Finding the length of the projection of the vector onto a line using parametric equation
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What is the length of the projection of the vector $(3, 4,-4)$ onto a line whose parametric equation is the following?
$$begin{aligned} x &= 2t + 1\ y &= -t + 3\ z &= t - 1end{aligned}$$
Hint: find a unit vector in the direction of the line and construction its projection operators.
linear-algebra analytic-geometry projection
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add a comment |
$begingroup$
What is the length of the projection of the vector $(3, 4,-4)$ onto a line whose parametric equation is the following?
$$begin{aligned} x &= 2t + 1\ y &= -t + 3\ z &= t - 1end{aligned}$$
Hint: find a unit vector in the direction of the line and construction its projection operators.
linear-algebra analytic-geometry projection
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2
$begingroup$
You mean length?
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– Fareed AF
Dec 15 '18 at 9:11
$begingroup$
What is the vector (3, 4-4) supposed to be? Do you mean the vector $(3, 4, 4)$?
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– The Pointer
Dec 15 '18 at 9:22
$begingroup$
I have corrected all of misspellings
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:26
add a comment |
$begingroup$
What is the length of the projection of the vector $(3, 4,-4)$ onto a line whose parametric equation is the following?
$$begin{aligned} x &= 2t + 1\ y &= -t + 3\ z &= t - 1end{aligned}$$
Hint: find a unit vector in the direction of the line and construction its projection operators.
linear-algebra analytic-geometry projection
$endgroup$
What is the length of the projection of the vector $(3, 4,-4)$ onto a line whose parametric equation is the following?
$$begin{aligned} x &= 2t + 1\ y &= -t + 3\ z &= t - 1end{aligned}$$
Hint: find a unit vector in the direction of the line and construction its projection operators.
linear-algebra analytic-geometry projection
linear-algebra analytic-geometry projection
edited Dec 15 '18 at 9:31
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked Dec 15 '18 at 9:08
Pınar OruçPınar Oruç
11
11
2
$begingroup$
You mean length?
$endgroup$
– Fareed AF
Dec 15 '18 at 9:11
$begingroup$
What is the vector (3, 4-4) supposed to be? Do you mean the vector $(3, 4, 4)$?
$endgroup$
– The Pointer
Dec 15 '18 at 9:22
$begingroup$
I have corrected all of misspellings
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:26
add a comment |
2
$begingroup$
You mean length?
$endgroup$
– Fareed AF
Dec 15 '18 at 9:11
$begingroup$
What is the vector (3, 4-4) supposed to be? Do you mean the vector $(3, 4, 4)$?
$endgroup$
– The Pointer
Dec 15 '18 at 9:22
$begingroup$
I have corrected all of misspellings
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:26
2
2
$begingroup$
You mean length?
$endgroup$
– Fareed AF
Dec 15 '18 at 9:11
$begingroup$
You mean length?
$endgroup$
– Fareed AF
Dec 15 '18 at 9:11
$begingroup$
What is the vector (3, 4-4) supposed to be? Do you mean the vector $(3, 4, 4)$?
$endgroup$
– The Pointer
Dec 15 '18 at 9:22
$begingroup$
What is the vector (3, 4-4) supposed to be? Do you mean the vector $(3, 4, 4)$?
$endgroup$
– The Pointer
Dec 15 '18 at 9:22
$begingroup$
I have corrected all of misspellings
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:26
$begingroup$
I have corrected all of misspellings
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:26
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint.
Your line is parallel to the vector $(2,-1,1)$, the unit vector along which is $displaystyle(frac2{sqrt6},frac{-1}{sqrt6},frac1{sqrt6})$. The projection of vector $vec A$ along $vec B$ is given by $displaystylefrac{vec Acdotvec B}{|vec B|}$.
$endgroup$
$begingroup$
Thank you for your help.
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
$begingroup$
You're welcome.
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:12
add a comment |
$begingroup$
To get the length of the projection of this vector on the line yes you first get a unit directing vector to the line.
$vec v(2,-1,1)$ is a directing vector the line its magnitude is $||vec v||=sqrt 6$
So a unit directing vector to the line will be $vec {v'} (frac{2}{sqrt 6},frac{-1}{sqrt 6},frac{1}{sqrt 6})$
Now just do the dot product $|(3,4,-4).vec {v'}|=|frac{6}{sqrt 6} -frac{4}{sqrt 6} -frac{4}{sqrt 6}|=frac{2}{sqrt 6}$
And thats the length of the projection.
Now why does that work
In general, $vec u.vec v= u.v.cos theta$ (where $theta$ is the angle between the 2 vectors)
Draw 2 vectors $vec u$ and $vec v$ and draw the projection of $vec v$ on $vec u$ for example.
You'll have a right triangle, and you'll find that $cos theta=frac{projection}{v}$ (I mean by 'projection' the length of the projection of $vec v$ on $vec u$)
Substitute above and you'll get that $vec u.vec v=u.{projection}$
But in your case you have a unit vector so $u=1$
Thats why $vec u.vec v={projection}$
$endgroup$
$begingroup$
I am sorry for my english because my mather tongue is Turkish. May i ask a question? Won’t i use parametrics equation?
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:37
$begingroup$
Don't worry it is fine. No there is no need to use parametric equations only the dot product, I'll try to explaim my answer more.
$endgroup$
– Fareed AF
Dec 15 '18 at 9:41
$begingroup$
Thank you for your help
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
add a comment |
$begingroup$
One parameterisation of the line is given by $$mathbf{x}(t) = tbegin{pmatrix} 2 \ -1 \ 1end{pmatrix} + begin{pmatrix}1 \ 3 \ 1 end{pmatrix}, t in mathbb{R}.$$ The geometrical interpretation of $mathbf{x}$ is that for every real number $t$, $mathbf{x}(t)$ corresponds to an arrow which points to a point on the line with its root at the origin in $mathbb{R}^3$.
In fact, the outputs of the function $mathbf{x}$ collectively point to every point on the line.
In order to find the length of the projection of $(3,4,-4)$ onto the line, we first find a unit vector which points along the line, and then take the magnitude of the component of $(3,4,-4)$ in the direction of that unit vector. To find a unit vector which points along the line, we find two different vectors which point to different points on the line, take their difference, and then normalise the result.
Finding two different vectors which point to the line is the same as finding two different outputs of $mathbf{x}$. For example, set $t = 0$, to obtain $$mathbf{x}(0) = begin{pmatrix} 1 \ 3 \ 1 end{pmatrix}.$$ Then we can for instance set $t = 1$, to obtain $$mathbf{x}(1) = begin{pmatrix} 2 \-1 \ 1end{pmatrix}+ begin{pmatrix}1 \ 3 \ 1 end{pmatrix} = begin{pmatrix} 3 \ 2 \ 2 end{pmatrix}.$$
Now, the vector pointing from $mathbf{x}(0)$ to $mathbf{x}(1)$ which we will call $mathbf{n}$ is given by $$mathbf{n} = mathbf{x}(1) - mathbf{x}(0) = begin{pmatrix} 2 \ -1 \ 1end{pmatrix}.$$ We then obtain the unit vector pointing along $mathbf{n}$ which we call $hat{mathbf{n}}$ by setting $$hat{mathbf{n}} = frac{mathbf{n}}{||mathbf{n}||} = frac{1}{sqrt{6}}begin{pmatrix} 2 \ -1 \ 1 end{pmatrix}.$$ The magnitude of the component of $(3,4,-4)$ in the direction of the line, i.e. the length of the projection of $(3,4,-4)$ onto the line, is given by $|(3,4,-4) cdot hat{mathbf{n}}| = frac{sqrt{6}}{3}.$
I suggest checking out this series of videos to help you nail the intuitions behind vectors, dot products and much more.
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint.
Your line is parallel to the vector $(2,-1,1)$, the unit vector along which is $displaystyle(frac2{sqrt6},frac{-1}{sqrt6},frac1{sqrt6})$. The projection of vector $vec A$ along $vec B$ is given by $displaystylefrac{vec Acdotvec B}{|vec B|}$.
$endgroup$
$begingroup$
Thank you for your help.
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
$begingroup$
You're welcome.
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:12
add a comment |
$begingroup$
Hint.
Your line is parallel to the vector $(2,-1,1)$, the unit vector along which is $displaystyle(frac2{sqrt6},frac{-1}{sqrt6},frac1{sqrt6})$. The projection of vector $vec A$ along $vec B$ is given by $displaystylefrac{vec Acdotvec B}{|vec B|}$.
$endgroup$
$begingroup$
Thank you for your help.
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
$begingroup$
You're welcome.
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:12
add a comment |
$begingroup$
Hint.
Your line is parallel to the vector $(2,-1,1)$, the unit vector along which is $displaystyle(frac2{sqrt6},frac{-1}{sqrt6},frac1{sqrt6})$. The projection of vector $vec A$ along $vec B$ is given by $displaystylefrac{vec Acdotvec B}{|vec B|}$.
$endgroup$
Hint.
Your line is parallel to the vector $(2,-1,1)$, the unit vector along which is $displaystyle(frac2{sqrt6},frac{-1}{sqrt6},frac1{sqrt6})$. The projection of vector $vec A$ along $vec B$ is given by $displaystylefrac{vec Acdotvec B}{|vec B|}$.
answered Dec 15 '18 at 9:26
Shubham JohriShubham Johri
5,204718
5,204718
$begingroup$
Thank you for your help.
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
$begingroup$
You're welcome.
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:12
add a comment |
$begingroup$
Thank you for your help.
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
$begingroup$
You're welcome.
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:12
$begingroup$
Thank you for your help.
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
$begingroup$
Thank you for your help.
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
$begingroup$
You're welcome.
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:12
$begingroup$
You're welcome.
$endgroup$
– Shubham Johri
Dec 15 '18 at 10:12
add a comment |
$begingroup$
To get the length of the projection of this vector on the line yes you first get a unit directing vector to the line.
$vec v(2,-1,1)$ is a directing vector the line its magnitude is $||vec v||=sqrt 6$
So a unit directing vector to the line will be $vec {v'} (frac{2}{sqrt 6},frac{-1}{sqrt 6},frac{1}{sqrt 6})$
Now just do the dot product $|(3,4,-4).vec {v'}|=|frac{6}{sqrt 6} -frac{4}{sqrt 6} -frac{4}{sqrt 6}|=frac{2}{sqrt 6}$
And thats the length of the projection.
Now why does that work
In general, $vec u.vec v= u.v.cos theta$ (where $theta$ is the angle between the 2 vectors)
Draw 2 vectors $vec u$ and $vec v$ and draw the projection of $vec v$ on $vec u$ for example.
You'll have a right triangle, and you'll find that $cos theta=frac{projection}{v}$ (I mean by 'projection' the length of the projection of $vec v$ on $vec u$)
Substitute above and you'll get that $vec u.vec v=u.{projection}$
But in your case you have a unit vector so $u=1$
Thats why $vec u.vec v={projection}$
$endgroup$
$begingroup$
I am sorry for my english because my mather tongue is Turkish. May i ask a question? Won’t i use parametrics equation?
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:37
$begingroup$
Don't worry it is fine. No there is no need to use parametric equations only the dot product, I'll try to explaim my answer more.
$endgroup$
– Fareed AF
Dec 15 '18 at 9:41
$begingroup$
Thank you for your help
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
add a comment |
$begingroup$
To get the length of the projection of this vector on the line yes you first get a unit directing vector to the line.
$vec v(2,-1,1)$ is a directing vector the line its magnitude is $||vec v||=sqrt 6$
So a unit directing vector to the line will be $vec {v'} (frac{2}{sqrt 6},frac{-1}{sqrt 6},frac{1}{sqrt 6})$
Now just do the dot product $|(3,4,-4).vec {v'}|=|frac{6}{sqrt 6} -frac{4}{sqrt 6} -frac{4}{sqrt 6}|=frac{2}{sqrt 6}$
And thats the length of the projection.
Now why does that work
In general, $vec u.vec v= u.v.cos theta$ (where $theta$ is the angle between the 2 vectors)
Draw 2 vectors $vec u$ and $vec v$ and draw the projection of $vec v$ on $vec u$ for example.
You'll have a right triangle, and you'll find that $cos theta=frac{projection}{v}$ (I mean by 'projection' the length of the projection of $vec v$ on $vec u$)
Substitute above and you'll get that $vec u.vec v=u.{projection}$
But in your case you have a unit vector so $u=1$
Thats why $vec u.vec v={projection}$
$endgroup$
$begingroup$
I am sorry for my english because my mather tongue is Turkish. May i ask a question? Won’t i use parametrics equation?
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:37
$begingroup$
Don't worry it is fine. No there is no need to use parametric equations only the dot product, I'll try to explaim my answer more.
$endgroup$
– Fareed AF
Dec 15 '18 at 9:41
$begingroup$
Thank you for your help
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
add a comment |
$begingroup$
To get the length of the projection of this vector on the line yes you first get a unit directing vector to the line.
$vec v(2,-1,1)$ is a directing vector the line its magnitude is $||vec v||=sqrt 6$
So a unit directing vector to the line will be $vec {v'} (frac{2}{sqrt 6},frac{-1}{sqrt 6},frac{1}{sqrt 6})$
Now just do the dot product $|(3,4,-4).vec {v'}|=|frac{6}{sqrt 6} -frac{4}{sqrt 6} -frac{4}{sqrt 6}|=frac{2}{sqrt 6}$
And thats the length of the projection.
Now why does that work
In general, $vec u.vec v= u.v.cos theta$ (where $theta$ is the angle between the 2 vectors)
Draw 2 vectors $vec u$ and $vec v$ and draw the projection of $vec v$ on $vec u$ for example.
You'll have a right triangle, and you'll find that $cos theta=frac{projection}{v}$ (I mean by 'projection' the length of the projection of $vec v$ on $vec u$)
Substitute above and you'll get that $vec u.vec v=u.{projection}$
But in your case you have a unit vector so $u=1$
Thats why $vec u.vec v={projection}$
$endgroup$
To get the length of the projection of this vector on the line yes you first get a unit directing vector to the line.
$vec v(2,-1,1)$ is a directing vector the line its magnitude is $||vec v||=sqrt 6$
So a unit directing vector to the line will be $vec {v'} (frac{2}{sqrt 6},frac{-1}{sqrt 6},frac{1}{sqrt 6})$
Now just do the dot product $|(3,4,-4).vec {v'}|=|frac{6}{sqrt 6} -frac{4}{sqrt 6} -frac{4}{sqrt 6}|=frac{2}{sqrt 6}$
And thats the length of the projection.
Now why does that work
In general, $vec u.vec v= u.v.cos theta$ (where $theta$ is the angle between the 2 vectors)
Draw 2 vectors $vec u$ and $vec v$ and draw the projection of $vec v$ on $vec u$ for example.
You'll have a right triangle, and you'll find that $cos theta=frac{projection}{v}$ (I mean by 'projection' the length of the projection of $vec v$ on $vec u$)
Substitute above and you'll get that $vec u.vec v=u.{projection}$
But in your case you have a unit vector so $u=1$
Thats why $vec u.vec v={projection}$
edited Dec 15 '18 at 10:46
answered Dec 15 '18 at 9:32
Fareed AFFareed AF
57112
57112
$begingroup$
I am sorry for my english because my mather tongue is Turkish. May i ask a question? Won’t i use parametrics equation?
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:37
$begingroup$
Don't worry it is fine. No there is no need to use parametric equations only the dot product, I'll try to explaim my answer more.
$endgroup$
– Fareed AF
Dec 15 '18 at 9:41
$begingroup$
Thank you for your help
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
add a comment |
$begingroup$
I am sorry for my english because my mather tongue is Turkish. May i ask a question? Won’t i use parametrics equation?
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:37
$begingroup$
Don't worry it is fine. No there is no need to use parametric equations only the dot product, I'll try to explaim my answer more.
$endgroup$
– Fareed AF
Dec 15 '18 at 9:41
$begingroup$
Thank you for your help
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
$begingroup$
I am sorry for my english because my mather tongue is Turkish. May i ask a question? Won’t i use parametrics equation?
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:37
$begingroup$
I am sorry for my english because my mather tongue is Turkish. May i ask a question? Won’t i use parametrics equation?
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:37
$begingroup$
Don't worry it is fine. No there is no need to use parametric equations only the dot product, I'll try to explaim my answer more.
$endgroup$
– Fareed AF
Dec 15 '18 at 9:41
$begingroup$
Don't worry it is fine. No there is no need to use parametric equations only the dot product, I'll try to explaim my answer more.
$endgroup$
– Fareed AF
Dec 15 '18 at 9:41
$begingroup$
Thank you for your help
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
$begingroup$
Thank you for your help
$endgroup$
– Pınar Oruç
Dec 15 '18 at 9:52
add a comment |
$begingroup$
One parameterisation of the line is given by $$mathbf{x}(t) = tbegin{pmatrix} 2 \ -1 \ 1end{pmatrix} + begin{pmatrix}1 \ 3 \ 1 end{pmatrix}, t in mathbb{R}.$$ The geometrical interpretation of $mathbf{x}$ is that for every real number $t$, $mathbf{x}(t)$ corresponds to an arrow which points to a point on the line with its root at the origin in $mathbb{R}^3$.
In fact, the outputs of the function $mathbf{x}$ collectively point to every point on the line.
In order to find the length of the projection of $(3,4,-4)$ onto the line, we first find a unit vector which points along the line, and then take the magnitude of the component of $(3,4,-4)$ in the direction of that unit vector. To find a unit vector which points along the line, we find two different vectors which point to different points on the line, take their difference, and then normalise the result.
Finding two different vectors which point to the line is the same as finding two different outputs of $mathbf{x}$. For example, set $t = 0$, to obtain $$mathbf{x}(0) = begin{pmatrix} 1 \ 3 \ 1 end{pmatrix}.$$ Then we can for instance set $t = 1$, to obtain $$mathbf{x}(1) = begin{pmatrix} 2 \-1 \ 1end{pmatrix}+ begin{pmatrix}1 \ 3 \ 1 end{pmatrix} = begin{pmatrix} 3 \ 2 \ 2 end{pmatrix}.$$
Now, the vector pointing from $mathbf{x}(0)$ to $mathbf{x}(1)$ which we will call $mathbf{n}$ is given by $$mathbf{n} = mathbf{x}(1) - mathbf{x}(0) = begin{pmatrix} 2 \ -1 \ 1end{pmatrix}.$$ We then obtain the unit vector pointing along $mathbf{n}$ which we call $hat{mathbf{n}}$ by setting $$hat{mathbf{n}} = frac{mathbf{n}}{||mathbf{n}||} = frac{1}{sqrt{6}}begin{pmatrix} 2 \ -1 \ 1 end{pmatrix}.$$ The magnitude of the component of $(3,4,-4)$ in the direction of the line, i.e. the length of the projection of $(3,4,-4)$ onto the line, is given by $|(3,4,-4) cdot hat{mathbf{n}}| = frac{sqrt{6}}{3}.$
I suggest checking out this series of videos to help you nail the intuitions behind vectors, dot products and much more.
$endgroup$
add a comment |
$begingroup$
One parameterisation of the line is given by $$mathbf{x}(t) = tbegin{pmatrix} 2 \ -1 \ 1end{pmatrix} + begin{pmatrix}1 \ 3 \ 1 end{pmatrix}, t in mathbb{R}.$$ The geometrical interpretation of $mathbf{x}$ is that for every real number $t$, $mathbf{x}(t)$ corresponds to an arrow which points to a point on the line with its root at the origin in $mathbb{R}^3$.
In fact, the outputs of the function $mathbf{x}$ collectively point to every point on the line.
In order to find the length of the projection of $(3,4,-4)$ onto the line, we first find a unit vector which points along the line, and then take the magnitude of the component of $(3,4,-4)$ in the direction of that unit vector. To find a unit vector which points along the line, we find two different vectors which point to different points on the line, take their difference, and then normalise the result.
Finding two different vectors which point to the line is the same as finding two different outputs of $mathbf{x}$. For example, set $t = 0$, to obtain $$mathbf{x}(0) = begin{pmatrix} 1 \ 3 \ 1 end{pmatrix}.$$ Then we can for instance set $t = 1$, to obtain $$mathbf{x}(1) = begin{pmatrix} 2 \-1 \ 1end{pmatrix}+ begin{pmatrix}1 \ 3 \ 1 end{pmatrix} = begin{pmatrix} 3 \ 2 \ 2 end{pmatrix}.$$
Now, the vector pointing from $mathbf{x}(0)$ to $mathbf{x}(1)$ which we will call $mathbf{n}$ is given by $$mathbf{n} = mathbf{x}(1) - mathbf{x}(0) = begin{pmatrix} 2 \ -1 \ 1end{pmatrix}.$$ We then obtain the unit vector pointing along $mathbf{n}$ which we call $hat{mathbf{n}}$ by setting $$hat{mathbf{n}} = frac{mathbf{n}}{||mathbf{n}||} = frac{1}{sqrt{6}}begin{pmatrix} 2 \ -1 \ 1 end{pmatrix}.$$ The magnitude of the component of $(3,4,-4)$ in the direction of the line, i.e. the length of the projection of $(3,4,-4)$ onto the line, is given by $|(3,4,-4) cdot hat{mathbf{n}}| = frac{sqrt{6}}{3}.$
I suggest checking out this series of videos to help you nail the intuitions behind vectors, dot products and much more.
$endgroup$
add a comment |
$begingroup$
One parameterisation of the line is given by $$mathbf{x}(t) = tbegin{pmatrix} 2 \ -1 \ 1end{pmatrix} + begin{pmatrix}1 \ 3 \ 1 end{pmatrix}, t in mathbb{R}.$$ The geometrical interpretation of $mathbf{x}$ is that for every real number $t$, $mathbf{x}(t)$ corresponds to an arrow which points to a point on the line with its root at the origin in $mathbb{R}^3$.
In fact, the outputs of the function $mathbf{x}$ collectively point to every point on the line.
In order to find the length of the projection of $(3,4,-4)$ onto the line, we first find a unit vector which points along the line, and then take the magnitude of the component of $(3,4,-4)$ in the direction of that unit vector. To find a unit vector which points along the line, we find two different vectors which point to different points on the line, take their difference, and then normalise the result.
Finding two different vectors which point to the line is the same as finding two different outputs of $mathbf{x}$. For example, set $t = 0$, to obtain $$mathbf{x}(0) = begin{pmatrix} 1 \ 3 \ 1 end{pmatrix}.$$ Then we can for instance set $t = 1$, to obtain $$mathbf{x}(1) = begin{pmatrix} 2 \-1 \ 1end{pmatrix}+ begin{pmatrix}1 \ 3 \ 1 end{pmatrix} = begin{pmatrix} 3 \ 2 \ 2 end{pmatrix}.$$
Now, the vector pointing from $mathbf{x}(0)$ to $mathbf{x}(1)$ which we will call $mathbf{n}$ is given by $$mathbf{n} = mathbf{x}(1) - mathbf{x}(0) = begin{pmatrix} 2 \ -1 \ 1end{pmatrix}.$$ We then obtain the unit vector pointing along $mathbf{n}$ which we call $hat{mathbf{n}}$ by setting $$hat{mathbf{n}} = frac{mathbf{n}}{||mathbf{n}||} = frac{1}{sqrt{6}}begin{pmatrix} 2 \ -1 \ 1 end{pmatrix}.$$ The magnitude of the component of $(3,4,-4)$ in the direction of the line, i.e. the length of the projection of $(3,4,-4)$ onto the line, is given by $|(3,4,-4) cdot hat{mathbf{n}}| = frac{sqrt{6}}{3}.$
I suggest checking out this series of videos to help you nail the intuitions behind vectors, dot products and much more.
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One parameterisation of the line is given by $$mathbf{x}(t) = tbegin{pmatrix} 2 \ -1 \ 1end{pmatrix} + begin{pmatrix}1 \ 3 \ 1 end{pmatrix}, t in mathbb{R}.$$ The geometrical interpretation of $mathbf{x}$ is that for every real number $t$, $mathbf{x}(t)$ corresponds to an arrow which points to a point on the line with its root at the origin in $mathbb{R}^3$.
In fact, the outputs of the function $mathbf{x}$ collectively point to every point on the line.
In order to find the length of the projection of $(3,4,-4)$ onto the line, we first find a unit vector which points along the line, and then take the magnitude of the component of $(3,4,-4)$ in the direction of that unit vector. To find a unit vector which points along the line, we find two different vectors which point to different points on the line, take their difference, and then normalise the result.
Finding two different vectors which point to the line is the same as finding two different outputs of $mathbf{x}$. For example, set $t = 0$, to obtain $$mathbf{x}(0) = begin{pmatrix} 1 \ 3 \ 1 end{pmatrix}.$$ Then we can for instance set $t = 1$, to obtain $$mathbf{x}(1) = begin{pmatrix} 2 \-1 \ 1end{pmatrix}+ begin{pmatrix}1 \ 3 \ 1 end{pmatrix} = begin{pmatrix} 3 \ 2 \ 2 end{pmatrix}.$$
Now, the vector pointing from $mathbf{x}(0)$ to $mathbf{x}(1)$ which we will call $mathbf{n}$ is given by $$mathbf{n} = mathbf{x}(1) - mathbf{x}(0) = begin{pmatrix} 2 \ -1 \ 1end{pmatrix}.$$ We then obtain the unit vector pointing along $mathbf{n}$ which we call $hat{mathbf{n}}$ by setting $$hat{mathbf{n}} = frac{mathbf{n}}{||mathbf{n}||} = frac{1}{sqrt{6}}begin{pmatrix} 2 \ -1 \ 1 end{pmatrix}.$$ The magnitude of the component of $(3,4,-4)$ in the direction of the line, i.e. the length of the projection of $(3,4,-4)$ onto the line, is given by $|(3,4,-4) cdot hat{mathbf{n}}| = frac{sqrt{6}}{3}.$
I suggest checking out this series of videos to help you nail the intuitions behind vectors, dot products and much more.
edited Dec 15 '18 at 17:14
answered Dec 15 '18 at 12:46
E-muE-mu
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You mean length?
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– Fareed AF
Dec 15 '18 at 9:11
$begingroup$
What is the vector (3, 4-4) supposed to be? Do you mean the vector $(3, 4, 4)$?
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– The Pointer
Dec 15 '18 at 9:22
$begingroup$
I have corrected all of misspellings
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– Pınar Oruç
Dec 15 '18 at 9:26