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do 3D perpendicular lines have negative reciprocal slopes?
Clash Royale CLAN TAG#URR8PPP
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it's stated in this answer that the directional vector is a nice analog for the slope in $3D$ space
$⟨x,y,z⟩=⟨x0,y0,z0⟩+t⟨mx,my,mz⟩$
now suppose I have two perpidicular lines $L1$ and $L2$ where $L_1$ is
$(a,b,c) + t (m_x,m_y,m_z)$.
The slopes of the perpendicular lines should be the negative reciprocals of each other thus I tried to define $L2$ as
$⟨x_2,y_2,z_2⟩ = (a,b,c) + t (-1/m_x,-1/m_y,-1/m_z)$
for many reasons, clearly that is not true, e.g. the dot product between the two directional vectors would equal $-3$ not $zero$, is the directional vector the same thing as the slope? do the perpendicular lines only have negative reciprocal slopes in $2D$ space? if not, what would be a nice analog to explain that fact in $3D$?
algebra-precalculus geometry
asked 2 hours ago
kk96kk
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up vote
2
down vote
favorite
it's stated in this answer that the directional vector is a nice analog for the slope in $3D$ space
$⟨x,y,z⟩=⟨x0,y0,z0⟩+t⟨mx,my,mz⟩$
now suppose I have two perpidicular lines $L1$ and $L2$ where $L_1$ is
$(a,b,c) + t (m_x,m_y,m_z)$.
The slopes of the perpendicular lines should be the negative reciprocals of each other thus I tried to define $L2$ as
$⟨x_2,y_2,z_2⟩ = (a,b,c) + t (-1/m_x,-1/m_y,-1/m_z)$
for many reasons, clearly that is not true, e.g. the dot product between the two directional vectors would equal $-3$ not $zero$, is the directional vector the same thing as the slope? do the perpendicular lines only have negative reciprocal slopes in $2D$ space? if not, what would be a nice analog to explain that fact in $3D$?
algebra-precalculus geometry
asked 2 hours ago
kk96kk
135
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
it's stated in this answer that the directional vector is a nice analog for the slope in $3D$ space
$⟨x,y,z⟩=⟨x0,y0,z0⟩+t⟨mx,my,mz⟩$
now suppose I have two perpidicular lines $L1$ and $L2$ where $L_1$ is
$(a,b,c) + t (m_x,m_y,m_z)$.
The slopes of the perpendicular lines should be the negative reciprocals of each other thus I tried to define $L2$ as
$⟨x_2,y_2,z_2⟩ = (a,b,c) + t (-1/m_x,-1/m_y,-1/m_z)$
for many reasons, clearly that is not true, e.g. the dot product between the two directional vectors would equal $-3$ not $zero$, is the directional vector the same thing as the slope? do the perpendicular lines only have negative reciprocal slopes in $2D$ space? if not, what would be a nice analog to explain that fact in $3D$?
algebra-precalculus geometry
asked 2 hours ago
kk96kk
135
it's stated in this answer that the directional vector is a nice analog for the slope in $3D$ space
$⟨x,y,z⟩=⟨x0,y0,z0⟩+t⟨mx,my,mz⟩$
now suppose I have two perpidicular lines $L1$ and $L2$ where $L_1$ is
$(a,b,c) + t (m_x,m_y,m_z)$.
The slopes of the perpendicular lines should be the negative reciprocals of each other thus I tried to define $L2$ as
$⟨x_2,y_2,z_2⟩ = (a,b,c) + t (-1/m_x,-1/m_y,-1/m_z)$
for many reasons, clearly that is not true, e.g. the dot product between the two directional vectors would equal $-3$ not $zero$, is the directional vector the same thing as the slope? do the perpendicular lines only have negative reciprocal slopes in $2D$ space? if not, what would be a nice analog to explain that fact in $3D$?
algebra-precalculus geometry
algebra-precalculus geometry
asked 2 hours ago
kk96kk
135
asked 2 hours ago
kk96kk
135
asked 2 hours ago
kk96kk
135
asked 2 hours ago
kk96kk
135
asked 2 hours ago
kk96kk
135
135
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2 Answers
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The "negative reciprocal" argument in the plane is just a restatement of the more natural dot product condition. In the plane a line has a direction vector too: it can be written as
$$
(x,y) = (a,b) + t(m_x, m_y) .
$$
The slope when the line is not parallel to the $y$=axis isis then $m_y/m_x$.
Now the assertion that the product of two slopes is $-1$ is equivalent to the assertion that the dot product of the two direction vectors is $0$.
answered 2 hours ago
Ethan Bolker
38.1k543101
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up vote
1
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Notice that in two dimensions, the slope (when defined) is the ratio of the components of the directional vector:
$$ langle x,yrangle = langle x_0, y_0rangle + tlangle m_x, m_yrangle land m_x neq 0
implies y = fracm_ym_x x + left(y_0 - fracm_ym_xx_0 right). $$
Note that flipping the signs of $m_x$ and $m_y$ simultaneously does nothing to the line at all.
The line given by
$langle x,yrangle = langle x_0, y_0rangle + tlangle -m_x, -m_yrangle$
is exactly the same as the line given by
$langle x,yrangle = langle x_0, y_0rangle + tlangle m_x, m_yrangle.$
The "negative reciprocal" rule is actually telling you to change
$fracm_ym_x$ to $-fracm_xm_y.$ A simple way to do this is to swap the two components and negate just one of them:
$langle x,yrangle = langle x_0, y_0rangle + tlangle m_y, -m_xrangle.$
Notice how the dot product shows that the new vector $langle m_y, -m_xrangle$
is perpendicular to the old vector $langle m_x, m_yrangle$:
$$
langle m_x, m_yrangle cdot langle m_y, -m_xrangle = m_x m_y + m_y(-m_x) = 0.
$$
In short, even in two dimensions, the rule is not to take the negative reciprocal of each component of the direction vector simultaneously.
In two dimensions there is only one line perpendicular to a given line,
but in three dimensions, there are many lines perpendicular to a given line.
We therefore have more choices for the direction vector for the perpendicular line.
Given a direction vector $langle m_x, m_y, m_zrangle,$
where $m_x$ and $m_y$ are not both zero,
one possible choice of a perpendicular vector is
$langle m_y, -m_x, 0rangle,$
which is an application of the negative reciprocal rule to the first two dimensions;
this still works, even in three dimensions, as long as we make sure not to allow the third dimension to make the dot product non-zero.
(That is why the third component of the perpendicular vector is zero.)
Another vector perpendicular to both of these is
$langle m_x m_z, m_y m_z, -(m_x^2 + m_y^2)rangle.$
answered 37 mins ago
David K
50.5k340113
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The "negative reciprocal" argument in the plane is just a restatement of the more natural dot product condition. In the plane a line has a direction vector too: it can be written as
$$
(x,y) = (a,b) + t(m_x, m_y) .
$$
The slope when the line is not parallel to the $y$=axis isis then $m_y/m_x$.
Now the assertion that the product of two slopes is $-1$ is equivalent to the assertion that the dot product of the two direction vectors is $0$.
answered 2 hours ago
Ethan Bolker
38.1k543101
add a comment |Â
up vote
2
down vote
The "negative reciprocal" argument in the plane is just a restatement of the more natural dot product condition. In the plane a line has a direction vector too: it can be written as
$$
(x,y) = (a,b) + t(m_x, m_y) .
$$
The slope when the line is not parallel to the $y$=axis isis then $m_y/m_x$.
Now the assertion that the product of two slopes is $-1$ is equivalent to the assertion that the dot product of the two direction vectors is $0$.
answered 2 hours ago
Ethan Bolker
38.1k543101
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The "negative reciprocal" argument in the plane is just a restatement of the more natural dot product condition. In the plane a line has a direction vector too: it can be written as
$$
(x,y) = (a,b) + t(m_x, m_y) .
$$
The slope when the line is not parallel to the $y$=axis isis then $m_y/m_x$.
Now the assertion that the product of two slopes is $-1$ is equivalent to the assertion that the dot product of the two direction vectors is $0$.
answered 2 hours ago
Ethan Bolker
38.1k543101
The "negative reciprocal" argument in the plane is just a restatement of the more natural dot product condition. In the plane a line has a direction vector too: it can be written as
$$
(x,y) = (a,b) + t(m_x, m_y) .
$$
The slope when the line is not parallel to the $y$=axis isis then $m_y/m_x$.
Now the assertion that the product of two slopes is $-1$ is equivalent to the assertion that the dot product of the two direction vectors is $0$.
answered 2 hours ago
Ethan Bolker
38.1k543101
answered 2 hours ago
Ethan Bolker
38.1k543101
answered 2 hours ago
Ethan Bolker
38.1k543101
answered 2 hours ago
Ethan Bolker
38.1k543101
38.1k543101
add a comment |Â
add a comment |Â
up vote
1
down vote
Notice that in two dimensions, the slope (when defined) is the ratio of the components of the directional vector:
$$ langle x,yrangle = langle x_0, y_0rangle + tlangle m_x, m_yrangle land m_x neq 0
implies y = fracm_ym_x x + left(y_0 - fracm_ym_xx_0 right). $$
Note that flipping the signs of $m_x$ and $m_y$ simultaneously does nothing to the line at all.
The line given by
$langle x,yrangle = langle x_0, y_0rangle + tlangle -m_x, -m_yrangle$
is exactly the same as the line given by
$langle x,yrangle = langle x_0, y_0rangle + tlangle m_x, m_yrangle.$
The "negative reciprocal" rule is actually telling you to change
$fracm_ym_x$ to $-fracm_xm_y.$ A simple way to do this is to swap the two components and negate just one of them:
$langle x,yrangle = langle x_0, y_0rangle + tlangle m_y, -m_xrangle.$
Notice how the dot product shows that the new vector $langle m_y, -m_xrangle$
is perpendicular to the old vector $langle m_x, m_yrangle$:
$$
langle m_x, m_yrangle cdot langle m_y, -m_xrangle = m_x m_y + m_y(-m_x) = 0.
$$
In short, even in two dimensions, the rule is not to take the negative reciprocal of each component of the direction vector simultaneously.
In two dimensions there is only one line perpendicular to a given line,
but in three dimensions, there are many lines perpendicular to a given line.
We therefore have more choices for the direction vector for the perpendicular line.
Given a direction vector $langle m_x, m_y, m_zrangle,$
where $m_x$ and $m_y$ are not both zero,
one possible choice of a perpendicular vector is
$langle m_y, -m_x, 0rangle,$
which is an application of the negative reciprocal rule to the first two dimensions;
this still works, even in three dimensions, as long as we make sure not to allow the third dimension to make the dot product non-zero.
(That is why the third component of the perpendicular vector is zero.)
Another vector perpendicular to both of these is
$langle m_x m_z, m_y m_z, -(m_x^2 + m_y^2)rangle.$
answered 37 mins ago
David K
50.5k340113
add a comment |Â
up vote
1
down vote
Notice that in two dimensions, the slope (when defined) is the ratio of the components of the directional vector:
$$ langle x,yrangle = langle x_0, y_0rangle + tlangle m_x, m_yrangle land m_x neq 0
implies y = fracm_ym_x x + left(y_0 - fracm_ym_xx_0 right). $$
Note that flipping the signs of $m_x$ and $m_y$ simultaneously does nothing to the line at all.
The line given by
$langle x,yrangle = langle x_0, y_0rangle + tlangle -m_x, -m_yrangle$
is exactly the same as the line given by
$langle x,yrangle = langle x_0, y_0rangle + tlangle m_x, m_yrangle.$
The "negative reciprocal" rule is actually telling you to change
$fracm_ym_x$ to $-fracm_xm_y.$ A simple way to do this is to swap the two components and negate just one of them:
$langle x,yrangle = langle x_0, y_0rangle + tlangle m_y, -m_xrangle.$
Notice how the dot product shows that the new vector $langle m_y, -m_xrangle$
is perpendicular to the old vector $langle m_x, m_yrangle$:
$$
langle m_x, m_yrangle cdot langle m_y, -m_xrangle = m_x m_y + m_y(-m_x) = 0.
$$
In short, even in two dimensions, the rule is not to take the negative reciprocal of each component of the direction vector simultaneously.
In two dimensions there is only one line perpendicular to a given line,
but in three dimensions, there are many lines perpendicular to a given line.
We therefore have more choices for the direction vector for the perpendicular line.
Given a direction vector $langle m_x, m_y, m_zrangle,$
where $m_x$ and $m_y$ are not both zero,
one possible choice of a perpendicular vector is
$langle m_y, -m_x, 0rangle,$
which is an application of the negative reciprocal rule to the first two dimensions;
this still works, even in three dimensions, as long as we make sure not to allow the third dimension to make the dot product non-zero.
(That is why the third component of the perpendicular vector is zero.)
Another vector perpendicular to both of these is
$langle m_x m_z, m_y m_z, -(m_x^2 + m_y^2)rangle.$
answered 37 mins ago
David K
50.5k340113
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Notice that in two dimensions, the slope (when defined) is the ratio of the components of the directional vector:
$$ langle x,yrangle = langle x_0, y_0rangle + tlangle m_x, m_yrangle land m_x neq 0
implies y = fracm_ym_x x + left(y_0 - fracm_ym_xx_0 right). $$
Note that flipping the signs of $m_x$ and $m_y$ simultaneously does nothing to the line at all.
The line given by
$langle x,yrangle = langle x_0, y_0rangle + tlangle -m_x, -m_yrangle$
is exactly the same as the line given by
$langle x,yrangle = langle x_0, y_0rangle + tlangle m_x, m_yrangle.$
The "negative reciprocal" rule is actually telling you to change
$fracm_ym_x$ to $-fracm_xm_y.$ A simple way to do this is to swap the two components and negate just one of them:
$langle x,yrangle = langle x_0, y_0rangle + tlangle m_y, -m_xrangle.$
Notice how the dot product shows that the new vector $langle m_y, -m_xrangle$
is perpendicular to the old vector $langle m_x, m_yrangle$:
$$
langle m_x, m_yrangle cdot langle m_y, -m_xrangle = m_x m_y + m_y(-m_x) = 0.
$$
In short, even in two dimensions, the rule is not to take the negative reciprocal of each component of the direction vector simultaneously.
In two dimensions there is only one line perpendicular to a given line,
but in three dimensions, there are many lines perpendicular to a given line.
We therefore have more choices for the direction vector for the perpendicular line.
Given a direction vector $langle m_x, m_y, m_zrangle,$
where $m_x$ and $m_y$ are not both zero,
one possible choice of a perpendicular vector is
$langle m_y, -m_x, 0rangle,$
which is an application of the negative reciprocal rule to the first two dimensions;
this still works, even in three dimensions, as long as we make sure not to allow the third dimension to make the dot product non-zero.
(That is why the third component of the perpendicular vector is zero.)
Another vector perpendicular to both of these is
$langle m_x m_z, m_y m_z, -(m_x^2 + m_y^2)rangle.$
answered 37 mins ago
David K
50.5k340113
Notice that in two dimensions, the slope (when defined) is the ratio of the components of the directional vector:
$$ langle x,yrangle = langle x_0, y_0rangle + tlangle m_x, m_yrangle land m_x neq 0
implies y = fracm_ym_x x + left(y_0 - fracm_ym_xx_0 right). $$
Note that flipping the signs of $m_x$ and $m_y$ simultaneously does nothing to the line at all.
The line given by
$langle x,yrangle = langle x_0, y_0rangle + tlangle -m_x, -m_yrangle$
is exactly the same as the line given by
$langle x,yrangle = langle x_0, y_0rangle + tlangle m_x, m_yrangle.$
The "negative reciprocal" rule is actually telling you to change
$fracm_ym_x$ to $-fracm_xm_y.$ A simple way to do this is to swap the two components and negate just one of them:
$langle x,yrangle = langle x_0, y_0rangle + tlangle m_y, -m_xrangle.$
Notice how the dot product shows that the new vector $langle m_y, -m_xrangle$
is perpendicular to the old vector $langle m_x, m_yrangle$:
$$
langle m_x, m_yrangle cdot langle m_y, -m_xrangle = m_x m_y + m_y(-m_x) = 0.
$$
In short, even in two dimensions, the rule is not to take the negative reciprocal of each component of the direction vector simultaneously.
In two dimensions there is only one line perpendicular to a given line,
but in three dimensions, there are many lines perpendicular to a given line.
We therefore have more choices for the direction vector for the perpendicular line.
Given a direction vector $langle m_x, m_y, m_zrangle,$
where $m_x$ and $m_y$ are not both zero,
one possible choice of a perpendicular vector is
$langle m_y, -m_x, 0rangle,$
which is an application of the negative reciprocal rule to the first two dimensions;
this still works, even in three dimensions, as long as we make sure not to allow the third dimension to make the dot product non-zero.
(That is why the third component of the perpendicular vector is zero.)
Another vector perpendicular to both of these is
$langle m_x m_z, m_y m_z, -(m_x^2 + m_y^2)rangle.$
answered 37 mins ago
David K
50.5k340113
answered 37 mins ago
David K
50.5k340113
answered 37 mins ago
David K
50.5k340113
answered 37 mins ago
David K
50.5k340113
50.5k340113
add a comment |Â
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