Mixing
Do these relationships hold true for all vectors?
Clash Royale CLAN TAG#URR8PPP
up vote
1
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$(u times v)·w$ is the negated version of $(u times w)·v$ and vice versa
$u times (v + w) = (u times v) + (u times w)$
I believe that these both hold true for all vectors, even when I substitute a zero vector into either u, v or w. But I am not 100% confident, could somebody please clarify. Thanks
vectors
edited Jul 30 at 12:15
xbh
1,0257
asked Jul 30 at 11:15
JohnLovesMaths
524
add a comment |Â
up vote
1
down vote
favorite
$(u times v)·w$ is the negated version of $(u times w)·v$ and vice versa
$u times (v + w) = (u times v) + (u times w)$
I believe that these both hold true for all vectors, even when I substitute a zero vector into either u, v or w. But I am not 100% confident, could somebody please clarify. Thanks
vectors
edited Jul 30 at 12:15
xbh
1,0257
asked Jul 30 at 11:15
JohnLovesMaths
524
Hint: Express the triple product as a determinant.
– amd
Jul 31 at 0:20
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$(u times v)·w$ is the negated version of $(u times w)·v$ and vice versa
$u times (v + w) = (u times v) + (u times w)$
I believe that these both hold true for all vectors, even when I substitute a zero vector into either u, v or w. But I am not 100% confident, could somebody please clarify. Thanks
vectors
edited Jul 30 at 12:15
xbh
1,0257
asked Jul 30 at 11:15
JohnLovesMaths
524
$(u times v)·w$ is the negated version of $(u times w)·v$ and vice versa
$u times (v + w) = (u times v) + (u times w)$
I believe that these both hold true for all vectors, even when I substitute a zero vector into either u, v or w. But I am not 100% confident, could somebody please clarify. Thanks
vectors
edited Jul 30 at 12:15
xbh
1,0257
asked Jul 30 at 11:15
JohnLovesMaths
524
edited Jul 30 at 12:15
xbh
1,0257
edited Jul 30 at 12:15
xbh
1,0257
edited Jul 30 at 12:15
xbh
1,0257
1,0257
asked Jul 30 at 11:15
JohnLovesMaths
524
asked Jul 30 at 11:15
JohnLovesMaths
524
asked Jul 30 at 11:15
JohnLovesMaths
524
524
Hint: Express the triple product as a determinant.
– amd
Jul 31 at 0:20
add a comment |Â
Hint: Express the triple product as a determinant.
– amd
Jul 31 at 0:20
Hint: Express the triple product as a determinant.
– amd
Jul 31 at 0:20
Hint: Express the triple product as a determinant.
– amd
Jul 31 at 0:20
add a comment |Â
2 Answers
2
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oldest
votes
up vote
0
down vote
Yes.
The first can be proven from anticommutativity of the cross product and associativity of the triple product:
$$(utimes v)cdot w = ucdot(vtimes w) = ucdot(-wtimes v) = -ucdot(wtimes v) = -(utimes w)cdot v$$
The second is a statement of "linearity", that the cross product distributes over addition.
In both cases, if $u = 0$, then all the products are $0$.
answered Jul 30 at 13:26
mr_e_man
748219
add a comment |Â
up vote
0
down vote
Let's start with
$u times (v + w) = (u times v) + (u times w)$
This is true, and this is one of the most important properties of vector's cross-product operation: it is linear. I guess the easiest way to make sure it's true is to write cross-product formula in coordinates form. F.e. if we have $a = b times (c + d)$ than:
$a_x = b_y (c_z+d_z) - b_z (c_y+d_y) = (b_y c_z - b_z c_y) + (b_y d_z - b_z d_y) $
and it's the same as sum of $x$ coordinates of two vectors: $b times c$ and $b times d$.
Situation is similar with other coordinates, and we have $a = b times (c + d) = b times c + b times d$
Now back to your first equality.
$(u times v)·w = - (u times w)·v$
This is also correct!
Note, that $|u times v|$ is the area of the parallelogram formed by vectors $u$ and $v$. The vector $u times v$ itself is orthogonal to this parallelogram. That means that $|(u times v)·w|$ is a volume of parallelepiped formed by vectors $u$, $v$ and $w$! So it must be $$|(u times v)·w| = |(u times w)·v| = V$$
Actually $(u times v)·w$ is a volume of oriented parallelepiped - it can be positive or negative depending on vectors and changes sign if you change the direction of one of the vector's to the opposite (very convenient property if you think about it!). It also changes sign if you switch any two vectors with each other. I'll leave the proof of this last property to you.
answered Jul 30 at 14:05
lesnik
1,245610
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes.
The first can be proven from anticommutativity of the cross product and associativity of the triple product:
$$(utimes v)cdot w = ucdot(vtimes w) = ucdot(-wtimes v) = -ucdot(wtimes v) = -(utimes w)cdot v$$
The second is a statement of "linearity", that the cross product distributes over addition.
In both cases, if $u = 0$, then all the products are $0$.
answered Jul 30 at 13:26
mr_e_man
748219
add a comment |Â
up vote
0
down vote
Yes.
The first can be proven from anticommutativity of the cross product and associativity of the triple product:
$$(utimes v)cdot w = ucdot(vtimes w) = ucdot(-wtimes v) = -ucdot(wtimes v) = -(utimes w)cdot v$$
The second is a statement of "linearity", that the cross product distributes over addition.
In both cases, if $u = 0$, then all the products are $0$.
answered Jul 30 at 13:26
mr_e_man
748219
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes.
The first can be proven from anticommutativity of the cross product and associativity of the triple product:
$$(utimes v)cdot w = ucdot(vtimes w) = ucdot(-wtimes v) = -ucdot(wtimes v) = -(utimes w)cdot v$$
The second is a statement of "linearity", that the cross product distributes over addition.
In both cases, if $u = 0$, then all the products are $0$.
answered Jul 30 at 13:26
mr_e_man
748219
Yes.
The first can be proven from anticommutativity of the cross product and associativity of the triple product:
$$(utimes v)cdot w = ucdot(vtimes w) = ucdot(-wtimes v) = -ucdot(wtimes v) = -(utimes w)cdot v$$
The second is a statement of "linearity", that the cross product distributes over addition.
In both cases, if $u = 0$, then all the products are $0$.
answered Jul 30 at 13:26
mr_e_man
748219
answered Jul 30 at 13:26
mr_e_man
748219
answered Jul 30 at 13:26
mr_e_man
748219
answered Jul 30 at 13:26
mr_e_man
748219
748219
add a comment |Â
add a comment |Â
up vote
0
down vote
Let's start with
$u times (v + w) = (u times v) + (u times w)$
This is true, and this is one of the most important properties of vector's cross-product operation: it is linear. I guess the easiest way to make sure it's true is to write cross-product formula in coordinates form. F.e. if we have $a = b times (c + d)$ than:
$a_x = b_y (c_z+d_z) - b_z (c_y+d_y) = (b_y c_z - b_z c_y) + (b_y d_z - b_z d_y) $
and it's the same as sum of $x$ coordinates of two vectors: $b times c$ and $b times d$.
Situation is similar with other coordinates, and we have $a = b times (c + d) = b times c + b times d$
Now back to your first equality.
$(u times v)·w = - (u times w)·v$
This is also correct!
Note, that $|u times v|$ is the area of the parallelogram formed by vectors $u$ and $v$. The vector $u times v$ itself is orthogonal to this parallelogram. That means that $|(u times v)·w|$ is a volume of parallelepiped formed by vectors $u$, $v$ and $w$! So it must be $$|(u times v)·w| = |(u times w)·v| = V$$
Actually $(u times v)·w$ is a volume of oriented parallelepiped - it can be positive or negative depending on vectors and changes sign if you change the direction of one of the vector's to the opposite (very convenient property if you think about it!). It also changes sign if you switch any two vectors with each other. I'll leave the proof of this last property to you.
answered Jul 30 at 14:05
lesnik
1,245610
add a comment |Â
up vote
0
down vote
Let's start with
$u times (v + w) = (u times v) + (u times w)$
This is true, and this is one of the most important properties of vector's cross-product operation: it is linear. I guess the easiest way to make sure it's true is to write cross-product formula in coordinates form. F.e. if we have $a = b times (c + d)$ than:
$a_x = b_y (c_z+d_z) - b_z (c_y+d_y) = (b_y c_z - b_z c_y) + (b_y d_z - b_z d_y) $
and it's the same as sum of $x$ coordinates of two vectors: $b times c$ and $b times d$.
Situation is similar with other coordinates, and we have $a = b times (c + d) = b times c + b times d$
Now back to your first equality.
$(u times v)·w = - (u times w)·v$
This is also correct!
Note, that $|u times v|$ is the area of the parallelogram formed by vectors $u$ and $v$. The vector $u times v$ itself is orthogonal to this parallelogram. That means that $|(u times v)·w|$ is a volume of parallelepiped formed by vectors $u$, $v$ and $w$! So it must be $$|(u times v)·w| = |(u times w)·v| = V$$
Actually $(u times v)·w$ is a volume of oriented parallelepiped - it can be positive or negative depending on vectors and changes sign if you change the direction of one of the vector's to the opposite (very convenient property if you think about it!). It also changes sign if you switch any two vectors with each other. I'll leave the proof of this last property to you.
answered Jul 30 at 14:05
lesnik
1,245610
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let's start with
$u times (v + w) = (u times v) + (u times w)$
This is true, and this is one of the most important properties of vector's cross-product operation: it is linear. I guess the easiest way to make sure it's true is to write cross-product formula in coordinates form. F.e. if we have $a = b times (c + d)$ than:
$a_x = b_y (c_z+d_z) - b_z (c_y+d_y) = (b_y c_z - b_z c_y) + (b_y d_z - b_z d_y) $
and it's the same as sum of $x$ coordinates of two vectors: $b times c$ and $b times d$.
Situation is similar with other coordinates, and we have $a = b times (c + d) = b times c + b times d$
Now back to your first equality.
$(u times v)·w = - (u times w)·v$
This is also correct!
Note, that $|u times v|$ is the area of the parallelogram formed by vectors $u$ and $v$. The vector $u times v$ itself is orthogonal to this parallelogram. That means that $|(u times v)·w|$ is a volume of parallelepiped formed by vectors $u$, $v$ and $w$! So it must be $$|(u times v)·w| = |(u times w)·v| = V$$
Actually $(u times v)·w$ is a volume of oriented parallelepiped - it can be positive or negative depending on vectors and changes sign if you change the direction of one of the vector's to the opposite (very convenient property if you think about it!). It also changes sign if you switch any two vectors with each other. I'll leave the proof of this last property to you.
answered Jul 30 at 14:05
lesnik
1,245610
Let's start with
$u times (v + w) = (u times v) + (u times w)$
This is true, and this is one of the most important properties of vector's cross-product operation: it is linear. I guess the easiest way to make sure it's true is to write cross-product formula in coordinates form. F.e. if we have $a = b times (c + d)$ than:
$a_x = b_y (c_z+d_z) - b_z (c_y+d_y) = (b_y c_z - b_z c_y) + (b_y d_z - b_z d_y) $
and it's the same as sum of $x$ coordinates of two vectors: $b times c$ and $b times d$.
Situation is similar with other coordinates, and we have $a = b times (c + d) = b times c + b times d$
Now back to your first equality.
$(u times v)·w = - (u times w)·v$
This is also correct!
Note, that $|u times v|$ is the area of the parallelogram formed by vectors $u$ and $v$. The vector $u times v$ itself is orthogonal to this parallelogram. That means that $|(u times v)·w|$ is a volume of parallelepiped formed by vectors $u$, $v$ and $w$! So it must be $$|(u times v)·w| = |(u times w)·v| = V$$
Actually $(u times v)·w$ is a volume of oriented parallelepiped - it can be positive or negative depending on vectors and changes sign if you change the direction of one of the vector's to the opposite (very convenient property if you think about it!). It also changes sign if you switch any two vectors with each other. I'll leave the proof of this last property to you.
answered Jul 30 at 14:05
lesnik
1,245610
answered Jul 30 at 14:05
lesnik
1,245610
answered Jul 30 at 14:05
lesnik
1,245610
answered Jul 30 at 14:05
lesnik
1,245610
1,245610
add a comment |Â
add a comment |Â
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Hint: Express the triple product as a determinant.
– amd
Jul 31 at 0:20