Calculate dot product without the use of angles
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In a book called Introduction to tensor analysis and the calculus of moving surfaces equation (2.6) gives a formula to calculate the dot product between two vectors in terms of length alone. That formula is the following:
$$
mathbf{U}cdot{V} = frac{left| mathbf{U} + mathbf{V} right|^2 - left| mathbf{U} - mathbf{V} right|^2}{4}
$$
I suppose then that the above shall be equal to
$$
mathbf{U}cdotmathbf{V} = |mathbf{U}||mathbf{V}|cos{alpha}
$$
where $alpha$ is the angle between $mathbf{U}$ and $mathbf{V}$. How can it be proved that both identities are equal?
vectors analytic-geometry inner-product-space
asked Dec 13 '18 at 21:18
jrojasqujrojasqu
209413
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add a comment |
$begingroup$
In a book called Introduction to tensor analysis and the calculus of moving surfaces equation (2.6) gives a formula to calculate the dot product between two vectors in terms of length alone. That formula is the following:
$$
mathbf{U}cdot{V} = frac{left| mathbf{U} + mathbf{V} right|^2 - left| mathbf{U} - mathbf{V} right|^2}{4}
$$
I suppose then that the above shall be equal to
$$
mathbf{U}cdotmathbf{V} = |mathbf{U}||mathbf{V}|cos{alpha}
$$
where $alpha$ is the angle between $mathbf{U}$ and $mathbf{V}$. How can it be proved that both identities are equal?
vectors analytic-geometry inner-product-space
asked Dec 13 '18 at 21:18
jrojasqujrojasqu
209413
$endgroup$
$begingroup$
I don't think you'd really prove both identities equal. It's more straightforward to show that each identity holds independently of the other, which gives you the equality.
$endgroup$
– platty
Dec 13 '18 at 21:21
3
$begingroup$
Hint : $$(a+b)^2 - (a-b)^2 = 4ab$$
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– rsadhvika
Dec 13 '18 at 21:23
$begingroup$
@platty Indeed, they hold independently of each other
$endgroup$
– jrojasqu
Dec 13 '18 at 21:38
add a comment |
$begingroup$
In a book called Introduction to tensor analysis and the calculus of moving surfaces equation (2.6) gives a formula to calculate the dot product between two vectors in terms of length alone. That formula is the following:
$$
mathbf{U}cdot{V} = frac{left| mathbf{U} + mathbf{V} right|^2 - left| mathbf{U} - mathbf{V} right|^2}{4}
$$
I suppose then that the above shall be equal to
$$
mathbf{U}cdotmathbf{V} = |mathbf{U}||mathbf{V}|cos{alpha}
$$
where $alpha$ is the angle between $mathbf{U}$ and $mathbf{V}$. How can it be proved that both identities are equal?
vectors analytic-geometry inner-product-space
asked Dec 13 '18 at 21:18
jrojasqujrojasqu
209413
$endgroup$
In a book called Introduction to tensor analysis and the calculus of moving surfaces equation (2.6) gives a formula to calculate the dot product between two vectors in terms of length alone. That formula is the following:
$$
mathbf{U}cdot{V} = frac{left| mathbf{U} + mathbf{V} right|^2 - left| mathbf{U} - mathbf{V} right|^2}{4}
$$
I suppose then that the above shall be equal to
$$
mathbf{U}cdotmathbf{V} = |mathbf{U}||mathbf{V}|cos{alpha}
$$
where $alpha$ is the angle between $mathbf{U}$ and $mathbf{V}$. How can it be proved that both identities are equal?
vectors analytic-geometry inner-product-space
vectors analytic-geometry inner-product-space
asked Dec 13 '18 at 21:18
jrojasqujrojasqu
209413
asked Dec 13 '18 at 21:18
jrojasqujrojasqu
209413
asked Dec 13 '18 at 21:18
jrojasqujrojasqu
209413
asked Dec 13 '18 at 21:18
jrojasqujrojasqu
209413
asked Dec 13 '18 at 21:18
jrojasqujrojasqu
209413
209413
$begingroup$
I don't think you'd really prove both identities equal. It's more straightforward to show that each identity holds independently of the other, which gives you the equality.
$endgroup$
– platty
Dec 13 '18 at 21:21
3
$begingroup$
Hint : $$(a+b)^2 - (a-b)^2 = 4ab$$
$endgroup$
– rsadhvika
Dec 13 '18 at 21:23
$begingroup$
@platty Indeed, they hold independently of each other
$endgroup$
– jrojasqu
Dec 13 '18 at 21:38
add a comment |
$begingroup$
I don't think you'd really prove both identities equal. It's more straightforward to show that each identity holds independently of the other, which gives you the equality.
$endgroup$
– platty
Dec 13 '18 at 21:21
3
$begingroup$
Hint : $$(a+b)^2 - (a-b)^2 = 4ab$$
$endgroup$
– rsadhvika
Dec 13 '18 at 21:23
$begingroup$
@platty Indeed, they hold independently of each other
$endgroup$
– jrojasqu
Dec 13 '18 at 21:38
$begingroup$
I don't think you'd really prove both identities equal. It's more straightforward to show that each identity holds independently of the other, which gives you the equality.
$endgroup$
– platty
Dec 13 '18 at 21:21
$begingroup$
I don't think you'd really prove both identities equal. It's more straightforward to show that each identity holds independently of the other, which gives you the equality.
$endgroup$
– platty
Dec 13 '18 at 21:21
3
3
$begingroup$
Hint : $$(a+b)^2 - (a-b)^2 = 4ab$$
$endgroup$
– rsadhvika
Dec 13 '18 at 21:23
$begingroup$
Hint : $$(a+b)^2 - (a-b)^2 = 4ab$$
$endgroup$
– rsadhvika
Dec 13 '18 at 21:23
$begingroup$
@platty Indeed, they hold independently of each other
$endgroup$
– jrojasqu
Dec 13 '18 at 21:38
$begingroup$
@platty Indeed, they hold independently of each other
$endgroup$
– jrojasqu
Dec 13 '18 at 21:38
add a comment |
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$begingroup$
I don't think you'd really prove both identities equal. It's more straightforward to show that each identity holds independently of the other, which gives you the equality.
$endgroup$
– platty
Dec 13 '18 at 21:21
3
$begingroup$
Hint : $$(a+b)^2 - (a-b)^2 = 4ab$$
$endgroup$
– rsadhvika
Dec 13 '18 at 21:23
$begingroup$
@platty Indeed, they hold independently of each other
$endgroup$
– jrojasqu
Dec 13 '18 at 21:38