Mixing
Find the angle between two planes using their normal vectors
$begingroup$
The angle between two intersecting planes is defined to be the angle between
their normal vectors. Find the angle between the planes $x – 2y + z = 0$ and $2x + 3y – 2z = 0$. Find the parametric equations of the line of intersection of the two planes above.
For the first plane I said $overrightarrow n_0 =langle 1, -2, 1 rangle$ and for the second plane $overrightarrow n_1 = langle 2, 3, -2 rangle$
Then using $cos(theta)= frac{mathbf A bullet mathbf B}{|mathbf A||mathbf B|}$ so $theta = cos^{-1}left(frac{mathbf A bullet mathbf B}{|mathbf A||mathbf B|}right)$
$mathbf A bullet mathbf B = -6$
$|mathbf A||mathbf B|= sqrt{102}$
$theta = cos^{-1}left(frac{-6}{sqrt{102}}right)$
Assuming this is correct so far, how do I find the parametric equations from here?
multivariable-calculus parametric plane-curves
edited Jan 20 at 4:40
Wolfgang
4,29743377
asked Nov 14 '15 at 18:19
hax0r_n_codehax0r_n_code
1,91374179
$endgroup$
|
show 2 more comments
$begingroup$
The angle between two intersecting planes is defined to be the angle between
their normal vectors. Find the angle between the planes $x – 2y + z = 0$ and $2x + 3y – 2z = 0$. Find the parametric equations of the line of intersection of the two planes above.
For the first plane I said $overrightarrow n_0 =langle 1, -2, 1 rangle$ and for the second plane $overrightarrow n_1 = langle 2, 3, -2 rangle$
Then using $cos(theta)= frac{mathbf A bullet mathbf B}{|mathbf A||mathbf B|}$ so $theta = cos^{-1}left(frac{mathbf A bullet mathbf B}{|mathbf A||mathbf B|}right)$
$mathbf A bullet mathbf B = -6$
$|mathbf A||mathbf B|= sqrt{102}$
$theta = cos^{-1}left(frac{-6}{sqrt{102}}right)$
Assuming this is correct so far, how do I find the parametric equations from here?
multivariable-calculus parametric plane-curves
edited Jan 20 at 4:40
Wolfgang
4,29743377
asked Nov 14 '15 at 18:19
hax0r_n_codehax0r_n_code
1,91374179
$endgroup$
$begingroup$
Hint: You have normals to both planes. The line of intersection lies in both planes. Can you think of a way to compute that line from the two normals? (BTW, as an unrelated note, to really complete the first part, what is $theta$, explicitly?)
$endgroup$
– rogerl
Nov 14 '15 at 18:23
$begingroup$
@rogerl $frac{3pi}{4}$?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:24
$begingroup$
Yes, that's right.
$endgroup$
– rogerl
Nov 14 '15 at 18:25
1
$begingroup$
How do you get $|n_0||n_1|$ to be $6sqrt2$, though? I get $sqrt 6cdotsqrt{17}$.
$endgroup$
– Henning Makholm
Nov 14 '15 at 18:27
$begingroup$
@HenningMakholm Yup I just caught that computational error.
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:28
|
show 2 more comments
$begingroup$
The angle between two intersecting planes is defined to be the angle between
their normal vectors. Find the angle between the planes $x – 2y + z = 0$ and $2x + 3y – 2z = 0$. Find the parametric equations of the line of intersection of the two planes above.
For the first plane I said $overrightarrow n_0 =langle 1, -2, 1 rangle$ and for the second plane $overrightarrow n_1 = langle 2, 3, -2 rangle$
Then using $cos(theta)= frac{mathbf A bullet mathbf B}{|mathbf A||mathbf B|}$ so $theta = cos^{-1}left(frac{mathbf A bullet mathbf B}{|mathbf A||mathbf B|}right)$
$mathbf A bullet mathbf B = -6$
$|mathbf A||mathbf B|= sqrt{102}$
$theta = cos^{-1}left(frac{-6}{sqrt{102}}right)$
Assuming this is correct so far, how do I find the parametric equations from here?
multivariable-calculus parametric plane-curves
edited Jan 20 at 4:40
Wolfgang
4,29743377
asked Nov 14 '15 at 18:19
hax0r_n_codehax0r_n_code
1,91374179
$endgroup$
The angle between two intersecting planes is defined to be the angle between
their normal vectors. Find the angle between the planes $x – 2y + z = 0$ and $2x + 3y – 2z = 0$. Find the parametric equations of the line of intersection of the two planes above.
For the first plane I said $overrightarrow n_0 =langle 1, -2, 1 rangle$ and for the second plane $overrightarrow n_1 = langle 2, 3, -2 rangle$
Then using $cos(theta)= frac{mathbf A bullet mathbf B}{|mathbf A||mathbf B|}$ so $theta = cos^{-1}left(frac{mathbf A bullet mathbf B}{|mathbf A||mathbf B|}right)$
$mathbf A bullet mathbf B = -6$
$|mathbf A||mathbf B|= sqrt{102}$
$theta = cos^{-1}left(frac{-6}{sqrt{102}}right)$
Assuming this is correct so far, how do I find the parametric equations from here?
multivariable-calculus parametric plane-curves
multivariable-calculus parametric plane-curves
edited Jan 20 at 4:40
Wolfgang
4,29743377
asked Nov 14 '15 at 18:19
hax0r_n_codehax0r_n_code
1,91374179
edited Jan 20 at 4:40
Wolfgang
4,29743377
asked Nov 14 '15 at 18:19
hax0r_n_codehax0r_n_code
1,91374179
edited Jan 20 at 4:40
Wolfgang
4,29743377
edited Jan 20 at 4:40
Wolfgang
4,29743377
edited Jan 20 at 4:40
Wolfgang
4,29743377
4,29743377
asked Nov 14 '15 at 18:19
hax0r_n_codehax0r_n_code
1,91374179
asked Nov 14 '15 at 18:19
hax0r_n_codehax0r_n_code
1,91374179
asked Nov 14 '15 at 18:19
hax0r_n_codehax0r_n_code
1,91374179
1,91374179
$begingroup$
Hint: You have normals to both planes. The line of intersection lies in both planes. Can you think of a way to compute that line from the two normals? (BTW, as an unrelated note, to really complete the first part, what is $theta$, explicitly?)
$endgroup$
– rogerl
Nov 14 '15 at 18:23
$begingroup$
@rogerl $frac{3pi}{4}$?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:24
$begingroup$
Yes, that's right.
$endgroup$
– rogerl
Nov 14 '15 at 18:25
1
$begingroup$
How do you get $|n_0||n_1|$ to be $6sqrt2$, though? I get $sqrt 6cdotsqrt{17}$.
$endgroup$
– Henning Makholm
Nov 14 '15 at 18:27
$begingroup$
@HenningMakholm Yup I just caught that computational error.
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:28
|
show 2 more comments
$begingroup$
Hint: You have normals to both planes. The line of intersection lies in both planes. Can you think of a way to compute that line from the two normals? (BTW, as an unrelated note, to really complete the first part, what is $theta$, explicitly?)
$endgroup$
– rogerl
Nov 14 '15 at 18:23
$begingroup$
@rogerl $frac{3pi}{4}$?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:24
$begingroup$
Yes, that's right.
$endgroup$
– rogerl
Nov 14 '15 at 18:25
1
$begingroup$
How do you get $|n_0||n_1|$ to be $6sqrt2$, though? I get $sqrt 6cdotsqrt{17}$.
$endgroup$
– Henning Makholm
Nov 14 '15 at 18:27
$begingroup$
@HenningMakholm Yup I just caught that computational error.
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:28
$begingroup$
Hint: You have normals to both planes. The line of intersection lies in both planes. Can you think of a way to compute that line from the two normals? (BTW, as an unrelated note, to really complete the first part, what is $theta$, explicitly?)
$endgroup$
– rogerl
Nov 14 '15 at 18:23
$begingroup$
Hint: You have normals to both planes. The line of intersection lies in both planes. Can you think of a way to compute that line from the two normals? (BTW, as an unrelated note, to really complete the first part, what is $theta$, explicitly?)
$endgroup$
– rogerl
Nov 14 '15 at 18:23
$begingroup$
@rogerl $frac{3pi}{4}$?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:24
$begingroup$
@rogerl $frac{3pi}{4}$?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:24
$begingroup$
Yes, that's right.
$endgroup$
– rogerl
Nov 14 '15 at 18:25
$begingroup$
Yes, that's right.
$endgroup$
– rogerl
Nov 14 '15 at 18:25
1
1
$begingroup$
How do you get $|n_0||n_1|$ to be $6sqrt2$, though? I get $sqrt 6cdotsqrt{17}$.
$endgroup$
– Henning Makholm
Nov 14 '15 at 18:27
$begingroup$
How do you get $|n_0||n_1|$ to be $6sqrt2$, though? I get $sqrt 6cdotsqrt{17}$.
$endgroup$
– Henning Makholm
Nov 14 '15 at 18:27
$begingroup$
@HenningMakholm Yup I just caught that computational error.
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:28
$begingroup$
@HenningMakholm Yup I just caught that computational error.
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:28
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
I don't think calculating the angle between two lines will help you to find the equation of the line of intersection of two lines.
It should be clear that the line of intersection is the line which is perpendicular to the normal of both the given planes.
Then the line will be along the cross product of the normal vector of both the planes.
$therefore$ doing simple calculation gives us
$$
vec{n_{0}}timesvec{n_{1}} = langle1,4,7rangle
$$
So now we know the direction in which the required line is pointing.
Also it is not hard to guess that $(0,0,0)$ lies in both the plane. Thus the equation of line should pass through it.
Hence the parametric equation of the line will become:-
$$
t = dfrac{t}{4} = dfrac{t}{7}
$$
with $t$ as a parameter.
answered Nov 14 '15 at 18:35
Prakhar GuptaPrakhar Gupta
381113
$endgroup$
$begingroup$
Thanks for the the help. Are you saying I calculated the angle between the two planes incorrectly?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:41
1
$begingroup$
No. Your method and accuracy are correct. Only that it won't help. There is no point in calculating angle to find the equation of the line of intersection of lines.
$endgroup$
– Prakhar Gupta
Nov 14 '15 at 18:45
$begingroup$
Oh, ok. You're saying this won't help me with finding the parametric equations? So that's why I need to use the cross product? The cross product produces the line and direction that intersects both planes. I then use that to produce the parametric equation and determine the parameter t?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:48
$begingroup$
I understand what you're saying, but finding the angle is also a part of the question. So I guess I'm confused as to if I needed to do this or not.
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:49
$begingroup$
$r(t)=langle 1, -2, 1 rangle + tlangle 1,4,7>$
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:55
|
show 4 more comments
$begingroup$
You can remember that $vec atimesvec b$ is perpendicular to both $vec a$ and $vec b$, so that $vec n_0timesvec n_1$ gives the direction of the intersection line. In addition, you know that both planes pass through $(0,0,0)$ and so does the intersection line.
answered Nov 14 '15 at 18:34
AretinoAretino
24.7k21444
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't think calculating the angle between two lines will help you to find the equation of the line of intersection of two lines.
It should be clear that the line of intersection is the line which is perpendicular to the normal of both the given planes.
Then the line will be along the cross product of the normal vector of both the planes.
$therefore$ doing simple calculation gives us
$$
vec{n_{0}}timesvec{n_{1}} = langle1,4,7rangle
$$
So now we know the direction in which the required line is pointing.
Also it is not hard to guess that $(0,0,0)$ lies in both the plane. Thus the equation of line should pass through it.
Hence the parametric equation of the line will become:-
$$
t = dfrac{t}{4} = dfrac{t}{7}
$$
with $t$ as a parameter.
answered Nov 14 '15 at 18:35
Prakhar GuptaPrakhar Gupta
381113
$endgroup$
$begingroup$
Thanks for the the help. Are you saying I calculated the angle between the two planes incorrectly?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:41
1
$begingroup$
No. Your method and accuracy are correct. Only that it won't help. There is no point in calculating angle to find the equation of the line of intersection of lines.
$endgroup$
– Prakhar Gupta
Nov 14 '15 at 18:45
$begingroup$
Oh, ok. You're saying this won't help me with finding the parametric equations? So that's why I need to use the cross product? The cross product produces the line and direction that intersects both planes. I then use that to produce the parametric equation and determine the parameter t?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:48
$begingroup$
I understand what you're saying, but finding the angle is also a part of the question. So I guess I'm confused as to if I needed to do this or not.
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:49
$begingroup$
$r(t)=langle 1, -2, 1 rangle + tlangle 1,4,7>$
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:55
|
show 4 more comments
$begingroup$
I don't think calculating the angle between two lines will help you to find the equation of the line of intersection of two lines.
It should be clear that the line of intersection is the line which is perpendicular to the normal of both the given planes.
Then the line will be along the cross product of the normal vector of both the planes.
$therefore$ doing simple calculation gives us
$$
vec{n_{0}}timesvec{n_{1}} = langle1,4,7rangle
$$
So now we know the direction in which the required line is pointing.
Also it is not hard to guess that $(0,0,0)$ lies in both the plane. Thus the equation of line should pass through it.
Hence the parametric equation of the line will become:-
$$
t = dfrac{t}{4} = dfrac{t}{7}
$$
with $t$ as a parameter.
answered Nov 14 '15 at 18:35
Prakhar GuptaPrakhar Gupta
381113
$endgroup$
$begingroup$
Thanks for the the help. Are you saying I calculated the angle between the two planes incorrectly?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:41
1
$begingroup$
No. Your method and accuracy are correct. Only that it won't help. There is no point in calculating angle to find the equation of the line of intersection of lines.
$endgroup$
– Prakhar Gupta
Nov 14 '15 at 18:45
$begingroup$
Oh, ok. You're saying this won't help me with finding the parametric equations? So that's why I need to use the cross product? The cross product produces the line and direction that intersects both planes. I then use that to produce the parametric equation and determine the parameter t?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:48
$begingroup$
I understand what you're saying, but finding the angle is also a part of the question. So I guess I'm confused as to if I needed to do this or not.
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:49
$begingroup$
$r(t)=langle 1, -2, 1 rangle + tlangle 1,4,7>$
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:55
|
show 4 more comments
$begingroup$
I don't think calculating the angle between two lines will help you to find the equation of the line of intersection of two lines.
It should be clear that the line of intersection is the line which is perpendicular to the normal of both the given planes.
Then the line will be along the cross product of the normal vector of both the planes.
$therefore$ doing simple calculation gives us
$$
vec{n_{0}}timesvec{n_{1}} = langle1,4,7rangle
$$
So now we know the direction in which the required line is pointing.
Also it is not hard to guess that $(0,0,0)$ lies in both the plane. Thus the equation of line should pass through it.
Hence the parametric equation of the line will become:-
$$
t = dfrac{t}{4} = dfrac{t}{7}
$$
with $t$ as a parameter.
answered Nov 14 '15 at 18:35
Prakhar GuptaPrakhar Gupta
381113
$endgroup$
I don't think calculating the angle between two lines will help you to find the equation of the line of intersection of two lines.
It should be clear that the line of intersection is the line which is perpendicular to the normal of both the given planes.
Then the line will be along the cross product of the normal vector of both the planes.
$therefore$ doing simple calculation gives us
$$
vec{n_{0}}timesvec{n_{1}} = langle1,4,7rangle
$$
So now we know the direction in which the required line is pointing.
Also it is not hard to guess that $(0,0,0)$ lies in both the plane. Thus the equation of line should pass through it.
Hence the parametric equation of the line will become:-
$$
t = dfrac{t}{4} = dfrac{t}{7}
$$
with $t$ as a parameter.
answered Nov 14 '15 at 18:35
Prakhar GuptaPrakhar Gupta
381113
answered Nov 14 '15 at 18:35
Prakhar GuptaPrakhar Gupta
381113
answered Nov 14 '15 at 18:35
Prakhar GuptaPrakhar Gupta
381113
answered Nov 14 '15 at 18:35
Prakhar GuptaPrakhar Gupta
381113
381113
$begingroup$
Thanks for the the help. Are you saying I calculated the angle between the two planes incorrectly?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:41
1
$begingroup$
No. Your method and accuracy are correct. Only that it won't help. There is no point in calculating angle to find the equation of the line of intersection of lines.
$endgroup$
– Prakhar Gupta
Nov 14 '15 at 18:45
$begingroup$
Oh, ok. You're saying this won't help me with finding the parametric equations? So that's why I need to use the cross product? The cross product produces the line and direction that intersects both planes. I then use that to produce the parametric equation and determine the parameter t?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:48
$begingroup$
I understand what you're saying, but finding the angle is also a part of the question. So I guess I'm confused as to if I needed to do this or not.
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:49
$begingroup$
$r(t)=langle 1, -2, 1 rangle + tlangle 1,4,7>$
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:55
|
show 4 more comments
$begingroup$
Thanks for the the help. Are you saying I calculated the angle between the two planes incorrectly?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:41
1
$begingroup$
No. Your method and accuracy are correct. Only that it won't help. There is no point in calculating angle to find the equation of the line of intersection of lines.
$endgroup$
– Prakhar Gupta
Nov 14 '15 at 18:45
$begingroup$
Oh, ok. You're saying this won't help me with finding the parametric equations? So that's why I need to use the cross product? The cross product produces the line and direction that intersects both planes. I then use that to produce the parametric equation and determine the parameter t?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:48
$begingroup$
I understand what you're saying, but finding the angle is also a part of the question. So I guess I'm confused as to if I needed to do this or not.
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:49
$begingroup$
$r(t)=langle 1, -2, 1 rangle + tlangle 1,4,7>$
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:55
$begingroup$
Thanks for the the help. Are you saying I calculated the angle between the two planes incorrectly?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:41
$begingroup$
Thanks for the the help. Are you saying I calculated the angle between the two planes incorrectly?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:41
1
1
$begingroup$
No. Your method and accuracy are correct. Only that it won't help. There is no point in calculating angle to find the equation of the line of intersection of lines.
$endgroup$
– Prakhar Gupta
Nov 14 '15 at 18:45
$begingroup$
No. Your method and accuracy are correct. Only that it won't help. There is no point in calculating angle to find the equation of the line of intersection of lines.
$endgroup$
– Prakhar Gupta
Nov 14 '15 at 18:45
$begingroup$
Oh, ok. You're saying this won't help me with finding the parametric equations? So that's why I need to use the cross product? The cross product produces the line and direction that intersects both planes. I then use that to produce the parametric equation and determine the parameter t?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:48
$begingroup$
Oh, ok. You're saying this won't help me with finding the parametric equations? So that's why I need to use the cross product? The cross product produces the line and direction that intersects both planes. I then use that to produce the parametric equation and determine the parameter t?
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:48
$begingroup$
I understand what you're saying, but finding the angle is also a part of the question. So I guess I'm confused as to if I needed to do this or not.
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:49
$begingroup$
I understand what you're saying, but finding the angle is also a part of the question. So I guess I'm confused as to if I needed to do this or not.
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:49
$begingroup$
$r(t)=langle 1, -2, 1 rangle + tlangle 1,4,7>$
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:55
$begingroup$
$r(t)=langle 1, -2, 1 rangle + tlangle 1,4,7>$
$endgroup$
– hax0r_n_code
Nov 14 '15 at 18:55
|
show 4 more comments
$begingroup$
You can remember that $vec atimesvec b$ is perpendicular to both $vec a$ and $vec b$, so that $vec n_0timesvec n_1$ gives the direction of the intersection line. In addition, you know that both planes pass through $(0,0,0)$ and so does the intersection line.
answered Nov 14 '15 at 18:34
AretinoAretino
24.7k21444
$endgroup$
add a comment |
$begingroup$
You can remember that $vec atimesvec b$ is perpendicular to both $vec a$ and $vec b$, so that $vec n_0timesvec n_1$ gives the direction of the intersection line. In addition, you know that both planes pass through $(0,0,0)$ and so does the intersection line.
answered Nov 14 '15 at 18:34
AretinoAretino
24.7k21444
$endgroup$
add a comment |
$begingroup$
You can remember that $vec atimesvec b$ is perpendicular to both $vec a$ and $vec b$, so that $vec n_0timesvec n_1$ gives the direction of the intersection line. In addition, you know that both planes pass through $(0,0,0)$ and so does the intersection line.
answered Nov 14 '15 at 18:34
AretinoAretino
24.7k21444
$endgroup$
You can remember that $vec atimesvec b$ is perpendicular to both $vec a$ and $vec b$, so that $vec n_0timesvec n_1$ gives the direction of the intersection line. In addition, you know that both planes pass through $(0,0,0)$ and so does the intersection line.
answered Nov 14 '15 at 18:34
AretinoAretino
24.7k21444
answered Nov 14 '15 at 18:34
AretinoAretino
24.7k21444
answered Nov 14 '15 at 18:34
AretinoAretino
24.7k21444
answered Nov 14 '15 at 18:34
AretinoAretino
24.7k21444
24.7k21444
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$begingroup$
Hint: You have normals to both planes. The line of intersection lies in both planes. Can you think of a way to compute that line from the two normals? (BTW, as an unrelated note, to really complete the first part, what is $theta$, explicitly?)
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– rogerl
Nov 14 '15 at 18:23
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@rogerl $frac{3pi}{4}$?
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– hax0r_n_code
Nov 14 '15 at 18:24
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Yes, that's right.
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– rogerl
Nov 14 '15 at 18:25
1
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How do you get $|n_0||n_1|$ to be $6sqrt2$, though? I get $sqrt 6cdotsqrt{17}$.
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– Henning Makholm
Nov 14 '15 at 18:27
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@HenningMakholm Yup I just caught that computational error.
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– hax0r_n_code
Nov 14 '15 at 18:28