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Determine the line $R$
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Determine the line $R$ that goes throught the point $p$, parallel to the plane $alpha$ and dissects the line $L$
Given $p=beginbmatrix2 \ 3 \ -1 endbmatrix, alpha equiv beginbmatrixx_1 \ x_2 \ x_3 endbmatrix= beginbmatrix0 \ -1 \ -1endbmatrix + rhobeginbmatrix1 \ 2 \ 2 endbmatrix + taubeginbmatrix5 \ 5 \ 6 endbmatrix$
and $L equiv begincases 5x_1 -x_2 -x_3 +2 = 0 \ 12x_1 -3x_2 -2x_3 +4=0 endcases$
I've determined the vector equation of $L$ by the Gauss-Jordan method which will result in $Lequiv beginbmatrix2 \ 4 \ 0 endbmatrix +sigma beginbmatrix1 \ 1\ 3 endbmatrix$.
To get the vector for $R$ you have to "scalar multiply" the vector of $L$ and $R$ which need to be equal to $0$. A possibility can be $beginbmatrix 0 \ -3 \ 1 endbmatrix$ but then I have to check parallelity with the plane $alpha$. With the Gauss-Jordan method I checked if the last vector is parallel to the plan $alpha$ which it is.
So I think $Requivbeginbmatrix2 \ 3\ -1 endbmatrix+ omicronbeginbmatrix0 \ -3 \ 1 endbmatrix$
Is this correct?
P.S.: I never had spatial geometry lessons. I had some books so mistakes can be made.
linear-algebra geometry euclidean-geometry
asked Jul 19 at 8:36
Anonymous I
804725
 |Â
show 2 more comments
up vote
2
down vote
favorite
Determine the line $R$ that goes throught the point $p$, parallel to the plane $alpha$ and dissects the line $L$
Given $p=beginbmatrix2 \ 3 \ -1 endbmatrix, alpha equiv beginbmatrixx_1 \ x_2 \ x_3 endbmatrix= beginbmatrix0 \ -1 \ -1endbmatrix + rhobeginbmatrix1 \ 2 \ 2 endbmatrix + taubeginbmatrix5 \ 5 \ 6 endbmatrix$
and $L equiv begincases 5x_1 -x_2 -x_3 +2 = 0 \ 12x_1 -3x_2 -2x_3 +4=0 endcases$
I've determined the vector equation of $L$ by the Gauss-Jordan method which will result in $Lequiv beginbmatrix2 \ 4 \ 0 endbmatrix +sigma beginbmatrix1 \ 1\ 3 endbmatrix$.
To get the vector for $R$ you have to "scalar multiply" the vector of $L$ and $R$ which need to be equal to $0$. A possibility can be $beginbmatrix 0 \ -3 \ 1 endbmatrix$ but then I have to check parallelity with the plane $alpha$. With the Gauss-Jordan method I checked if the last vector is parallel to the plan $alpha$ which it is.
So I think $Requivbeginbmatrix2 \ 3\ -1 endbmatrix+ omicronbeginbmatrix0 \ -3 \ 1 endbmatrix$
Is this correct?
P.S.: I never had spatial geometry lessons. I had some books so mistakes can be made.
linear-algebra geometry euclidean-geometry
asked Jul 19 at 8:36
Anonymous I
804725
The vector (parametric) equation for L seems uncorrect.
– gimusi
Jul 19 at 8:38
@gimusi : Can you point out the mistake I've made exactly?
– Anonymous I
Jul 19 at 8:40
Vector (2,4,0) doesn't satisfy the cartesian equations.
– gimusi
Jul 19 at 8:44
What about $beginbmatrix-2 \ -4 \ 0 endbmatrix$?
– Anonymous I
Jul 19 at 8:46
1
You can check from the cartesian equations that it doesn't work.
– gimusi
Jul 19 at 9:00
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Determine the line $R$ that goes throught the point $p$, parallel to the plane $alpha$ and dissects the line $L$
Given $p=beginbmatrix2 \ 3 \ -1 endbmatrix, alpha equiv beginbmatrixx_1 \ x_2 \ x_3 endbmatrix= beginbmatrix0 \ -1 \ -1endbmatrix + rhobeginbmatrix1 \ 2 \ 2 endbmatrix + taubeginbmatrix5 \ 5 \ 6 endbmatrix$
and $L equiv begincases 5x_1 -x_2 -x_3 +2 = 0 \ 12x_1 -3x_2 -2x_3 +4=0 endcases$
I've determined the vector equation of $L$ by the Gauss-Jordan method which will result in $Lequiv beginbmatrix2 \ 4 \ 0 endbmatrix +sigma beginbmatrix1 \ 1\ 3 endbmatrix$.
To get the vector for $R$ you have to "scalar multiply" the vector of $L$ and $R$ which need to be equal to $0$. A possibility can be $beginbmatrix 0 \ -3 \ 1 endbmatrix$ but then I have to check parallelity with the plane $alpha$. With the Gauss-Jordan method I checked if the last vector is parallel to the plan $alpha$ which it is.
So I think $Requivbeginbmatrix2 \ 3\ -1 endbmatrix+ omicronbeginbmatrix0 \ -3 \ 1 endbmatrix$
Is this correct?
P.S.: I never had spatial geometry lessons. I had some books so mistakes can be made.
linear-algebra geometry euclidean-geometry
asked Jul 19 at 8:36
Anonymous I
804725
Determine the line $R$ that goes throught the point $p$, parallel to the plane $alpha$ and dissects the line $L$
Given $p=beginbmatrix2 \ 3 \ -1 endbmatrix, alpha equiv beginbmatrixx_1 \ x_2 \ x_3 endbmatrix= beginbmatrix0 \ -1 \ -1endbmatrix + rhobeginbmatrix1 \ 2 \ 2 endbmatrix + taubeginbmatrix5 \ 5 \ 6 endbmatrix$
and $L equiv begincases 5x_1 -x_2 -x_3 +2 = 0 \ 12x_1 -3x_2 -2x_3 +4=0 endcases$
I've determined the vector equation of $L$ by the Gauss-Jordan method which will result in $Lequiv beginbmatrix2 \ 4 \ 0 endbmatrix +sigma beginbmatrix1 \ 1\ 3 endbmatrix$.
To get the vector for $R$ you have to "scalar multiply" the vector of $L$ and $R$ which need to be equal to $0$. A possibility can be $beginbmatrix 0 \ -3 \ 1 endbmatrix$ but then I have to check parallelity with the plane $alpha$. With the Gauss-Jordan method I checked if the last vector is parallel to the plan $alpha$ which it is.
So I think $Requivbeginbmatrix2 \ 3\ -1 endbmatrix+ omicronbeginbmatrix0 \ -3 \ 1 endbmatrix$
Is this correct?
P.S.: I never had spatial geometry lessons. I had some books so mistakes can be made.
linear-algebra geometry euclidean-geometry
asked Jul 19 at 8:36
Anonymous I
804725
edited Jul 19 at 9:05
asked Jul 19 at 8:36
Anonymous I
804725
asked Jul 19 at 8:36
Anonymous I
804725
asked Jul 19 at 8:36
Anonymous I
804725
804725
The vector (parametric) equation for L seems uncorrect.
– gimusi
Jul 19 at 8:38
@gimusi : Can you point out the mistake I've made exactly?
– Anonymous I
Jul 19 at 8:40
Vector (2,4,0) doesn't satisfy the cartesian equations.
– gimusi
Jul 19 at 8:44
What about $beginbmatrix-2 \ -4 \ 0 endbmatrix$?
– Anonymous I
Jul 19 at 8:46
1
You can check from the cartesian equations that it doesn't work.
– gimusi
Jul 19 at 9:00
 |Â
show 2 more comments
The vector (parametric) equation for L seems uncorrect.
– gimusi
Jul 19 at 8:38
@gimusi : Can you point out the mistake I've made exactly?
– Anonymous I
Jul 19 at 8:40
Vector (2,4,0) doesn't satisfy the cartesian equations.
– gimusi
Jul 19 at 8:44
What about $beginbmatrix-2 \ -4 \ 0 endbmatrix$?
– Anonymous I
Jul 19 at 8:46
1
You can check from the cartesian equations that it doesn't work.
– gimusi
Jul 19 at 9:00
The vector (parametric) equation for L seems uncorrect.
– gimusi
Jul 19 at 8:38
The vector (parametric) equation for L seems uncorrect.
– gimusi
Jul 19 at 8:38
@gimusi : Can you point out the mistake I've made exactly?
– Anonymous I
Jul 19 at 8:40
@gimusi : Can you point out the mistake I've made exactly?
– Anonymous I
Jul 19 at 8:40
Vector (2,4,0) doesn't satisfy the cartesian equations.
– gimusi
Jul 19 at 8:44
Vector (2,4,0) doesn't satisfy the cartesian equations.
– gimusi
Jul 19 at 8:44
What about $beginbmatrix-2 \ -4 \ 0 endbmatrix$?
– Anonymous I
Jul 19 at 8:46
What about $beginbmatrix-2 \ -4 \ 0 endbmatrix$?
– Anonymous I
Jul 19 at 8:46
1
1
You can check from the cartesian equations that it doesn't work.
– gimusi
Jul 19 at 9:00
You can check from the cartesian equations that it doesn't work.
– gimusi
Jul 19 at 9:00
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Being parallel to the plane is equivalent to the condition that the line is perpendicular to the plane's normal (vector).
Given $alpha equiv beginbmatrixx_1 \ x_2 \ x_3 endbmatrix= beginbmatrix0 \ -1 \ -1endbmatrix + rhocolorbluebeginbmatrix1 \ 2 \ 2 endbmatrix + taucolorpurplebeginbmatrix5 \ 5 \ 6 endbmatrix$
Either converting this to a cartesian equation or computing the following cross product gives you a normal vector of $alpha$:
$$vec n_alpha = colorbluebeginbmatrix1 \ 2 \ 2 endbmatrix times colorpurplebeginbmatrix5 \ 5 \ 6 endbmatrix=colorredbeginbmatrix2 \ 4 \ -5 endbmatrix$$
and $L equiv begincases 5x_1 -x_2 -x_3 +2 = 0 \ 12x_1 -3x_2 -2x_3 +4=0 endcases$
I've determined the vector equation of $L$ by the Gauss-Jordan method which will result in $Lequiv beginbmatrix2 \ 4 \ 0 endbmatrix +sigma beginbmatrix1 \ 1\ 3 endbmatrix$.
Maybe something went wrong in the calculations. A vector equation of $L$ is given by:
$$Lequiv beginbmatrix0 \ 0 \ 2 endbmatrix +sigma beginbmatrix1 \ 2 \ 3 endbmatrix tag$*$$$
Now a direction vector of $R$, through $p$ and a point on $L$, is given by their difference:
$$underbracebeginbmatrix0 \ 0 \ 2 endbmatrix +sigma beginbmatrix1 \ 2 \ 3 endbmatrix_mboxpoint on $L$-underbracebeginbmatrix2 \ 3 \ -1 endbmatrix_p=colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix$$
Recall that we want this to be orthogonal to $vec n_alpha$, so:
$$colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix cdot colorredbeginbmatrix2 \ 4 \ -5 endbmatrix = 0 iff sigma = ldots$$
This $sigma$ gives you the point on $L$ where $R$ passes through and thus, together with $p$, completely determines $R$. So with this value for $sigma$, you have:
$$Requiv beginbmatrix2 \ 3 \ -1 endbmatrix + mu colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix$$
Addition after comments. I wouldn't focus on the first coordinate but on the other two:
- $R_2 to R_2-3R_1$
- $R_1 to R_1+R_2$
$$beginbmatrix 5 & -1 & -1 & -2 \ 12 & -3 & -2 & -4 endbmatrix
sim beginbmatrix 5 & -1 & -1 & -2 \ -3 & 0 & 1 & 2 endbmatrix
sim beginbmatrix 2 & -1 & 0 & 0 \ -3 & 0 & 1 & 2 endbmatrix$$
And from here it's easy to see you arrive at $(*)$, solving in terms of the first coordinate.
answered Jul 19 at 9:02
StackTD
20.1k1443
I'll check it now. I maybe used something incorrectly.
– Anonymous I
Jul 19 at 9:06
1
Oké, I'll understand your answer. I just need to find out where I made a mistake. Thanks for the answer.
– Anonymous I
Jul 19 at 9:08
Alright; you're welcome!
– StackTD
Jul 19 at 9:11
Now I get $Lequiv beginbmatrix-2 \ 4 \ 0 endbmatrix+ omicron beginbmatrix1 \ -2 \ 3 endbmatrix$. My mind is letting me down. :(
– Anonymous I
Jul 19 at 9:17
That's not correct but it's hard to tell where it goes wrong in your calculations...
– StackTD
Jul 19 at 9:18
 |Â
show 9 more comments
up vote
1
down vote
Here are a couple of alternative approaches. Note that the pair of equations that define $L$ are the equations of planes, that is, $L$ is described as the intersection of two planes.
For $R$ to be parallel to $alpha$, its direction vector must be some nontrivial linear combination of the generators of $alpha$, so we know that a parametric equation of $R$ is $p+lambda(rho[1,2,2]^T+tau[5,5,6]^T)$ for some as yet undetermined $rho$ and $tau$. The other condition on $R$ is that it intersects $L$, i.e., that it has a common intersection with both of the defining planes. Substituting into their equations produces the system $$lambda(rho+14tau)+10 = 0 \ lambda(2rho+33tau)+21 = 0.$$ This system is underdetermined, which makes sense because any nonzero scalar multiple of a line’s direction vector is also a direction vector for the line. Set $lambda$ to some convenient value such as $1$ and solve the resulting system for $rho$ and $tau$.
What you’re essentially doing above is finding the intersection of three planes: the two that define $L$ and a plane through $p$ that is parallel to $alpha$. The latter plane consists of all of the lines through $p$ that are parallel to $alpha$. You could use this idea to solve the problem is a slightly different way, which I’ll illustrate using homogeneous coordinates. If the equation of a plane is $ax+by+cz+d=0$, it can be represented by the homogeneous vector $mathbfpi = [a,b,c,d]^T$, so that the equation of the plane becomes $mathbfpi^Tmathbf x = 0$ (i.e., the dot product of the plane vector and coordinates of a point on the plane). The parallel plane to $alpha$ through $p$ is then a null vector $mathbfpi$ of the matrix $$beginbmatrix2&3&-1&1\1&2&2&0\5&5&6&0endbmatrix.$$ (Think about how this relates to the condition $mathbfpi^Tmathbf x=0$.) The rows of this matrix are just the point $p$ and the two generators of $alpha$—in projective-geometric terms, these are three points on the plane, two of which are at infinity. You can compute this null vector in any of the usual ways, such as Gaussian elimination. If you’ve done it correctly, the resulting space should be one-dimensional.†One possibility is $mathbfpi = [2,4,-5,-21]^T$. To compute the intersection of $pi$ and $L$, use the same procedure: assemble the vectors that represent the three planes into a matrix and find a null vector of it, then convert to inhomogeneous Cartesian coordinates by dividing through by the last component of the vector. This gives you another point on $R$, from which I trust you know how to generate an equation of this line.
†You can also compute the plane’s representative vector by taking the cross product of the two generators of $alpha$ to get the first three components of $mathbfpi$ and setting its last component to the negative of the dot product of this normal vector with $p$. This is a direct analogue of one of the ways you’d find the point-normal equation of the plane. However, this method is specific to three-dimensional space, while the null space method works in spaces of any dimension.
answered Jul 19 at 20:49
amd
25.9k2943
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Being parallel to the plane is equivalent to the condition that the line is perpendicular to the plane's normal (vector).
Given $alpha equiv beginbmatrixx_1 \ x_2 \ x_3 endbmatrix= beginbmatrix0 \ -1 \ -1endbmatrix + rhocolorbluebeginbmatrix1 \ 2 \ 2 endbmatrix + taucolorpurplebeginbmatrix5 \ 5 \ 6 endbmatrix$
Either converting this to a cartesian equation or computing the following cross product gives you a normal vector of $alpha$:
$$vec n_alpha = colorbluebeginbmatrix1 \ 2 \ 2 endbmatrix times colorpurplebeginbmatrix5 \ 5 \ 6 endbmatrix=colorredbeginbmatrix2 \ 4 \ -5 endbmatrix$$
and $L equiv begincases 5x_1 -x_2 -x_3 +2 = 0 \ 12x_1 -3x_2 -2x_3 +4=0 endcases$
I've determined the vector equation of $L$ by the Gauss-Jordan method which will result in $Lequiv beginbmatrix2 \ 4 \ 0 endbmatrix +sigma beginbmatrix1 \ 1\ 3 endbmatrix$.
Maybe something went wrong in the calculations. A vector equation of $L$ is given by:
$$Lequiv beginbmatrix0 \ 0 \ 2 endbmatrix +sigma beginbmatrix1 \ 2 \ 3 endbmatrix tag$*$$$
Now a direction vector of $R$, through $p$ and a point on $L$, is given by their difference:
$$underbracebeginbmatrix0 \ 0 \ 2 endbmatrix +sigma beginbmatrix1 \ 2 \ 3 endbmatrix_mboxpoint on $L$-underbracebeginbmatrix2 \ 3 \ -1 endbmatrix_p=colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix$$
Recall that we want this to be orthogonal to $vec n_alpha$, so:
$$colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix cdot colorredbeginbmatrix2 \ 4 \ -5 endbmatrix = 0 iff sigma = ldots$$
This $sigma$ gives you the point on $L$ where $R$ passes through and thus, together with $p$, completely determines $R$. So with this value for $sigma$, you have:
$$Requiv beginbmatrix2 \ 3 \ -1 endbmatrix + mu colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix$$
Addition after comments. I wouldn't focus on the first coordinate but on the other two:
- $R_2 to R_2-3R_1$
- $R_1 to R_1+R_2$
$$beginbmatrix 5 & -1 & -1 & -2 \ 12 & -3 & -2 & -4 endbmatrix
sim beginbmatrix 5 & -1 & -1 & -2 \ -3 & 0 & 1 & 2 endbmatrix
sim beginbmatrix 2 & -1 & 0 & 0 \ -3 & 0 & 1 & 2 endbmatrix$$
And from here it's easy to see you arrive at $(*)$, solving in terms of the first coordinate.
answered Jul 19 at 9:02
StackTD
20.1k1443
I'll check it now. I maybe used something incorrectly.
– Anonymous I
Jul 19 at 9:06
1
Oké, I'll understand your answer. I just need to find out where I made a mistake. Thanks for the answer.
– Anonymous I
Jul 19 at 9:08
Alright; you're welcome!
– StackTD
Jul 19 at 9:11
Now I get $Lequiv beginbmatrix-2 \ 4 \ 0 endbmatrix+ omicron beginbmatrix1 \ -2 \ 3 endbmatrix$. My mind is letting me down. :(
– Anonymous I
Jul 19 at 9:17
That's not correct but it's hard to tell where it goes wrong in your calculations...
– StackTD
Jul 19 at 9:18
 |Â
show 9 more comments
up vote
2
down vote
accepted
Being parallel to the plane is equivalent to the condition that the line is perpendicular to the plane's normal (vector).
Given $alpha equiv beginbmatrixx_1 \ x_2 \ x_3 endbmatrix= beginbmatrix0 \ -1 \ -1endbmatrix + rhocolorbluebeginbmatrix1 \ 2 \ 2 endbmatrix + taucolorpurplebeginbmatrix5 \ 5 \ 6 endbmatrix$
Either converting this to a cartesian equation or computing the following cross product gives you a normal vector of $alpha$:
$$vec n_alpha = colorbluebeginbmatrix1 \ 2 \ 2 endbmatrix times colorpurplebeginbmatrix5 \ 5 \ 6 endbmatrix=colorredbeginbmatrix2 \ 4 \ -5 endbmatrix$$
and $L equiv begincases 5x_1 -x_2 -x_3 +2 = 0 \ 12x_1 -3x_2 -2x_3 +4=0 endcases$
I've determined the vector equation of $L$ by the Gauss-Jordan method which will result in $Lequiv beginbmatrix2 \ 4 \ 0 endbmatrix +sigma beginbmatrix1 \ 1\ 3 endbmatrix$.
Maybe something went wrong in the calculations. A vector equation of $L$ is given by:
$$Lequiv beginbmatrix0 \ 0 \ 2 endbmatrix +sigma beginbmatrix1 \ 2 \ 3 endbmatrix tag$*$$$
Now a direction vector of $R$, through $p$ and a point on $L$, is given by their difference:
$$underbracebeginbmatrix0 \ 0 \ 2 endbmatrix +sigma beginbmatrix1 \ 2 \ 3 endbmatrix_mboxpoint on $L$-underbracebeginbmatrix2 \ 3 \ -1 endbmatrix_p=colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix$$
Recall that we want this to be orthogonal to $vec n_alpha$, so:
$$colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix cdot colorredbeginbmatrix2 \ 4 \ -5 endbmatrix = 0 iff sigma = ldots$$
This $sigma$ gives you the point on $L$ where $R$ passes through and thus, together with $p$, completely determines $R$. So with this value for $sigma$, you have:
$$Requiv beginbmatrix2 \ 3 \ -1 endbmatrix + mu colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix$$
Addition after comments. I wouldn't focus on the first coordinate but on the other two:
- $R_2 to R_2-3R_1$
- $R_1 to R_1+R_2$
$$beginbmatrix 5 & -1 & -1 & -2 \ 12 & -3 & -2 & -4 endbmatrix
sim beginbmatrix 5 & -1 & -1 & -2 \ -3 & 0 & 1 & 2 endbmatrix
sim beginbmatrix 2 & -1 & 0 & 0 \ -3 & 0 & 1 & 2 endbmatrix$$
And from here it's easy to see you arrive at $(*)$, solving in terms of the first coordinate.
answered Jul 19 at 9:02
StackTD
20.1k1443
I'll check it now. I maybe used something incorrectly.
– Anonymous I
Jul 19 at 9:06
1
Oké, I'll understand your answer. I just need to find out where I made a mistake. Thanks for the answer.
– Anonymous I
Jul 19 at 9:08
Alright; you're welcome!
– StackTD
Jul 19 at 9:11
Now I get $Lequiv beginbmatrix-2 \ 4 \ 0 endbmatrix+ omicron beginbmatrix1 \ -2 \ 3 endbmatrix$. My mind is letting me down. :(
– Anonymous I
Jul 19 at 9:17
That's not correct but it's hard to tell where it goes wrong in your calculations...
– StackTD
Jul 19 at 9:18
 |Â
show 9 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Being parallel to the plane is equivalent to the condition that the line is perpendicular to the plane's normal (vector).
Given $alpha equiv beginbmatrixx_1 \ x_2 \ x_3 endbmatrix= beginbmatrix0 \ -1 \ -1endbmatrix + rhocolorbluebeginbmatrix1 \ 2 \ 2 endbmatrix + taucolorpurplebeginbmatrix5 \ 5 \ 6 endbmatrix$
Either converting this to a cartesian equation or computing the following cross product gives you a normal vector of $alpha$:
$$vec n_alpha = colorbluebeginbmatrix1 \ 2 \ 2 endbmatrix times colorpurplebeginbmatrix5 \ 5 \ 6 endbmatrix=colorredbeginbmatrix2 \ 4 \ -5 endbmatrix$$
and $L equiv begincases 5x_1 -x_2 -x_3 +2 = 0 \ 12x_1 -3x_2 -2x_3 +4=0 endcases$
I've determined the vector equation of $L$ by the Gauss-Jordan method which will result in $Lequiv beginbmatrix2 \ 4 \ 0 endbmatrix +sigma beginbmatrix1 \ 1\ 3 endbmatrix$.
Maybe something went wrong in the calculations. A vector equation of $L$ is given by:
$$Lequiv beginbmatrix0 \ 0 \ 2 endbmatrix +sigma beginbmatrix1 \ 2 \ 3 endbmatrix tag$*$$$
Now a direction vector of $R$, through $p$ and a point on $L$, is given by their difference:
$$underbracebeginbmatrix0 \ 0 \ 2 endbmatrix +sigma beginbmatrix1 \ 2 \ 3 endbmatrix_mboxpoint on $L$-underbracebeginbmatrix2 \ 3 \ -1 endbmatrix_p=colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix$$
Recall that we want this to be orthogonal to $vec n_alpha$, so:
$$colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix cdot colorredbeginbmatrix2 \ 4 \ -5 endbmatrix = 0 iff sigma = ldots$$
This $sigma$ gives you the point on $L$ where $R$ passes through and thus, together with $p$, completely determines $R$. So with this value for $sigma$, you have:
$$Requiv beginbmatrix2 \ 3 \ -1 endbmatrix + mu colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix$$
Addition after comments. I wouldn't focus on the first coordinate but on the other two:
- $R_2 to R_2-3R_1$
- $R_1 to R_1+R_2$
$$beginbmatrix 5 & -1 & -1 & -2 \ 12 & -3 & -2 & -4 endbmatrix
sim beginbmatrix 5 & -1 & -1 & -2 \ -3 & 0 & 1 & 2 endbmatrix
sim beginbmatrix 2 & -1 & 0 & 0 \ -3 & 0 & 1 & 2 endbmatrix$$
And from here it's easy to see you arrive at $(*)$, solving in terms of the first coordinate.
answered Jul 19 at 9:02
StackTD
20.1k1443
Being parallel to the plane is equivalent to the condition that the line is perpendicular to the plane's normal (vector).
Given $alpha equiv beginbmatrixx_1 \ x_2 \ x_3 endbmatrix= beginbmatrix0 \ -1 \ -1endbmatrix + rhocolorbluebeginbmatrix1 \ 2 \ 2 endbmatrix + taucolorpurplebeginbmatrix5 \ 5 \ 6 endbmatrix$
Either converting this to a cartesian equation or computing the following cross product gives you a normal vector of $alpha$:
$$vec n_alpha = colorbluebeginbmatrix1 \ 2 \ 2 endbmatrix times colorpurplebeginbmatrix5 \ 5 \ 6 endbmatrix=colorredbeginbmatrix2 \ 4 \ -5 endbmatrix$$
and $L equiv begincases 5x_1 -x_2 -x_3 +2 = 0 \ 12x_1 -3x_2 -2x_3 +4=0 endcases$
I've determined the vector equation of $L$ by the Gauss-Jordan method which will result in $Lequiv beginbmatrix2 \ 4 \ 0 endbmatrix +sigma beginbmatrix1 \ 1\ 3 endbmatrix$.
Maybe something went wrong in the calculations. A vector equation of $L$ is given by:
$$Lequiv beginbmatrix0 \ 0 \ 2 endbmatrix +sigma beginbmatrix1 \ 2 \ 3 endbmatrix tag$*$$$
Now a direction vector of $R$, through $p$ and a point on $L$, is given by their difference:
$$underbracebeginbmatrix0 \ 0 \ 2 endbmatrix +sigma beginbmatrix1 \ 2 \ 3 endbmatrix_mboxpoint on $L$-underbracebeginbmatrix2 \ 3 \ -1 endbmatrix_p=colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix$$
Recall that we want this to be orthogonal to $vec n_alpha$, so:
$$colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix cdot colorredbeginbmatrix2 \ 4 \ -5 endbmatrix = 0 iff sigma = ldots$$
This $sigma$ gives you the point on $L$ where $R$ passes through and thus, together with $p$, completely determines $R$. So with this value for $sigma$, you have:
$$Requiv beginbmatrix2 \ 3 \ -1 endbmatrix + mu colorgreenbeginbmatrix sigma-2 \ 2sigma-3 \ 3sigma+1 endbmatrix$$
Addition after comments. I wouldn't focus on the first coordinate but on the other two:
- $R_2 to R_2-3R_1$
- $R_1 to R_1+R_2$
$$beginbmatrix 5 & -1 & -1 & -2 \ 12 & -3 & -2 & -4 endbmatrix
sim beginbmatrix 5 & -1 & -1 & -2 \ -3 & 0 & 1 & 2 endbmatrix
sim beginbmatrix 2 & -1 & 0 & 0 \ -3 & 0 & 1 & 2 endbmatrix$$
And from here it's easy to see you arrive at $(*)$, solving in terms of the first coordinate.
answered Jul 19 at 9:02
StackTD
20.1k1443
edited Jul 19 at 9:46
answered Jul 19 at 9:02
StackTD
20.1k1443
answered Jul 19 at 9:02
StackTD
20.1k1443
answered Jul 19 at 9:02
StackTD
20.1k1443
20.1k1443
I'll check it now. I maybe used something incorrectly.
– Anonymous I
Jul 19 at 9:06
1
Oké, I'll understand your answer. I just need to find out where I made a mistake. Thanks for the answer.
– Anonymous I
Jul 19 at 9:08
Alright; you're welcome!
– StackTD
Jul 19 at 9:11
Now I get $Lequiv beginbmatrix-2 \ 4 \ 0 endbmatrix+ omicron beginbmatrix1 \ -2 \ 3 endbmatrix$. My mind is letting me down. :(
– Anonymous I
Jul 19 at 9:17
That's not correct but it's hard to tell where it goes wrong in your calculations...
– StackTD
Jul 19 at 9:18
 |Â
show 9 more comments
I'll check it now. I maybe used something incorrectly.
– Anonymous I
Jul 19 at 9:06
1
Oké, I'll understand your answer. I just need to find out where I made a mistake. Thanks for the answer.
– Anonymous I
Jul 19 at 9:08
Alright; you're welcome!
– StackTD
Jul 19 at 9:11
Now I get $Lequiv beginbmatrix-2 \ 4 \ 0 endbmatrix+ omicron beginbmatrix1 \ -2 \ 3 endbmatrix$. My mind is letting me down. :(
– Anonymous I
Jul 19 at 9:17
That's not correct but it's hard to tell where it goes wrong in your calculations...
– StackTD
Jul 19 at 9:18
I'll check it now. I maybe used something incorrectly.
– Anonymous I
Jul 19 at 9:06
I'll check it now. I maybe used something incorrectly.
– Anonymous I
Jul 19 at 9:06
1
1
Oké, I'll understand your answer. I just need to find out where I made a mistake. Thanks for the answer.
– Anonymous I
Jul 19 at 9:08
Oké, I'll understand your answer. I just need to find out where I made a mistake. Thanks for the answer.
– Anonymous I
Jul 19 at 9:08
Alright; you're welcome!
– StackTD
Jul 19 at 9:11
Alright; you're welcome!
– StackTD
Jul 19 at 9:11
Now I get $Lequiv beginbmatrix-2 \ 4 \ 0 endbmatrix+ omicron beginbmatrix1 \ -2 \ 3 endbmatrix$. My mind is letting me down. :(
– Anonymous I
Jul 19 at 9:17
Now I get $Lequiv beginbmatrix-2 \ 4 \ 0 endbmatrix+ omicron beginbmatrix1 \ -2 \ 3 endbmatrix$. My mind is letting me down. :(
– Anonymous I
Jul 19 at 9:17
That's not correct but it's hard to tell where it goes wrong in your calculations...
– StackTD
Jul 19 at 9:18
That's not correct but it's hard to tell where it goes wrong in your calculations...
– StackTD
Jul 19 at 9:18
 |Â
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up vote
1
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Here are a couple of alternative approaches. Note that the pair of equations that define $L$ are the equations of planes, that is, $L$ is described as the intersection of two planes.
For $R$ to be parallel to $alpha$, its direction vector must be some nontrivial linear combination of the generators of $alpha$, so we know that a parametric equation of $R$ is $p+lambda(rho[1,2,2]^T+tau[5,5,6]^T)$ for some as yet undetermined $rho$ and $tau$. The other condition on $R$ is that it intersects $L$, i.e., that it has a common intersection with both of the defining planes. Substituting into their equations produces the system $$lambda(rho+14tau)+10 = 0 \ lambda(2rho+33tau)+21 = 0.$$ This system is underdetermined, which makes sense because any nonzero scalar multiple of a line’s direction vector is also a direction vector for the line. Set $lambda$ to some convenient value such as $1$ and solve the resulting system for $rho$ and $tau$.
What you’re essentially doing above is finding the intersection of three planes: the two that define $L$ and a plane through $p$ that is parallel to $alpha$. The latter plane consists of all of the lines through $p$ that are parallel to $alpha$. You could use this idea to solve the problem is a slightly different way, which I’ll illustrate using homogeneous coordinates. If the equation of a plane is $ax+by+cz+d=0$, it can be represented by the homogeneous vector $mathbfpi = [a,b,c,d]^T$, so that the equation of the plane becomes $mathbfpi^Tmathbf x = 0$ (i.e., the dot product of the plane vector and coordinates of a point on the plane). The parallel plane to $alpha$ through $p$ is then a null vector $mathbfpi$ of the matrix $$beginbmatrix2&3&-1&1\1&2&2&0\5&5&6&0endbmatrix.$$ (Think about how this relates to the condition $mathbfpi^Tmathbf x=0$.) The rows of this matrix are just the point $p$ and the two generators of $alpha$—in projective-geometric terms, these are three points on the plane, two of which are at infinity. You can compute this null vector in any of the usual ways, such as Gaussian elimination. If you’ve done it correctly, the resulting space should be one-dimensional.†One possibility is $mathbfpi = [2,4,-5,-21]^T$. To compute the intersection of $pi$ and $L$, use the same procedure: assemble the vectors that represent the three planes into a matrix and find a null vector of it, then convert to inhomogeneous Cartesian coordinates by dividing through by the last component of the vector. This gives you another point on $R$, from which I trust you know how to generate an equation of this line.
†You can also compute the plane’s representative vector by taking the cross product of the two generators of $alpha$ to get the first three components of $mathbfpi$ and setting its last component to the negative of the dot product of this normal vector with $p$. This is a direct analogue of one of the ways you’d find the point-normal equation of the plane. However, this method is specific to three-dimensional space, while the null space method works in spaces of any dimension.
answered Jul 19 at 20:49
amd
25.9k2943
add a comment |Â
up vote
1
down vote
Here are a couple of alternative approaches. Note that the pair of equations that define $L$ are the equations of planes, that is, $L$ is described as the intersection of two planes.
For $R$ to be parallel to $alpha$, its direction vector must be some nontrivial linear combination of the generators of $alpha$, so we know that a parametric equation of $R$ is $p+lambda(rho[1,2,2]^T+tau[5,5,6]^T)$ for some as yet undetermined $rho$ and $tau$. The other condition on $R$ is that it intersects $L$, i.e., that it has a common intersection with both of the defining planes. Substituting into their equations produces the system $$lambda(rho+14tau)+10 = 0 \ lambda(2rho+33tau)+21 = 0.$$ This system is underdetermined, which makes sense because any nonzero scalar multiple of a line’s direction vector is also a direction vector for the line. Set $lambda$ to some convenient value such as $1$ and solve the resulting system for $rho$ and $tau$.
What you’re essentially doing above is finding the intersection of three planes: the two that define $L$ and a plane through $p$ that is parallel to $alpha$. The latter plane consists of all of the lines through $p$ that are parallel to $alpha$. You could use this idea to solve the problem is a slightly different way, which I’ll illustrate using homogeneous coordinates. If the equation of a plane is $ax+by+cz+d=0$, it can be represented by the homogeneous vector $mathbfpi = [a,b,c,d]^T$, so that the equation of the plane becomes $mathbfpi^Tmathbf x = 0$ (i.e., the dot product of the plane vector and coordinates of a point on the plane). The parallel plane to $alpha$ through $p$ is then a null vector $mathbfpi$ of the matrix $$beginbmatrix2&3&-1&1\1&2&2&0\5&5&6&0endbmatrix.$$ (Think about how this relates to the condition $mathbfpi^Tmathbf x=0$.) The rows of this matrix are just the point $p$ and the two generators of $alpha$—in projective-geometric terms, these are three points on the plane, two of which are at infinity. You can compute this null vector in any of the usual ways, such as Gaussian elimination. If you’ve done it correctly, the resulting space should be one-dimensional.†One possibility is $mathbfpi = [2,4,-5,-21]^T$. To compute the intersection of $pi$ and $L$, use the same procedure: assemble the vectors that represent the three planes into a matrix and find a null vector of it, then convert to inhomogeneous Cartesian coordinates by dividing through by the last component of the vector. This gives you another point on $R$, from which I trust you know how to generate an equation of this line.
†You can also compute the plane’s representative vector by taking the cross product of the two generators of $alpha$ to get the first three components of $mathbfpi$ and setting its last component to the negative of the dot product of this normal vector with $p$. This is a direct analogue of one of the ways you’d find the point-normal equation of the plane. However, this method is specific to three-dimensional space, while the null space method works in spaces of any dimension.
answered Jul 19 at 20:49
amd
25.9k2943
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here are a couple of alternative approaches. Note that the pair of equations that define $L$ are the equations of planes, that is, $L$ is described as the intersection of two planes.
For $R$ to be parallel to $alpha$, its direction vector must be some nontrivial linear combination of the generators of $alpha$, so we know that a parametric equation of $R$ is $p+lambda(rho[1,2,2]^T+tau[5,5,6]^T)$ for some as yet undetermined $rho$ and $tau$. The other condition on $R$ is that it intersects $L$, i.e., that it has a common intersection with both of the defining planes. Substituting into their equations produces the system $$lambda(rho+14tau)+10 = 0 \ lambda(2rho+33tau)+21 = 0.$$ This system is underdetermined, which makes sense because any nonzero scalar multiple of a line’s direction vector is also a direction vector for the line. Set $lambda$ to some convenient value such as $1$ and solve the resulting system for $rho$ and $tau$.
What you’re essentially doing above is finding the intersection of three planes: the two that define $L$ and a plane through $p$ that is parallel to $alpha$. The latter plane consists of all of the lines through $p$ that are parallel to $alpha$. You could use this idea to solve the problem is a slightly different way, which I’ll illustrate using homogeneous coordinates. If the equation of a plane is $ax+by+cz+d=0$, it can be represented by the homogeneous vector $mathbfpi = [a,b,c,d]^T$, so that the equation of the plane becomes $mathbfpi^Tmathbf x = 0$ (i.e., the dot product of the plane vector and coordinates of a point on the plane). The parallel plane to $alpha$ through $p$ is then a null vector $mathbfpi$ of the matrix $$beginbmatrix2&3&-1&1\1&2&2&0\5&5&6&0endbmatrix.$$ (Think about how this relates to the condition $mathbfpi^Tmathbf x=0$.) The rows of this matrix are just the point $p$ and the two generators of $alpha$—in projective-geometric terms, these are three points on the plane, two of which are at infinity. You can compute this null vector in any of the usual ways, such as Gaussian elimination. If you’ve done it correctly, the resulting space should be one-dimensional.†One possibility is $mathbfpi = [2,4,-5,-21]^T$. To compute the intersection of $pi$ and $L$, use the same procedure: assemble the vectors that represent the three planes into a matrix and find a null vector of it, then convert to inhomogeneous Cartesian coordinates by dividing through by the last component of the vector. This gives you another point on $R$, from which I trust you know how to generate an equation of this line.
†You can also compute the plane’s representative vector by taking the cross product of the two generators of $alpha$ to get the first three components of $mathbfpi$ and setting its last component to the negative of the dot product of this normal vector with $p$. This is a direct analogue of one of the ways you’d find the point-normal equation of the plane. However, this method is specific to three-dimensional space, while the null space method works in spaces of any dimension.
answered Jul 19 at 20:49
amd
25.9k2943
Here are a couple of alternative approaches. Note that the pair of equations that define $L$ are the equations of planes, that is, $L$ is described as the intersection of two planes.
For $R$ to be parallel to $alpha$, its direction vector must be some nontrivial linear combination of the generators of $alpha$, so we know that a parametric equation of $R$ is $p+lambda(rho[1,2,2]^T+tau[5,5,6]^T)$ for some as yet undetermined $rho$ and $tau$. The other condition on $R$ is that it intersects $L$, i.e., that it has a common intersection with both of the defining planes. Substituting into their equations produces the system $$lambda(rho+14tau)+10 = 0 \ lambda(2rho+33tau)+21 = 0.$$ This system is underdetermined, which makes sense because any nonzero scalar multiple of a line’s direction vector is also a direction vector for the line. Set $lambda$ to some convenient value such as $1$ and solve the resulting system for $rho$ and $tau$.
What you’re essentially doing above is finding the intersection of three planes: the two that define $L$ and a plane through $p$ that is parallel to $alpha$. The latter plane consists of all of the lines through $p$ that are parallel to $alpha$. You could use this idea to solve the problem is a slightly different way, which I’ll illustrate using homogeneous coordinates. If the equation of a plane is $ax+by+cz+d=0$, it can be represented by the homogeneous vector $mathbfpi = [a,b,c,d]^T$, so that the equation of the plane becomes $mathbfpi^Tmathbf x = 0$ (i.e., the dot product of the plane vector and coordinates of a point on the plane). The parallel plane to $alpha$ through $p$ is then a null vector $mathbfpi$ of the matrix $$beginbmatrix2&3&-1&1\1&2&2&0\5&5&6&0endbmatrix.$$ (Think about how this relates to the condition $mathbfpi^Tmathbf x=0$.) The rows of this matrix are just the point $p$ and the two generators of $alpha$—in projective-geometric terms, these are three points on the plane, two of which are at infinity. You can compute this null vector in any of the usual ways, such as Gaussian elimination. If you’ve done it correctly, the resulting space should be one-dimensional.†One possibility is $mathbfpi = [2,4,-5,-21]^T$. To compute the intersection of $pi$ and $L$, use the same procedure: assemble the vectors that represent the three planes into a matrix and find a null vector of it, then convert to inhomogeneous Cartesian coordinates by dividing through by the last component of the vector. This gives you another point on $R$, from which I trust you know how to generate an equation of this line.
†You can also compute the plane’s representative vector by taking the cross product of the two generators of $alpha$ to get the first three components of $mathbfpi$ and setting its last component to the negative of the dot product of this normal vector with $p$. This is a direct analogue of one of the ways you’d find the point-normal equation of the plane. However, this method is specific to three-dimensional space, while the null space method works in spaces of any dimension.
answered Jul 19 at 20:49
amd
25.9k2943
edited Jul 19 at 21:57
answered Jul 19 at 20:49
amd
25.9k2943
answered Jul 19 at 20:49
amd
25.9k2943
answered Jul 19 at 20:49
amd
25.9k2943
25.9k2943
add a comment |Â
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The vector (parametric) equation for L seems uncorrect.
– gimusi
Jul 19 at 8:38
@gimusi : Can you point out the mistake I've made exactly?
– Anonymous I
Jul 19 at 8:40
Vector (2,4,0) doesn't satisfy the cartesian equations.
– gimusi
Jul 19 at 8:44
What about $beginbmatrix-2 \ -4 \ 0 endbmatrix$?
– Anonymous I
Jul 19 at 8:46
1
You can check from the cartesian equations that it doesn't work.
– gimusi
Jul 19 at 9:00