles
made by cutting theparallelogram in half along a diagonal (regard the
chosen diagonal as the side opposite the “angle” being used in the Law
of Cosines) to establish the polarization identity.
After some Googling I got some intution (although not a clear one) to use dot product of v and w, the thing is dot product operation has not been introduced yet. Is there any way to use Law of Cosines to get Polarization Identity without using dot product?
I am including a screenshot with the entire problem and a diagram for better illustration.
linear-algebra
asked Jan 22 at 18:06
MitoMito
31
$endgroup$
I am teaching myself some basic linear algebra using MATH51 textbook and I got to a problem which states following:
The parallelogram in R3 formed by two 3-vectors v and w. The vectors
on the diagonals of the parallelogram are v - w and v + w.
For the case n = 2, apply the Law of Cosines to each of the triangles
made by cutting theparallelogram in half along a diagonal (regard the
chosen diagonal as the side opposite the “angle” being used in the Law
of Cosines) to establish the polarization identity.
After some Googling I got some intution (although not a clear one) to use dot product of v and w, the thing is dot product operation has not been introduced yet. Is there any way to use Law of Cosines to get Polarization Identity without using dot product?
I am including a screenshot with the entire problem and a diagram for better illustration.
linear-algebra
linear-algebra
asked Jan 22 at 18:06
MitoMito
31
asked Jan 22 at 18:06
MitoMito
31
asked Jan 22 at 18:06
MitoMito
31
asked Jan 22 at 18:06
MitoMito
31
asked Jan 22 at 18:06
MitoMito
31
31
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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@Mito
What is called "polarization identity" in your book is usually called parallelogram identity (the polarization identity has to do with recovering the dot product from the length). You have two triangles. One has sides $||v||$, $||w||$ and $||v+w||$ and the other one has sides $||v||$, $||w||$ and $||v-w||$. Since they form a parallelogram, the angles opposite to $||v-w||$ and $||v+w||$ are supplementary so they have opposite cosines. When you apply the law of cosines you get
$$
||v-w||^2=||v||^2+||w||^2-2||v|||w||cos A
$$
and
$$
||v+w||^2=||v||^2+||w||^2-2||v|||w||cos (pi - A)=||v||^2+||w||^2+2||v|||w||cos A
$$
By adding these two you get what you need.
answered Jan 22 at 18:15
GReyesGReyes
1,93515
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
@Mito
What is called "polarization identity" in your book is usually called parallelogram identity (the polarization identity has to do with recovering the dot product from the length). You have two triangles. One has sides $||v||$, $||w||$ and $||v+w||$ and the other one has sides $||v||$, $||w||$ and $||v-w||$. Since they form a parallelogram, the angles opposite to $||v-w||$ and $||v+w||$ are supplementary so they have opposite cosines. When you apply the law of cosines you get
$$
||v-w||^2=||v||^2+||w||^2-2||v|||w||cos A
$$
and
$$
||v+w||^2=||v||^2+||w||^2-2||v|||w||cos (pi - A)=||v||^2+||w||^2+2||v|||w||cos A
$$
By adding these two you get what you need.
answered Jan 22 at 18:15
GReyesGReyes
1,93515
$endgroup$
add a comment |
$begingroup$
@Mito
What is called "polarization identity" in your book is usually called parallelogram identity (the polarization identity has to do with recovering the dot product from the length). You have two triangles. One has sides $||v||$, $||w||$ and $||v+w||$ and the other one has sides $||v||$, $||w||$ and $||v-w||$. Since they form a parallelogram, the angles opposite to $||v-w||$ and $||v+w||$ are supplementary so they have opposite cosines. When you apply the law of cosines you get
$$
||v-w||^2=||v||^2+||w||^2-2||v|||w||cos A
$$
and
$$
||v+w||^2=||v||^2+||w||^2-2||v|||w||cos (pi - A)=||v||^2+||w||^2+2||v|||w||cos A
$$
By adding these two you get what you need.
answered Jan 22 at 18:15
GReyesGReyes
1,93515
$endgroup$
add a comment |
$begingroup$
@Mito
What is called "polarization identity" in your book is usually called parallelogram identity (the polarization identity has to do with recovering the dot product from the length). You have two triangles. One has sides $||v||$, $||w||$ and $||v+w||$ and the other one has sides $||v||$, $||w||$ and $||v-w||$. Since they form a parallelogram, the angles opposite to $||v-w||$ and $||v+w||$ are supplementary so they have opposite cosines. When you apply the law of cosines you get
$$
||v-w||^2=||v||^2+||w||^2-2||v|||w||cos A
$$
and
$$
||v+w||^2=||v||^2+||w||^2-2||v|||w||cos (pi - A)=||v||^2+||w||^2+2||v|||w||cos A
$$
By adding these two you get what you need.
answered Jan 22 at 18:15
GReyesGReyes
1,93515
$endgroup$
@Mito
What is called "polarization identity" in your book is usually called parallelogram identity (the polarization identity has to do with recovering the dot product from the length). You have two triangles. One has sides $||v||$, $||w||$ and $||v+w||$ and the other one has sides $||v||$, $||w||$ and $||v-w||$. Since they form a parallelogram, the angles opposite to $||v-w||$ and $||v+w||$ are supplementary so they have opposite cosines. When you apply the law of cosines you get
$$
||v-w||^2=||v||^2+||w||^2-2||v|||w||cos A
$$
and
$$
||v+w||^2=||v||^2+||w||^2-2||v|||w||cos (pi - A)=||v||^2+||w||^2+2||v|||w||cos A
$$
By adding these two you get what you need.
answered Jan 22 at 18:15
GReyesGReyes
1,93515
answered Jan 22 at 18:15
GReyesGReyes
1,93515
answered Jan 22 at 18:15
GReyesGReyes
1,93515
answered Jan 22 at 18:15
GReyesGReyes
1,93515
1,93515
add a comment |
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