Mixing
What makes two lines in 3-space perpendicular?
$begingroup$
I need to find the equations for the line which passes through the point $(-5,3)$ and is perpendicular to the line $<3+t,-5-2t>$.
I know I can get the direction vector $<1,-2>$ out of the line equation. If I needed a parallel line I would just use the direction vector and the given point to construct a vector parametric equation for the line.
But, I am not sure what it means for two lines to be perpendicular with respect to their direction vector. My textbook (Stewart's Calculus 5th edition) does not define this, and I don't have any other calculus references for the course.
My intuition is to find the vector that is orthogonal to the direction vector, but I'm not sure if this is correct.
I set my direction vector as $vec{a}$ and said $vec{a}cdotvec{b} = vec{0}$, and solving this using the dot production definition gives me $vec{b} = <-1,2>$. Is this the vector I use to define the perpendicular line?
calculus vectors
asked Jan 30 at 1:19
jeanquiltjeanquilt
301213
$endgroup$
add a comment |
$begingroup$
I need to find the equations for the line which passes through the point $(-5,3)$ and is perpendicular to the line $<3+t,-5-2t>$.
I know I can get the direction vector $<1,-2>$ out of the line equation. If I needed a parallel line I would just use the direction vector and the given point to construct a vector parametric equation for the line.
But, I am not sure what it means for two lines to be perpendicular with respect to their direction vector. My textbook (Stewart's Calculus 5th edition) does not define this, and I don't have any other calculus references for the course.
My intuition is to find the vector that is orthogonal to the direction vector, but I'm not sure if this is correct.
I set my direction vector as $vec{a}$ and said $vec{a}cdotvec{b} = vec{0}$, and solving this using the dot production definition gives me $vec{b} = <-1,2>$. Is this the vector I use to define the perpendicular line?
calculus vectors
asked Jan 30 at 1:19
jeanquiltjeanquilt
301213
$endgroup$
add a comment |
$begingroup$
I need to find the equations for the line which passes through the point $(-5,3)$ and is perpendicular to the line $<3+t,-5-2t>$.
I know I can get the direction vector $<1,-2>$ out of the line equation. If I needed a parallel line I would just use the direction vector and the given point to construct a vector parametric equation for the line.
But, I am not sure what it means for two lines to be perpendicular with respect to their direction vector. My textbook (Stewart's Calculus 5th edition) does not define this, and I don't have any other calculus references for the course.
My intuition is to find the vector that is orthogonal to the direction vector, but I'm not sure if this is correct.
I set my direction vector as $vec{a}$ and said $vec{a}cdotvec{b} = vec{0}$, and solving this using the dot production definition gives me $vec{b} = <-1,2>$. Is this the vector I use to define the perpendicular line?
calculus vectors
asked Jan 30 at 1:19
jeanquiltjeanquilt
301213
$endgroup$
I need to find the equations for the line which passes through the point $(-5,3)$ and is perpendicular to the line $<3+t,-5-2t>$.
I know I can get the direction vector $<1,-2>$ out of the line equation. If I needed a parallel line I would just use the direction vector and the given point to construct a vector parametric equation for the line.
But, I am not sure what it means for two lines to be perpendicular with respect to their direction vector. My textbook (Stewart's Calculus 5th edition) does not define this, and I don't have any other calculus references for the course.
My intuition is to find the vector that is orthogonal to the direction vector, but I'm not sure if this is correct.
I set my direction vector as $vec{a}$ and said $vec{a}cdotvec{b} = vec{0}$, and solving this using the dot production definition gives me $vec{b} = <-1,2>$. Is this the vector I use to define the perpendicular line?
calculus vectors
calculus vectors
asked Jan 30 at 1:19
jeanquiltjeanquilt
301213
asked Jan 30 at 1:19
jeanquiltjeanquilt
301213
asked Jan 30 at 1:19
jeanquiltjeanquilt
301213
asked Jan 30 at 1:19
jeanquiltjeanquilt
301213
asked Jan 30 at 1:19
jeanquiltjeanquilt
301213
301213
add a comment |
add a comment |
1 Answer
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$begingroup$
You are correct that you should find a vector that is orthogonal to $<1, -2>$, however the vector you calculated as $vec{b}$ does not seem to satisfy that requirement:
$$
vec{a}cdotvec{b} = (1)(-1) + (-2)(2) = -5 neq 0
$$
One $vec{b}$ that satisfies the requirement would be $<1, 1/2>$, since we would have
$$
vec{a}cdotvec{b} = (1)(1) + (-2)(1/2) = 1 - 1 = 0
$$
as desired. You can then use that vector (or any scalar multiple of it) as the direction vector for your line.
answered Jan 30 at 1:27
AlexAlex
52538
$endgroup$
$begingroup$
So, if I my knowledge of high school plane geometry is correct, two lines are perpendicular if their slopes are negative reciprocals. So, in 3D, I see that principle still applies: I can take the negative reciprocal of each component in the direction vector and get the perpendicular direction vector?
$endgroup$
– jeanquilt
Jan 30 at 1:31
1
$begingroup$
For one thing, it's important to note that we're still talking about 2D here: the line you gave above is only on a plane. To answer your question, not quite. If you try that here, you'd get $<-1, 1/2>$ which is not orthogonal to $<1, -2>$. However, you can switch the entries and flip the sign on one: $<2, 1>$ is orthogonal to $<1, -2>$.
$endgroup$
– Alex
Jan 30 at 1:37
$begingroup$
So, if I find any one orthogonal vector, will all of the scalar multiples of that vector be orthogonal to the original direction as well? I notice that $<2,1>$ is equal to $2*<1,1/2>$. I think I've constructed a rough proof that this is true but I wanted to double check in case I made a mathematical mistake.
$endgroup$
– jeanquilt
Jan 30 at 1:51
1
$begingroup$
Yes, that's true. Think about it geometrically: two vectors on the plane are orthogonal if there is a right angle between them. Making any vector $c$ times longer (i.e. multiplying it by scalar $c$) does not change its direction, so it won't change the angle.
$endgroup$
– Alex
Jan 30 at 16:19
add a comment |
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1 Answer
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active
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1 Answer
1
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$begingroup$
You are correct that you should find a vector that is orthogonal to $<1, -2>$, however the vector you calculated as $vec{b}$ does not seem to satisfy that requirement:
$$
vec{a}cdotvec{b} = (1)(-1) + (-2)(2) = -5 neq 0
$$
One $vec{b}$ that satisfies the requirement would be $<1, 1/2>$, since we would have
$$
vec{a}cdotvec{b} = (1)(1) + (-2)(1/2) = 1 - 1 = 0
$$
as desired. You can then use that vector (or any scalar multiple of it) as the direction vector for your line.
answered Jan 30 at 1:27
AlexAlex
52538
$endgroup$
$begingroup$
So, if I my knowledge of high school plane geometry is correct, two lines are perpendicular if their slopes are negative reciprocals. So, in 3D, I see that principle still applies: I can take the negative reciprocal of each component in the direction vector and get the perpendicular direction vector?
$endgroup$
– jeanquilt
Jan 30 at 1:31
1
$begingroup$
For one thing, it's important to note that we're still talking about 2D here: the line you gave above is only on a plane. To answer your question, not quite. If you try that here, you'd get $<-1, 1/2>$ which is not orthogonal to $<1, -2>$. However, you can switch the entries and flip the sign on one: $<2, 1>$ is orthogonal to $<1, -2>$.
$endgroup$
– Alex
Jan 30 at 1:37
$begingroup$
So, if I find any one orthogonal vector, will all of the scalar multiples of that vector be orthogonal to the original direction as well? I notice that $<2,1>$ is equal to $2*<1,1/2>$. I think I've constructed a rough proof that this is true but I wanted to double check in case I made a mathematical mistake.
$endgroup$
– jeanquilt
Jan 30 at 1:51
1
$begingroup$
Yes, that's true. Think about it geometrically: two vectors on the plane are orthogonal if there is a right angle between them. Making any vector $c$ times longer (i.e. multiplying it by scalar $c$) does not change its direction, so it won't change the angle.
$endgroup$
– Alex
Jan 30 at 16:19
add a comment |
$begingroup$
You are correct that you should find a vector that is orthogonal to $<1, -2>$, however the vector you calculated as $vec{b}$ does not seem to satisfy that requirement:
$$
vec{a}cdotvec{b} = (1)(-1) + (-2)(2) = -5 neq 0
$$
One $vec{b}$ that satisfies the requirement would be $<1, 1/2>$, since we would have
$$
vec{a}cdotvec{b} = (1)(1) + (-2)(1/2) = 1 - 1 = 0
$$
as desired. You can then use that vector (or any scalar multiple of it) as the direction vector for your line.
answered Jan 30 at 1:27
AlexAlex
52538
$endgroup$
$begingroup$
So, if I my knowledge of high school plane geometry is correct, two lines are perpendicular if their slopes are negative reciprocals. So, in 3D, I see that principle still applies: I can take the negative reciprocal of each component in the direction vector and get the perpendicular direction vector?
$endgroup$
– jeanquilt
Jan 30 at 1:31
1
$begingroup$
For one thing, it's important to note that we're still talking about 2D here: the line you gave above is only on a plane. To answer your question, not quite. If you try that here, you'd get $<-1, 1/2>$ which is not orthogonal to $<1, -2>$. However, you can switch the entries and flip the sign on one: $<2, 1>$ is orthogonal to $<1, -2>$.
$endgroup$
– Alex
Jan 30 at 1:37
$begingroup$
So, if I find any one orthogonal vector, will all of the scalar multiples of that vector be orthogonal to the original direction as well? I notice that $<2,1>$ is equal to $2*<1,1/2>$. I think I've constructed a rough proof that this is true but I wanted to double check in case I made a mathematical mistake.
$endgroup$
– jeanquilt
Jan 30 at 1:51
1
$begingroup$
Yes, that's true. Think about it geometrically: two vectors on the plane are orthogonal if there is a right angle between them. Making any vector $c$ times longer (i.e. multiplying it by scalar $c$) does not change its direction, so it won't change the angle.
$endgroup$
– Alex
Jan 30 at 16:19
add a comment |
$begingroup$
You are correct that you should find a vector that is orthogonal to $<1, -2>$, however the vector you calculated as $vec{b}$ does not seem to satisfy that requirement:
$$
vec{a}cdotvec{b} = (1)(-1) + (-2)(2) = -5 neq 0
$$
One $vec{b}$ that satisfies the requirement would be $<1, 1/2>$, since we would have
$$
vec{a}cdotvec{b} = (1)(1) + (-2)(1/2) = 1 - 1 = 0
$$
as desired. You can then use that vector (or any scalar multiple of it) as the direction vector for your line.
answered Jan 30 at 1:27
AlexAlex
52538
$endgroup$
You are correct that you should find a vector that is orthogonal to $<1, -2>$, however the vector you calculated as $vec{b}$ does not seem to satisfy that requirement:
$$
vec{a}cdotvec{b} = (1)(-1) + (-2)(2) = -5 neq 0
$$
One $vec{b}$ that satisfies the requirement would be $<1, 1/2>$, since we would have
$$
vec{a}cdotvec{b} = (1)(1) + (-2)(1/2) = 1 - 1 = 0
$$
as desired. You can then use that vector (or any scalar multiple of it) as the direction vector for your line.
answered Jan 30 at 1:27
AlexAlex
52538
answered Jan 30 at 1:27
AlexAlex
52538
answered Jan 30 at 1:27
AlexAlex
52538
answered Jan 30 at 1:27
AlexAlex
52538
52538
$begingroup$
So, if I my knowledge of high school plane geometry is correct, two lines are perpendicular if their slopes are negative reciprocals. So, in 3D, I see that principle still applies: I can take the negative reciprocal of each component in the direction vector and get the perpendicular direction vector?
$endgroup$
– jeanquilt
Jan 30 at 1:31
1
$begingroup$
For one thing, it's important to note that we're still talking about 2D here: the line you gave above is only on a plane. To answer your question, not quite. If you try that here, you'd get $<-1, 1/2>$ which is not orthogonal to $<1, -2>$. However, you can switch the entries and flip the sign on one: $<2, 1>$ is orthogonal to $<1, -2>$.
$endgroup$
– Alex
Jan 30 at 1:37
$begingroup$
So, if I find any one orthogonal vector, will all of the scalar multiples of that vector be orthogonal to the original direction as well? I notice that $<2,1>$ is equal to $2*<1,1/2>$. I think I've constructed a rough proof that this is true but I wanted to double check in case I made a mathematical mistake.
$endgroup$
– jeanquilt
Jan 30 at 1:51
1
$begingroup$
Yes, that's true. Think about it geometrically: two vectors on the plane are orthogonal if there is a right angle between them. Making any vector $c$ times longer (i.e. multiplying it by scalar $c$) does not change its direction, so it won't change the angle.
$endgroup$
– Alex
Jan 30 at 16:19
add a comment |
$begingroup$
So, if I my knowledge of high school plane geometry is correct, two lines are perpendicular if their slopes are negative reciprocals. So, in 3D, I see that principle still applies: I can take the negative reciprocal of each component in the direction vector and get the perpendicular direction vector?
$endgroup$
– jeanquilt
Jan 30 at 1:31
1
$begingroup$
For one thing, it's important to note that we're still talking about 2D here: the line you gave above is only on a plane. To answer your question, not quite. If you try that here, you'd get $<-1, 1/2>$ which is not orthogonal to $<1, -2>$. However, you can switch the entries and flip the sign on one: $<2, 1>$ is orthogonal to $<1, -2>$.
$endgroup$
– Alex
Jan 30 at 1:37
$begingroup$
So, if I find any one orthogonal vector, will all of the scalar multiples of that vector be orthogonal to the original direction as well? I notice that $<2,1>$ is equal to $2*<1,1/2>$. I think I've constructed a rough proof that this is true but I wanted to double check in case I made a mathematical mistake.
$endgroup$
– jeanquilt
Jan 30 at 1:51
1
$begingroup$
Yes, that's true. Think about it geometrically: two vectors on the plane are orthogonal if there is a right angle between them. Making any vector $c$ times longer (i.e. multiplying it by scalar $c$) does not change its direction, so it won't change the angle.
$endgroup$
– Alex
Jan 30 at 16:19
$begingroup$
So, if I my knowledge of high school plane geometry is correct, two lines are perpendicular if their slopes are negative reciprocals. So, in 3D, I see that principle still applies: I can take the negative reciprocal of each component in the direction vector and get the perpendicular direction vector?
$endgroup$
– jeanquilt
Jan 30 at 1:31
$begingroup$
So, if I my knowledge of high school plane geometry is correct, two lines are perpendicular if their slopes are negative reciprocals. So, in 3D, I see that principle still applies: I can take the negative reciprocal of each component in the direction vector and get the perpendicular direction vector?
$endgroup$
– jeanquilt
Jan 30 at 1:31
1
1
$begingroup$
For one thing, it's important to note that we're still talking about 2D here: the line you gave above is only on a plane. To answer your question, not quite. If you try that here, you'd get $<-1, 1/2>$ which is not orthogonal to $<1, -2>$. However, you can switch the entries and flip the sign on one: $<2, 1>$ is orthogonal to $<1, -2>$.
$endgroup$
– Alex
Jan 30 at 1:37
$begingroup$
For one thing, it's important to note that we're still talking about 2D here: the line you gave above is only on a plane. To answer your question, not quite. If you try that here, you'd get $<-1, 1/2>$ which is not orthogonal to $<1, -2>$. However, you can switch the entries and flip the sign on one: $<2, 1>$ is orthogonal to $<1, -2>$.
$endgroup$
– Alex
Jan 30 at 1:37
$begingroup$
So, if I find any one orthogonal vector, will all of the scalar multiples of that vector be orthogonal to the original direction as well? I notice that $<2,1>$ is equal to $2*<1,1/2>$. I think I've constructed a rough proof that this is true but I wanted to double check in case I made a mathematical mistake.
$endgroup$
– jeanquilt
Jan 30 at 1:51
$begingroup$
So, if I find any one orthogonal vector, will all of the scalar multiples of that vector be orthogonal to the original direction as well? I notice that $<2,1>$ is equal to $2*<1,1/2>$. I think I've constructed a rough proof that this is true but I wanted to double check in case I made a mathematical mistake.
$endgroup$
– jeanquilt
Jan 30 at 1:51
1
1
$begingroup$
Yes, that's true. Think about it geometrically: two vectors on the plane are orthogonal if there is a right angle between them. Making any vector $c$ times longer (i.e. multiplying it by scalar $c$) does not change its direction, so it won't change the angle.
$endgroup$
– Alex
Jan 30 at 16:19
$begingroup$
Yes, that's true. Think about it geometrically: two vectors on the plane are orthogonal if there is a right angle between them. Making any vector $c$ times longer (i.e. multiplying it by scalar $c$) does not change its direction, so it won't change the angle.
$endgroup$
– Alex
Jan 30 at 16:19
add a comment |
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