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Equation of the line passing through $(3,-2,-5)$ and $(3,-2,6)$
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Find the Cartesian equation of the line passing through $(3,-2,-5)$ and $(3,-2,6)$ in $3$D.
The equation of the line through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by
$$
vecr=veca+lambda(vecb-veca)
$$
where $veca=x_1hati+y_1hatj+z_1hatk$ and $vecb-veca=(x_2-x_1)hati+(y_2-y_1)hatj+(z_2-z_1)hatk$
$$
fracx-x_1x_2-x_1=fracy-y_1y_2-y_1=fracz-z_1z_2-z_1
$$
For the line through the points $(3,-2,-5)$ and $(3,-2,6)$ is
$$lambda=
boxedfracx-30=fracy+20=fracz+511
$$
Is it the correct solution and how do I make sense of the final equation of the line through the given points ?
vectors 3d
asked Jul 29 at 8:42
ss1729
1,393520
add a comment |Â
up vote
0
down vote
favorite
Find the Cartesian equation of the line passing through $(3,-2,-5)$ and $(3,-2,6)$ in $3$D.
The equation of the line through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by
$$
vecr=veca+lambda(vecb-veca)
$$
where $veca=x_1hati+y_1hatj+z_1hatk$ and $vecb-veca=(x_2-x_1)hati+(y_2-y_1)hatj+(z_2-z_1)hatk$
$$
fracx-x_1x_2-x_1=fracy-y_1y_2-y_1=fracz-z_1z_2-z_1
$$
For the line through the points $(3,-2,-5)$ and $(3,-2,6)$ is
$$lambda=
boxedfracx-30=fracy+20=fracz+511
$$
Is it the correct solution and how do I make sense of the final equation of the line through the given points ?
vectors 3d
asked Jul 29 at 8:42
ss1729
1,393520
There can't be a single equation for a line in dimension $3$.
– Bernard
Jul 29 at 9:03
@Bernard Its 2 equations really.
– Sorfosh
Jul 29 at 9:28
@Sorfosh: Yes, of course, but the O.P. was abouttheequation…
– Bernard
Jul 29 at 9:30
ofcause its two equations. but my doubt is in the equation there is 0 in the denominator, and how do I make sense of it ?
– ss1729
Jul 29 at 9:31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find the Cartesian equation of the line passing through $(3,-2,-5)$ and $(3,-2,6)$ in $3$D.
The equation of the line through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by
$$
vecr=veca+lambda(vecb-veca)
$$
where $veca=x_1hati+y_1hatj+z_1hatk$ and $vecb-veca=(x_2-x_1)hati+(y_2-y_1)hatj+(z_2-z_1)hatk$
$$
fracx-x_1x_2-x_1=fracy-y_1y_2-y_1=fracz-z_1z_2-z_1
$$
For the line through the points $(3,-2,-5)$ and $(3,-2,6)$ is
$$lambda=
boxedfracx-30=fracy+20=fracz+511
$$
Is it the correct solution and how do I make sense of the final equation of the line through the given points ?
vectors 3d
asked Jul 29 at 8:42
ss1729
1,393520
Find the Cartesian equation of the line passing through $(3,-2,-5)$ and $(3,-2,6)$ in $3$D.
The equation of the line through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by
$$
vecr=veca+lambda(vecb-veca)
$$
where $veca=x_1hati+y_1hatj+z_1hatk$ and $vecb-veca=(x_2-x_1)hati+(y_2-y_1)hatj+(z_2-z_1)hatk$
$$
fracx-x_1x_2-x_1=fracy-y_1y_2-y_1=fracz-z_1z_2-z_1
$$
For the line through the points $(3,-2,-5)$ and $(3,-2,6)$ is
$$lambda=
boxedfracx-30=fracy+20=fracz+511
$$
Is it the correct solution and how do I make sense of the final equation of the line through the given points ?
vectors 3d
asked Jul 29 at 8:42
ss1729
1,393520
edited Jul 29 at 9:29
asked Jul 29 at 8:42
ss1729
1,393520
asked Jul 29 at 8:42
ss1729
1,393520
asked Jul 29 at 8:42
ss1729
1,393520
1,393520
There can't be a single equation for a line in dimension $3$.
– Bernard
Jul 29 at 9:03
@Bernard Its 2 equations really.
– Sorfosh
Jul 29 at 9:28
@Sorfosh: Yes, of course, but the O.P. was abouttheequation…
– Bernard
Jul 29 at 9:30
ofcause its two equations. but my doubt is in the equation there is 0 in the denominator, and how do I make sense of it ?
– ss1729
Jul 29 at 9:31
add a comment |Â
There can't be a single equation for a line in dimension $3$.
– Bernard
Jul 29 at 9:03
@Bernard Its 2 equations really.
– Sorfosh
Jul 29 at 9:28
@Sorfosh: Yes, of course, but the O.P. was abouttheequation…
– Bernard
Jul 29 at 9:30
ofcause its two equations. but my doubt is in the equation there is 0 in the denominator, and how do I make sense of it ?
– ss1729
Jul 29 at 9:31
There can't be a single equation for a line in dimension $3$.
– Bernard
Jul 29 at 9:03
There can't be a single equation for a line in dimension $3$.
– Bernard
Jul 29 at 9:03
@Bernard Its 2 equations really.
– Sorfosh
Jul 29 at 9:28
@Bernard Its 2 equations really.
– Sorfosh
Jul 29 at 9:28
@Sorfosh: Yes, of course, but the O.P. was about
the equation…– Bernard
Jul 29 at 9:30
@Sorfosh: Yes, of course, but the O.P. was about
the equation…– Bernard
Jul 29 at 9:30
ofcause its two equations. but my doubt is in the equation there is 0 in the denominator, and how do I make sense of it ?
– ss1729
Jul 29 at 9:31
ofcause its two equations. but my doubt is in the equation there is 0 in the denominator, and how do I make sense of it ?
– ss1729
Jul 29 at 9:31
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
1
down vote
accepted
I don't think we should divide by zero.
Observe the first two coordinates of the points.
They satisfy $x=3$ and $y=-2$.
The line is the intersection of $x=3$ and $y=-2$.
The formula is only used when $x_1 ne x_2$, $y_1 ne y_2$, and $z_1 ne z_2$.
answered Jul 29 at 8:55
Siong Thye Goh
76.9k134794
thats whr my doubt is. how do we justify dividing by 0 in the equation ?
– ss1729
Jul 29 at 9:25
hmm.. we dont' divide by zero. If the values are the same in $x$ coordiante, we know the lines falls on $x=x_0$.
– Siong Thye Goh
Jul 29 at 9:29
I think the problem is you do not understand where the equation of the line comes from. Your goal is to eliminate the parameter $t$, so you really just do nothing if the constant in front of it is $0$
– Sorfosh
Jul 29 at 9:30
This article might be of your interest, notice that they impose the condition that denominator is non-zero. Also, the article later describe what to do when the values are equal in a particular dimension.
– Siong Thye Goh
Jul 29 at 9:49
1
Yes, $x=3, y=-2$. Books are written by humans. Not everything that is written is correct.
– Siong Thye Goh
Jul 30 at 3:33
 |Â
show 1 more comment
up vote
1
down vote
Set $A=(3,-2,-5)$, $B=(3,-2,6)$. As any point on the line $(AB)$ is a barycenter $lambda A+(1-lambda)B$ of $A$ and $B$, the coordinates of a point $M$ of this line are obtained by projections on the axes of this relation:
begincases
x_M=lambdacdot 2+(1-lambda)2=2, \
y_M=lambda(-3)+(1-lambda)(-3)=-3,\
z_M=-5lambda+6(1-lambda)=6-11lambda.
endcases
This is a parametric representation, and we see that $z_M$ can take any real value, so the cartesian equations are:
begincases
x=2, \
y=-3.
endcases
answered Jul 29 at 9:17
Bernard
110k635102
add a comment |Â
up vote
1
down vote
In $2$-D, how would you find the line passing through $(3,-5)$ and $(3,6)$? Note that the formula
$$fracx-x_1x_2-x_1=fracy-y_1y_2-y_1 textor fracx-x_1y-y_1=fracx_2-x_1y_2-y_1$$
shows the slope of the line. Thus:
$$fracx-3y+5=frac011 Rightarrow x-3=0 Rightarrow x=3.$$
Indeed, it is a vertical line $x=3$ ($y$ can be any value).
Similarly, the line passing through $(3,-2,-5)$ and $(3,-2,6)$ is a vertical line $x=3,y=-2$ ($z$ can be any value).
answered Jul 29 at 13:33
farruhota
13.5k2632
add a comment |Â
up vote
0
down vote
By your work $vec(0,0,1)$ is parallel to the line, which gives the answer:
$$(x,y,z)=(3,-2,-5)+t(0,0,1).$$
answered Jul 29 at 9:22
Michael Rozenberg
87.7k1578180
add a comment |Â
up vote
0
down vote
You have $$x=3, y=-2$$
For $z$ may use parametric form $$z=-5+11t$$
Thus your equation is $$r(t) = (3,-2,-5+11t)$$
answered Jul 29 at 10:01
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I don't think we should divide by zero.
Observe the first two coordinates of the points.
They satisfy $x=3$ and $y=-2$.
The line is the intersection of $x=3$ and $y=-2$.
The formula is only used when $x_1 ne x_2$, $y_1 ne y_2$, and $z_1 ne z_2$.
answered Jul 29 at 8:55
Siong Thye Goh
76.9k134794
thats whr my doubt is. how do we justify dividing by 0 in the equation ?
– ss1729
Jul 29 at 9:25
hmm.. we dont' divide by zero. If the values are the same in $x$ coordiante, we know the lines falls on $x=x_0$.
– Siong Thye Goh
Jul 29 at 9:29
I think the problem is you do not understand where the equation of the line comes from. Your goal is to eliminate the parameter $t$, so you really just do nothing if the constant in front of it is $0$
– Sorfosh
Jul 29 at 9:30
This article might be of your interest, notice that they impose the condition that denominator is non-zero. Also, the article later describe what to do when the values are equal in a particular dimension.
– Siong Thye Goh
Jul 29 at 9:49
1
Yes, $x=3, y=-2$. Books are written by humans. Not everything that is written is correct.
– Siong Thye Goh
Jul 30 at 3:33
 |Â
show 1 more comment
up vote
1
down vote
accepted
I don't think we should divide by zero.
Observe the first two coordinates of the points.
They satisfy $x=3$ and $y=-2$.
The line is the intersection of $x=3$ and $y=-2$.
The formula is only used when $x_1 ne x_2$, $y_1 ne y_2$, and $z_1 ne z_2$.
answered Jul 29 at 8:55
Siong Thye Goh
76.9k134794
thats whr my doubt is. how do we justify dividing by 0 in the equation ?
– ss1729
Jul 29 at 9:25
hmm.. we dont' divide by zero. If the values are the same in $x$ coordiante, we know the lines falls on $x=x_0$.
– Siong Thye Goh
Jul 29 at 9:29
I think the problem is you do not understand where the equation of the line comes from. Your goal is to eliminate the parameter $t$, so you really just do nothing if the constant in front of it is $0$
– Sorfosh
Jul 29 at 9:30
This article might be of your interest, notice that they impose the condition that denominator is non-zero. Also, the article later describe what to do when the values are equal in a particular dimension.
– Siong Thye Goh
Jul 29 at 9:49
1
Yes, $x=3, y=-2$. Books are written by humans. Not everything that is written is correct.
– Siong Thye Goh
Jul 30 at 3:33
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I don't think we should divide by zero.
Observe the first two coordinates of the points.
They satisfy $x=3$ and $y=-2$.
The line is the intersection of $x=3$ and $y=-2$.
The formula is only used when $x_1 ne x_2$, $y_1 ne y_2$, and $z_1 ne z_2$.
answered Jul 29 at 8:55
Siong Thye Goh
76.9k134794
I don't think we should divide by zero.
Observe the first two coordinates of the points.
They satisfy $x=3$ and $y=-2$.
The line is the intersection of $x=3$ and $y=-2$.
The formula is only used when $x_1 ne x_2$, $y_1 ne y_2$, and $z_1 ne z_2$.
answered Jul 29 at 8:55
Siong Thye Goh
76.9k134794
edited Jul 29 at 9:44
answered Jul 29 at 8:55
Siong Thye Goh
76.9k134794
answered Jul 29 at 8:55
Siong Thye Goh
76.9k134794
answered Jul 29 at 8:55
Siong Thye Goh
76.9k134794
76.9k134794
thats whr my doubt is. how do we justify dividing by 0 in the equation ?
– ss1729
Jul 29 at 9:25
hmm.. we dont' divide by zero. If the values are the same in $x$ coordiante, we know the lines falls on $x=x_0$.
– Siong Thye Goh
Jul 29 at 9:29
I think the problem is you do not understand where the equation of the line comes from. Your goal is to eliminate the parameter $t$, so you really just do nothing if the constant in front of it is $0$
– Sorfosh
Jul 29 at 9:30
This article might be of your interest, notice that they impose the condition that denominator is non-zero. Also, the article later describe what to do when the values are equal in a particular dimension.
– Siong Thye Goh
Jul 29 at 9:49
1
Yes, $x=3, y=-2$. Books are written by humans. Not everything that is written is correct.
– Siong Thye Goh
Jul 30 at 3:33
 |Â
show 1 more comment
thats whr my doubt is. how do we justify dividing by 0 in the equation ?
– ss1729
Jul 29 at 9:25
hmm.. we dont' divide by zero. If the values are the same in $x$ coordiante, we know the lines falls on $x=x_0$.
– Siong Thye Goh
Jul 29 at 9:29
I think the problem is you do not understand where the equation of the line comes from. Your goal is to eliminate the parameter $t$, so you really just do nothing if the constant in front of it is $0$
– Sorfosh
Jul 29 at 9:30
This article might be of your interest, notice that they impose the condition that denominator is non-zero. Also, the article later describe what to do when the values are equal in a particular dimension.
– Siong Thye Goh
Jul 29 at 9:49
1
Yes, $x=3, y=-2$. Books are written by humans. Not everything that is written is correct.
– Siong Thye Goh
Jul 30 at 3:33
thats whr my doubt is. how do we justify dividing by 0 in the equation ?
– ss1729
Jul 29 at 9:25
thats whr my doubt is. how do we justify dividing by 0 in the equation ?
– ss1729
Jul 29 at 9:25
hmm.. we dont' divide by zero. If the values are the same in $x$ coordiante, we know the lines falls on $x=x_0$.
– Siong Thye Goh
Jul 29 at 9:29
hmm.. we dont' divide by zero. If the values are the same in $x$ coordiante, we know the lines falls on $x=x_0$.
– Siong Thye Goh
Jul 29 at 9:29
I think the problem is you do not understand where the equation of the line comes from. Your goal is to eliminate the parameter $t$, so you really just do nothing if the constant in front of it is $0$
– Sorfosh
Jul 29 at 9:30
I think the problem is you do not understand where the equation of the line comes from. Your goal is to eliminate the parameter $t$, so you really just do nothing if the constant in front of it is $0$
– Sorfosh
Jul 29 at 9:30
This article might be of your interest, notice that they impose the condition that denominator is non-zero. Also, the article later describe what to do when the values are equal in a particular dimension.
– Siong Thye Goh
Jul 29 at 9:49
This article might be of your interest, notice that they impose the condition that denominator is non-zero. Also, the article later describe what to do when the values are equal in a particular dimension.
– Siong Thye Goh
Jul 29 at 9:49
1
1
Yes, $x=3, y=-2$. Books are written by humans. Not everything that is written is correct.
– Siong Thye Goh
Jul 30 at 3:33
Yes, $x=3, y=-2$. Books are written by humans. Not everything that is written is correct.
– Siong Thye Goh
Jul 30 at 3:33
 |Â
show 1 more comment
up vote
1
down vote
Set $A=(3,-2,-5)$, $B=(3,-2,6)$. As any point on the line $(AB)$ is a barycenter $lambda A+(1-lambda)B$ of $A$ and $B$, the coordinates of a point $M$ of this line are obtained by projections on the axes of this relation:
begincases
x_M=lambdacdot 2+(1-lambda)2=2, \
y_M=lambda(-3)+(1-lambda)(-3)=-3,\
z_M=-5lambda+6(1-lambda)=6-11lambda.
endcases
This is a parametric representation, and we see that $z_M$ can take any real value, so the cartesian equations are:
begincases
x=2, \
y=-3.
endcases
answered Jul 29 at 9:17
Bernard
110k635102
add a comment |Â
up vote
1
down vote
Set $A=(3,-2,-5)$, $B=(3,-2,6)$. As any point on the line $(AB)$ is a barycenter $lambda A+(1-lambda)B$ of $A$ and $B$, the coordinates of a point $M$ of this line are obtained by projections on the axes of this relation:
begincases
x_M=lambdacdot 2+(1-lambda)2=2, \
y_M=lambda(-3)+(1-lambda)(-3)=-3,\
z_M=-5lambda+6(1-lambda)=6-11lambda.
endcases
This is a parametric representation, and we see that $z_M$ can take any real value, so the cartesian equations are:
begincases
x=2, \
y=-3.
endcases
answered Jul 29 at 9:17
Bernard
110k635102
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Set $A=(3,-2,-5)$, $B=(3,-2,6)$. As any point on the line $(AB)$ is a barycenter $lambda A+(1-lambda)B$ of $A$ and $B$, the coordinates of a point $M$ of this line are obtained by projections on the axes of this relation:
begincases
x_M=lambdacdot 2+(1-lambda)2=2, \
y_M=lambda(-3)+(1-lambda)(-3)=-3,\
z_M=-5lambda+6(1-lambda)=6-11lambda.
endcases
This is a parametric representation, and we see that $z_M$ can take any real value, so the cartesian equations are:
begincases
x=2, \
y=-3.
endcases
answered Jul 29 at 9:17
Bernard
110k635102
Set $A=(3,-2,-5)$, $B=(3,-2,6)$. As any point on the line $(AB)$ is a barycenter $lambda A+(1-lambda)B$ of $A$ and $B$, the coordinates of a point $M$ of this line are obtained by projections on the axes of this relation:
begincases
x_M=lambdacdot 2+(1-lambda)2=2, \
y_M=lambda(-3)+(1-lambda)(-3)=-3,\
z_M=-5lambda+6(1-lambda)=6-11lambda.
endcases
This is a parametric representation, and we see that $z_M$ can take any real value, so the cartesian equations are:
begincases
x=2, \
y=-3.
endcases
answered Jul 29 at 9:17
Bernard
110k635102
answered Jul 29 at 9:17
Bernard
110k635102
answered Jul 29 at 9:17
Bernard
110k635102
answered Jul 29 at 9:17
Bernard
110k635102
110k635102
add a comment |Â
add a comment |Â
up vote
1
down vote
In $2$-D, how would you find the line passing through $(3,-5)$ and $(3,6)$? Note that the formula
$$fracx-x_1x_2-x_1=fracy-y_1y_2-y_1 textor fracx-x_1y-y_1=fracx_2-x_1y_2-y_1$$
shows the slope of the line. Thus:
$$fracx-3y+5=frac011 Rightarrow x-3=0 Rightarrow x=3.$$
Indeed, it is a vertical line $x=3$ ($y$ can be any value).
Similarly, the line passing through $(3,-2,-5)$ and $(3,-2,6)$ is a vertical line $x=3,y=-2$ ($z$ can be any value).
answered Jul 29 at 13:33
farruhota
13.5k2632
add a comment |Â
up vote
1
down vote
In $2$-D, how would you find the line passing through $(3,-5)$ and $(3,6)$? Note that the formula
$$fracx-x_1x_2-x_1=fracy-y_1y_2-y_1 textor fracx-x_1y-y_1=fracx_2-x_1y_2-y_1$$
shows the slope of the line. Thus:
$$fracx-3y+5=frac011 Rightarrow x-3=0 Rightarrow x=3.$$
Indeed, it is a vertical line $x=3$ ($y$ can be any value).
Similarly, the line passing through $(3,-2,-5)$ and $(3,-2,6)$ is a vertical line $x=3,y=-2$ ($z$ can be any value).
answered Jul 29 at 13:33
farruhota
13.5k2632
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In $2$-D, how would you find the line passing through $(3,-5)$ and $(3,6)$? Note that the formula
$$fracx-x_1x_2-x_1=fracy-y_1y_2-y_1 textor fracx-x_1y-y_1=fracx_2-x_1y_2-y_1$$
shows the slope of the line. Thus:
$$fracx-3y+5=frac011 Rightarrow x-3=0 Rightarrow x=3.$$
Indeed, it is a vertical line $x=3$ ($y$ can be any value).
Similarly, the line passing through $(3,-2,-5)$ and $(3,-2,6)$ is a vertical line $x=3,y=-2$ ($z$ can be any value).
answered Jul 29 at 13:33
farruhota
13.5k2632
In $2$-D, how would you find the line passing through $(3,-5)$ and $(3,6)$? Note that the formula
$$fracx-x_1x_2-x_1=fracy-y_1y_2-y_1 textor fracx-x_1y-y_1=fracx_2-x_1y_2-y_1$$
shows the slope of the line. Thus:
$$fracx-3y+5=frac011 Rightarrow x-3=0 Rightarrow x=3.$$
Indeed, it is a vertical line $x=3$ ($y$ can be any value).
Similarly, the line passing through $(3,-2,-5)$ and $(3,-2,6)$ is a vertical line $x=3,y=-2$ ($z$ can be any value).
answered Jul 29 at 13:33
farruhota
13.5k2632
answered Jul 29 at 13:33
farruhota
13.5k2632
answered Jul 29 at 13:33
farruhota
13.5k2632
answered Jul 29 at 13:33
farruhota
13.5k2632
13.5k2632
add a comment |Â
add a comment |Â
up vote
0
down vote
By your work $vec(0,0,1)$ is parallel to the line, which gives the answer:
$$(x,y,z)=(3,-2,-5)+t(0,0,1).$$
answered Jul 29 at 9:22
Michael Rozenberg
87.7k1578180
add a comment |Â
up vote
0
down vote
By your work $vec(0,0,1)$ is parallel to the line, which gives the answer:
$$(x,y,z)=(3,-2,-5)+t(0,0,1).$$
answered Jul 29 at 9:22
Michael Rozenberg
87.7k1578180
add a comment |Â
up vote
0
down vote
up vote
0
down vote
By your work $vec(0,0,1)$ is parallel to the line, which gives the answer:
$$(x,y,z)=(3,-2,-5)+t(0,0,1).$$
answered Jul 29 at 9:22
Michael Rozenberg
87.7k1578180
By your work $vec(0,0,1)$ is parallel to the line, which gives the answer:
$$(x,y,z)=(3,-2,-5)+t(0,0,1).$$
answered Jul 29 at 9:22
Michael Rozenberg
87.7k1578180
answered Jul 29 at 9:22
Michael Rozenberg
87.7k1578180
answered Jul 29 at 9:22
Michael Rozenberg
87.7k1578180
answered Jul 29 at 9:22
Michael Rozenberg
87.7k1578180
87.7k1578180
add a comment |Â
add a comment |Â
up vote
0
down vote
You have $$x=3, y=-2$$
For $z$ may use parametric form $$z=-5+11t$$
Thus your equation is $$r(t) = (3,-2,-5+11t)$$
answered Jul 29 at 10:01
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
0
down vote
You have $$x=3, y=-2$$
For $z$ may use parametric form $$z=-5+11t$$
Thus your equation is $$r(t) = (3,-2,-5+11t)$$
answered Jul 29 at 10:01
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You have $$x=3, y=-2$$
For $z$ may use parametric form $$z=-5+11t$$
Thus your equation is $$r(t) = (3,-2,-5+11t)$$
answered Jul 29 at 10:01
Mohammad Riazi-Kermani
27.3k41851
You have $$x=3, y=-2$$
For $z$ may use parametric form $$z=-5+11t$$
Thus your equation is $$r(t) = (3,-2,-5+11t)$$
answered Jul 29 at 10:01
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 10:01
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 10:01
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 10:01
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
add a comment |Â
add a comment |Â
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There can't be a single equation for a line in dimension $3$.
– Bernard
Jul 29 at 9:03
@Bernard Its 2 equations really.
– Sorfosh
Jul 29 at 9:28
@Sorfosh: Yes, of course, but the O.P. was about
theequation…– Bernard
Jul 29 at 9:30
ofcause its two equations. but my doubt is in the equation there is 0 in the denominator, and how do I make sense of it ?
– ss1729
Jul 29 at 9:31