Mixing
Determining the general perpendicular normal
$begingroup$
Given a vector $n in mathbb R^3$, how can I find (any) perpendicular normal $x$, such that $n^Tcdot x=0$ for all possible values of $n$ (i.e. with no edge cases)
linear-algebra
edited Jan 8 at 15:36
Andrei
11.8k21026
asked Jan 8 at 15:16
MercuryMercury
14115
$endgroup$
add a comment |
$begingroup$
Given a vector $n in mathbb R^3$, how can I find (any) perpendicular normal $x$, such that $n^Tcdot x=0$ for all possible values of $n$ (i.e. with no edge cases)
linear-algebra
edited Jan 8 at 15:36
Andrei
11.8k21026
asked Jan 8 at 15:16
MercuryMercury
14115
$endgroup$
1
$begingroup$
$n^Tx = 0$ is a linear equation in $x$, so use your preferred method for solving linear equations.
$endgroup$
– user3482749
Jan 8 at 15:23
add a comment |
$begingroup$
Given a vector $n in mathbb R^3$, how can I find (any) perpendicular normal $x$, such that $n^Tcdot x=0$ for all possible values of $n$ (i.e. with no edge cases)
linear-algebra
edited Jan 8 at 15:36
Andrei
11.8k21026
asked Jan 8 at 15:16
MercuryMercury
14115
$endgroup$
Given a vector $n in mathbb R^3$, how can I find (any) perpendicular normal $x$, such that $n^Tcdot x=0$ for all possible values of $n$ (i.e. with no edge cases)
linear-algebra
linear-algebra
edited Jan 8 at 15:36
Andrei
11.8k21026
asked Jan 8 at 15:16
MercuryMercury
14115
edited Jan 8 at 15:36
Andrei
11.8k21026
asked Jan 8 at 15:16
MercuryMercury
14115
edited Jan 8 at 15:36
Andrei
11.8k21026
edited Jan 8 at 15:36
Andrei
11.8k21026
edited Jan 8 at 15:36
Andrei
11.8k21026
11.8k21026
asked Jan 8 at 15:16
MercuryMercury
14115
asked Jan 8 at 15:16
MercuryMercury
14115
asked Jan 8 at 15:16
MercuryMercury
14115
14115
1
$begingroup$
$n^Tx = 0$ is a linear equation in $x$, so use your preferred method for solving linear equations.
$endgroup$
– user3482749
Jan 8 at 15:23
add a comment |
1
$begingroup$
$n^Tx = 0$ is a linear equation in $x$, so use your preferred method for solving linear equations.
$endgroup$
– user3482749
Jan 8 at 15:23
1
1
$begingroup$
$n^Tx = 0$ is a linear equation in $x$, so use your preferred method for solving linear equations.
$endgroup$
– user3482749
Jan 8 at 15:23
$begingroup$
$n^Tx = 0$ is a linear equation in $x$, so use your preferred method for solving linear equations.
$endgroup$
– user3482749
Jan 8 at 15:23
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
I’m not sure what you mean by “no edge cases” here. Most methods require at least some case analysis to avoid a degenerate case. Bob Werner’s answer does give a nifty case-free algorithm that looks like it will generalize to higher-dimensional spaces, but it seems like overkill for the problem in $mathbb R^3$.
Given a vector $n=(a,b,c)inmathbb R^3$, its cross products with the standard basis vectors $$ntimes(1,0,0) = (0,c,-b) \ ntimes(0,1,0) = (-c,0,a) \ ntimes(0,0,1) = (b,-a,0)$$ are all orthogonal to $n$. If $n$ is nonzero, then at least two of these are nonzero. If you want all of the orthogonal vectors to $n$, choose two of the above vectors that are linearly independent: their span is the set of vectors orthogonal to $n$, but this is simply the plane through the origin with $n$ as a normal.
answered Jan 8 at 19:57
amdamd
29.7k21050
$endgroup$
$begingroup$
I have written a (2nd) solution which is alongside yours, but by using the (classical) operator $n_{times}$ allowing to avoid to distinguish cases.
$endgroup$
– Jean Marie
Jan 10 at 18:59
$begingroup$
@JeanMarie Isn’t there an implicit case analysis in the computation of the column space of $n_times$?
$endgroup$
– amd
Jan 10 at 19:03
1
$begingroup$
On the computational side, I agree, but on the theoretical side this "formula" permits to express things : for example a vector which is orthogonal to $n_1$ and $n_2$ belongs to $range(X_{n_1})cap range(X_{n_2})$.
$endgroup$
– Jean Marie
Jan 10 at 19:08
$begingroup$
@JeanMarie A fair point. Of course, one can express the same thing without introducing $n_times$ via $span(n)^perp$ (not coincidentally, the first method in your answer).
$endgroup$
– amd
Jan 10 at 20:02
add a comment |
$begingroup$
Levent Kitis posted a solution on UseNet sci.math. The first component of the vector is altered with the vector's norm, then a symmetric orthogonal Householder matrix is formed using the altered vector. The first row of the Householder matrix parallels the original vector, and the other two rows are perpendicular to it:
Let the given normal vector $b$ have components $b_1, b_2, b_3$
$b = (b_1, b_2, b_3)$
and let $B$ denote the magnitude of $b$.
Here is an algorithm that finds two unit vectors perpendicular to
$b$ and perpendicular to each other:
(1) Let $v = (b_1 + B, b_2, b_3)$ or $v = (b_1 - B, b_2, b_3)$ whichever makes
the first component bigger in absolute value.
(2) Form the three by three matrix $H$
$H = I - 2 v v^T/(v^T v)$
where $I$ is the identity matrix and $T$ denotes the transpose.
The first row of $H$ is a unit vector parallel to $b$.
The other two rows are unit vectors perpendicular to $b$
and perpendicular to each other.
Explicitly, with $V$ set equal to the square of the magnitude of $v$
$V = v_1^2 + v_2^2 + v_3^2$
the first row of $H$ is
$[ 1 - 2 v_1^2/V,; -2 v_1 v_2/V,; -2 v_1 v_3/V ]$
The second row is
$[ -2 v_1 v_2/V,; 1 - 2 v_2^2 /V,; -2 v_2 v_3/V ]$
and the third row
$[ -2 v_1 v_3/V,; -2 v_2 v_3/V,; 1 - 2 v_3^2/V ]$
answered Jan 8 at 17:57
Bob WernerBob Werner
1111
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add a comment |
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From vector $n$, find the largest component, in absolute value. If the others are not zero, set it to $0$, to get $n'$. Otherwise also set one of the other components to $1$. Then orthogonalize $$x=n'-frac{(ncdot n')}{|n|^2}n$$
answered Jan 8 at 15:45
AndreiAndrei
11.8k21026
$endgroup$
$begingroup$
what about corner case of (1,1,1)/sqrt(3)?
$endgroup$
– Mercury
Jan 8 at 16:18
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The idea is to have a different vector. I will change the answer. You can set the largest component to 0
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– Andrei
Jan 8 at 22:30
add a comment |
$begingroup$
1st method : Working with coordinates, let $n=(a,b,c)^T$. One at least of its coordinates is non zero (otherwise it would be the null vector).
Let us assume it is $a neq 0$.
Let the generic vector orthogonal to $n$ be $u=(x,y,z)^T$ with $$ax+by+cz=0,$$
meaning that $x=-(b/a)y-(c/a)z$. Thus the general orthogonal vector has parametric representation :
$$begin{pmatrix}x\y\zend{pmatrix}
=begin{pmatrix}-(b/a)y-(c/a)z\y\zend{pmatrix}
=y begin{pmatrix}-b/a\1\0end{pmatrix}+ z begin{pmatrix}-c/a\0\1end{pmatrix}$$
for any real $y,z$, generating evidently a 2-dimensional space.
2nd method : if you are looking for a "closed form" formula, valid for any $n$, without distinguishing cases, here is one. Consider the skew-symmetric matrix :
$$X_n:=left(begin{array}{rrr}0&-c&b\c&0&-a\-b&a&0end{array}right) text{with} n=(a,b,c)^Ttag{1}$$
then, the normal plane to $n$ is the range space $Range(X_n)$.
Why that ? Because $X_n$ is the matrix naturally associated with the cross product with vector $n$, as proven by the following computation :
$$text{for} v:=(x,y,z)^T, X_n v=left(begin{array}{rrr}0&-c&b\c&0&-a\-b&a&0end{array}right)left(begin{array}{r}x\y\zend{array}right)=underbrace{left(begin{array}{r}bz-cy\cx-az\ay-bxend{array}right)}_{n times v}$$
Remark : operator $X_{n}$ is classical. See for example http://www.blackmesapress.com/CrossProduct.htm.
answered Jan 8 at 15:40
Jean MarieJean Marie
29.4k42050
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I have provided a second solution.
$endgroup$
– Jean Marie
Jan 10 at 18:51
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
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$begingroup$
I’m not sure what you mean by “no edge cases” here. Most methods require at least some case analysis to avoid a degenerate case. Bob Werner’s answer does give a nifty case-free algorithm that looks like it will generalize to higher-dimensional spaces, but it seems like overkill for the problem in $mathbb R^3$.
Given a vector $n=(a,b,c)inmathbb R^3$, its cross products with the standard basis vectors $$ntimes(1,0,0) = (0,c,-b) \ ntimes(0,1,0) = (-c,0,a) \ ntimes(0,0,1) = (b,-a,0)$$ are all orthogonal to $n$. If $n$ is nonzero, then at least two of these are nonzero. If you want all of the orthogonal vectors to $n$, choose two of the above vectors that are linearly independent: their span is the set of vectors orthogonal to $n$, but this is simply the plane through the origin with $n$ as a normal.
answered Jan 8 at 19:57
amdamd
29.7k21050
$endgroup$
$begingroup$
I have written a (2nd) solution which is alongside yours, but by using the (classical) operator $n_{times}$ allowing to avoid to distinguish cases.
$endgroup$
– Jean Marie
Jan 10 at 18:59
$begingroup$
@JeanMarie Isn’t there an implicit case analysis in the computation of the column space of $n_times$?
$endgroup$
– amd
Jan 10 at 19:03
1
$begingroup$
On the computational side, I agree, but on the theoretical side this "formula" permits to express things : for example a vector which is orthogonal to $n_1$ and $n_2$ belongs to $range(X_{n_1})cap range(X_{n_2})$.
$endgroup$
– Jean Marie
Jan 10 at 19:08
$begingroup$
@JeanMarie A fair point. Of course, one can express the same thing without introducing $n_times$ via $span(n)^perp$ (not coincidentally, the first method in your answer).
$endgroup$
– amd
Jan 10 at 20:02
add a comment |
$begingroup$
I’m not sure what you mean by “no edge cases” here. Most methods require at least some case analysis to avoid a degenerate case. Bob Werner’s answer does give a nifty case-free algorithm that looks like it will generalize to higher-dimensional spaces, but it seems like overkill for the problem in $mathbb R^3$.
Given a vector $n=(a,b,c)inmathbb R^3$, its cross products with the standard basis vectors $$ntimes(1,0,0) = (0,c,-b) \ ntimes(0,1,0) = (-c,0,a) \ ntimes(0,0,1) = (b,-a,0)$$ are all orthogonal to $n$. If $n$ is nonzero, then at least two of these are nonzero. If you want all of the orthogonal vectors to $n$, choose two of the above vectors that are linearly independent: their span is the set of vectors orthogonal to $n$, but this is simply the plane through the origin with $n$ as a normal.
answered Jan 8 at 19:57
amdamd
29.7k21050
$endgroup$
$begingroup$
I have written a (2nd) solution which is alongside yours, but by using the (classical) operator $n_{times}$ allowing to avoid to distinguish cases.
$endgroup$
– Jean Marie
Jan 10 at 18:59
$begingroup$
@JeanMarie Isn’t there an implicit case analysis in the computation of the column space of $n_times$?
$endgroup$
– amd
Jan 10 at 19:03
1
$begingroup$
On the computational side, I agree, but on the theoretical side this "formula" permits to express things : for example a vector which is orthogonal to $n_1$ and $n_2$ belongs to $range(X_{n_1})cap range(X_{n_2})$.
$endgroup$
– Jean Marie
Jan 10 at 19:08
$begingroup$
@JeanMarie A fair point. Of course, one can express the same thing without introducing $n_times$ via $span(n)^perp$ (not coincidentally, the first method in your answer).
$endgroup$
– amd
Jan 10 at 20:02
add a comment |
$begingroup$
I’m not sure what you mean by “no edge cases” here. Most methods require at least some case analysis to avoid a degenerate case. Bob Werner’s answer does give a nifty case-free algorithm that looks like it will generalize to higher-dimensional spaces, but it seems like overkill for the problem in $mathbb R^3$.
Given a vector $n=(a,b,c)inmathbb R^3$, its cross products with the standard basis vectors $$ntimes(1,0,0) = (0,c,-b) \ ntimes(0,1,0) = (-c,0,a) \ ntimes(0,0,1) = (b,-a,0)$$ are all orthogonal to $n$. If $n$ is nonzero, then at least two of these are nonzero. If you want all of the orthogonal vectors to $n$, choose two of the above vectors that are linearly independent: their span is the set of vectors orthogonal to $n$, but this is simply the plane through the origin with $n$ as a normal.
answered Jan 8 at 19:57
amdamd
29.7k21050
$endgroup$
I’m not sure what you mean by “no edge cases” here. Most methods require at least some case analysis to avoid a degenerate case. Bob Werner’s answer does give a nifty case-free algorithm that looks like it will generalize to higher-dimensional spaces, but it seems like overkill for the problem in $mathbb R^3$.
Given a vector $n=(a,b,c)inmathbb R^3$, its cross products with the standard basis vectors $$ntimes(1,0,0) = (0,c,-b) \ ntimes(0,1,0) = (-c,0,a) \ ntimes(0,0,1) = (b,-a,0)$$ are all orthogonal to $n$. If $n$ is nonzero, then at least two of these are nonzero. If you want all of the orthogonal vectors to $n$, choose two of the above vectors that are linearly independent: their span is the set of vectors orthogonal to $n$, but this is simply the plane through the origin with $n$ as a normal.
answered Jan 8 at 19:57
amdamd
29.7k21050
answered Jan 8 at 19:57
amdamd
29.7k21050
answered Jan 8 at 19:57
amdamd
29.7k21050
answered Jan 8 at 19:57
amdamd
29.7k21050
29.7k21050
$begingroup$
I have written a (2nd) solution which is alongside yours, but by using the (classical) operator $n_{times}$ allowing to avoid to distinguish cases.
$endgroup$
– Jean Marie
Jan 10 at 18:59
$begingroup$
@JeanMarie Isn’t there an implicit case analysis in the computation of the column space of $n_times$?
$endgroup$
– amd
Jan 10 at 19:03
1
$begingroup$
On the computational side, I agree, but on the theoretical side this "formula" permits to express things : for example a vector which is orthogonal to $n_1$ and $n_2$ belongs to $range(X_{n_1})cap range(X_{n_2})$.
$endgroup$
– Jean Marie
Jan 10 at 19:08
$begingroup$
@JeanMarie A fair point. Of course, one can express the same thing without introducing $n_times$ via $span(n)^perp$ (not coincidentally, the first method in your answer).
$endgroup$
– amd
Jan 10 at 20:02
add a comment |
$begingroup$
I have written a (2nd) solution which is alongside yours, but by using the (classical) operator $n_{times}$ allowing to avoid to distinguish cases.
$endgroup$
– Jean Marie
Jan 10 at 18:59
$begingroup$
@JeanMarie Isn’t there an implicit case analysis in the computation of the column space of $n_times$?
$endgroup$
– amd
Jan 10 at 19:03
1
$begingroup$
On the computational side, I agree, but on the theoretical side this "formula" permits to express things : for example a vector which is orthogonal to $n_1$ and $n_2$ belongs to $range(X_{n_1})cap range(X_{n_2})$.
$endgroup$
– Jean Marie
Jan 10 at 19:08
$begingroup$
@JeanMarie A fair point. Of course, one can express the same thing without introducing $n_times$ via $span(n)^perp$ (not coincidentally, the first method in your answer).
$endgroup$
– amd
Jan 10 at 20:02
$begingroup$
I have written a (2nd) solution which is alongside yours, but by using the (classical) operator $n_{times}$ allowing to avoid to distinguish cases.
$endgroup$
– Jean Marie
Jan 10 at 18:59
$begingroup$
I have written a (2nd) solution which is alongside yours, but by using the (classical) operator $n_{times}$ allowing to avoid to distinguish cases.
$endgroup$
– Jean Marie
Jan 10 at 18:59
$begingroup$
@JeanMarie Isn’t there an implicit case analysis in the computation of the column space of $n_times$?
$endgroup$
– amd
Jan 10 at 19:03
$begingroup$
@JeanMarie Isn’t there an implicit case analysis in the computation of the column space of $n_times$?
$endgroup$
– amd
Jan 10 at 19:03
1
1
$begingroup$
On the computational side, I agree, but on the theoretical side this "formula" permits to express things : for example a vector which is orthogonal to $n_1$ and $n_2$ belongs to $range(X_{n_1})cap range(X_{n_2})$.
$endgroup$
– Jean Marie
Jan 10 at 19:08
$begingroup$
On the computational side, I agree, but on the theoretical side this "formula" permits to express things : for example a vector which is orthogonal to $n_1$ and $n_2$ belongs to $range(X_{n_1})cap range(X_{n_2})$.
$endgroup$
– Jean Marie
Jan 10 at 19:08
$begingroup$
@JeanMarie A fair point. Of course, one can express the same thing without introducing $n_times$ via $span(n)^perp$ (not coincidentally, the first method in your answer).
$endgroup$
– amd
Jan 10 at 20:02
$begingroup$
@JeanMarie A fair point. Of course, one can express the same thing without introducing $n_times$ via $span(n)^perp$ (not coincidentally, the first method in your answer).
$endgroup$
– amd
Jan 10 at 20:02
add a comment |
$begingroup$
Levent Kitis posted a solution on UseNet sci.math. The first component of the vector is altered with the vector's norm, then a symmetric orthogonal Householder matrix is formed using the altered vector. The first row of the Householder matrix parallels the original vector, and the other two rows are perpendicular to it:
Let the given normal vector $b$ have components $b_1, b_2, b_3$
$b = (b_1, b_2, b_3)$
and let $B$ denote the magnitude of $b$.
Here is an algorithm that finds two unit vectors perpendicular to
$b$ and perpendicular to each other:
(1) Let $v = (b_1 + B, b_2, b_3)$ or $v = (b_1 - B, b_2, b_3)$ whichever makes
the first component bigger in absolute value.
(2) Form the three by three matrix $H$
$H = I - 2 v v^T/(v^T v)$
where $I$ is the identity matrix and $T$ denotes the transpose.
The first row of $H$ is a unit vector parallel to $b$.
The other two rows are unit vectors perpendicular to $b$
and perpendicular to each other.
Explicitly, with $V$ set equal to the square of the magnitude of $v$
$V = v_1^2 + v_2^2 + v_3^2$
the first row of $H$ is
$[ 1 - 2 v_1^2/V,; -2 v_1 v_2/V,; -2 v_1 v_3/V ]$
The second row is
$[ -2 v_1 v_2/V,; 1 - 2 v_2^2 /V,; -2 v_2 v_3/V ]$
and the third row
$[ -2 v_1 v_3/V,; -2 v_2 v_3/V,; 1 - 2 v_3^2/V ]$
answered Jan 8 at 17:57
Bob WernerBob Werner
1111
$endgroup$
add a comment |
$begingroup$
Levent Kitis posted a solution on UseNet sci.math. The first component of the vector is altered with the vector's norm, then a symmetric orthogonal Householder matrix is formed using the altered vector. The first row of the Householder matrix parallels the original vector, and the other two rows are perpendicular to it:
Let the given normal vector $b$ have components $b_1, b_2, b_3$
$b = (b_1, b_2, b_3)$
and let $B$ denote the magnitude of $b$.
Here is an algorithm that finds two unit vectors perpendicular to
$b$ and perpendicular to each other:
(1) Let $v = (b_1 + B, b_2, b_3)$ or $v = (b_1 - B, b_2, b_3)$ whichever makes
the first component bigger in absolute value.
(2) Form the three by three matrix $H$
$H = I - 2 v v^T/(v^T v)$
where $I$ is the identity matrix and $T$ denotes the transpose.
The first row of $H$ is a unit vector parallel to $b$.
The other two rows are unit vectors perpendicular to $b$
and perpendicular to each other.
Explicitly, with $V$ set equal to the square of the magnitude of $v$
$V = v_1^2 + v_2^2 + v_3^2$
the first row of $H$ is
$[ 1 - 2 v_1^2/V,; -2 v_1 v_2/V,; -2 v_1 v_3/V ]$
The second row is
$[ -2 v_1 v_2/V,; 1 - 2 v_2^2 /V,; -2 v_2 v_3/V ]$
and the third row
$[ -2 v_1 v_3/V,; -2 v_2 v_3/V,; 1 - 2 v_3^2/V ]$
answered Jan 8 at 17:57
Bob WernerBob Werner
1111
$endgroup$
add a comment |
$begingroup$
Levent Kitis posted a solution on UseNet sci.math. The first component of the vector is altered with the vector's norm, then a symmetric orthogonal Householder matrix is formed using the altered vector. The first row of the Householder matrix parallels the original vector, and the other two rows are perpendicular to it:
Let the given normal vector $b$ have components $b_1, b_2, b_3$
$b = (b_1, b_2, b_3)$
and let $B$ denote the magnitude of $b$.
Here is an algorithm that finds two unit vectors perpendicular to
$b$ and perpendicular to each other:
(1) Let $v = (b_1 + B, b_2, b_3)$ or $v = (b_1 - B, b_2, b_3)$ whichever makes
the first component bigger in absolute value.
(2) Form the three by three matrix $H$
$H = I - 2 v v^T/(v^T v)$
where $I$ is the identity matrix and $T$ denotes the transpose.
The first row of $H$ is a unit vector parallel to $b$.
The other two rows are unit vectors perpendicular to $b$
and perpendicular to each other.
Explicitly, with $V$ set equal to the square of the magnitude of $v$
$V = v_1^2 + v_2^2 + v_3^2$
the first row of $H$ is
$[ 1 - 2 v_1^2/V,; -2 v_1 v_2/V,; -2 v_1 v_3/V ]$
The second row is
$[ -2 v_1 v_2/V,; 1 - 2 v_2^2 /V,; -2 v_2 v_3/V ]$
and the third row
$[ -2 v_1 v_3/V,; -2 v_2 v_3/V,; 1 - 2 v_3^2/V ]$
answered Jan 8 at 17:57
Bob WernerBob Werner
1111
$endgroup$
Levent Kitis posted a solution on UseNet sci.math. The first component of the vector is altered with the vector's norm, then a symmetric orthogonal Householder matrix is formed using the altered vector. The first row of the Householder matrix parallels the original vector, and the other two rows are perpendicular to it:
Let the given normal vector $b$ have components $b_1, b_2, b_3$
$b = (b_1, b_2, b_3)$
and let $B$ denote the magnitude of $b$.
Here is an algorithm that finds two unit vectors perpendicular to
$b$ and perpendicular to each other:
(1) Let $v = (b_1 + B, b_2, b_3)$ or $v = (b_1 - B, b_2, b_3)$ whichever makes
the first component bigger in absolute value.
(2) Form the three by three matrix $H$
$H = I - 2 v v^T/(v^T v)$
where $I$ is the identity matrix and $T$ denotes the transpose.
The first row of $H$ is a unit vector parallel to $b$.
The other two rows are unit vectors perpendicular to $b$
and perpendicular to each other.
Explicitly, with $V$ set equal to the square of the magnitude of $v$
$V = v_1^2 + v_2^2 + v_3^2$
the first row of $H$ is
$[ 1 - 2 v_1^2/V,; -2 v_1 v_2/V,; -2 v_1 v_3/V ]$
The second row is
$[ -2 v_1 v_2/V,; 1 - 2 v_2^2 /V,; -2 v_2 v_3/V ]$
and the third row
$[ -2 v_1 v_3/V,; -2 v_2 v_3/V,; 1 - 2 v_3^2/V ]$
answered Jan 8 at 17:57
Bob WernerBob Werner
1111
answered Jan 8 at 17:57
Bob WernerBob Werner
1111
answered Jan 8 at 17:57
Bob WernerBob Werner
1111
answered Jan 8 at 17:57
Bob WernerBob Werner
1111
1111
add a comment |
add a comment |
$begingroup$
From vector $n$, find the largest component, in absolute value. If the others are not zero, set it to $0$, to get $n'$. Otherwise also set one of the other components to $1$. Then orthogonalize $$x=n'-frac{(ncdot n')}{|n|^2}n$$
answered Jan 8 at 15:45
AndreiAndrei
11.8k21026
$endgroup$
$begingroup$
what about corner case of (1,1,1)/sqrt(3)?
$endgroup$
– Mercury
Jan 8 at 16:18
$begingroup$
The idea is to have a different vector. I will change the answer. You can set the largest component to 0
$endgroup$
– Andrei
Jan 8 at 22:30
add a comment |
$begingroup$
From vector $n$, find the largest component, in absolute value. If the others are not zero, set it to $0$, to get $n'$. Otherwise also set one of the other components to $1$. Then orthogonalize $$x=n'-frac{(ncdot n')}{|n|^2}n$$
answered Jan 8 at 15:45
AndreiAndrei
11.8k21026
$endgroup$
$begingroup$
what about corner case of (1,1,1)/sqrt(3)?
$endgroup$
– Mercury
Jan 8 at 16:18
$begingroup$
The idea is to have a different vector. I will change the answer. You can set the largest component to 0
$endgroup$
– Andrei
Jan 8 at 22:30
add a comment |
$begingroup$
From vector $n$, find the largest component, in absolute value. If the others are not zero, set it to $0$, to get $n'$. Otherwise also set one of the other components to $1$. Then orthogonalize $$x=n'-frac{(ncdot n')}{|n|^2}n$$
answered Jan 8 at 15:45
AndreiAndrei
11.8k21026
$endgroup$
From vector $n$, find the largest component, in absolute value. If the others are not zero, set it to $0$, to get $n'$. Otherwise also set one of the other components to $1$. Then orthogonalize $$x=n'-frac{(ncdot n')}{|n|^2}n$$
answered Jan 8 at 15:45
AndreiAndrei
11.8k21026
edited Jan 8 at 22:30
answered Jan 8 at 15:45
AndreiAndrei
11.8k21026
answered Jan 8 at 15:45
AndreiAndrei
11.8k21026
answered Jan 8 at 15:45
AndreiAndrei
11.8k21026
11.8k21026
$begingroup$
what about corner case of (1,1,1)/sqrt(3)?
$endgroup$
– Mercury
Jan 8 at 16:18
$begingroup$
The idea is to have a different vector. I will change the answer. You can set the largest component to 0
$endgroup$
– Andrei
Jan 8 at 22:30
add a comment |
$begingroup$
what about corner case of (1,1,1)/sqrt(3)?
$endgroup$
– Mercury
Jan 8 at 16:18
$begingroup$
The idea is to have a different vector. I will change the answer. You can set the largest component to 0
$endgroup$
– Andrei
Jan 8 at 22:30
$begingroup$
what about corner case of (1,1,1)/sqrt(3)?
$endgroup$
– Mercury
Jan 8 at 16:18
$begingroup$
what about corner case of (1,1,1)/sqrt(3)?
$endgroup$
– Mercury
Jan 8 at 16:18
$begingroup$
The idea is to have a different vector. I will change the answer. You can set the largest component to 0
$endgroup$
– Andrei
Jan 8 at 22:30
$begingroup$
The idea is to have a different vector. I will change the answer. You can set the largest component to 0
$endgroup$
– Andrei
Jan 8 at 22:30
add a comment |
$begingroup$
1st method : Working with coordinates, let $n=(a,b,c)^T$. One at least of its coordinates is non zero (otherwise it would be the null vector).
Let us assume it is $a neq 0$.
Let the generic vector orthogonal to $n$ be $u=(x,y,z)^T$ with $$ax+by+cz=0,$$
meaning that $x=-(b/a)y-(c/a)z$. Thus the general orthogonal vector has parametric representation :
$$begin{pmatrix}x\y\zend{pmatrix}
=begin{pmatrix}-(b/a)y-(c/a)z\y\zend{pmatrix}
=y begin{pmatrix}-b/a\1\0end{pmatrix}+ z begin{pmatrix}-c/a\0\1end{pmatrix}$$
for any real $y,z$, generating evidently a 2-dimensional space.
2nd method : if you are looking for a "closed form" formula, valid for any $n$, without distinguishing cases, here is one. Consider the skew-symmetric matrix :
$$X_n:=left(begin{array}{rrr}0&-c&b\c&0&-a\-b&a&0end{array}right) text{with} n=(a,b,c)^Ttag{1}$$
then, the normal plane to $n$ is the range space $Range(X_n)$.
Why that ? Because $X_n$ is the matrix naturally associated with the cross product with vector $n$, as proven by the following computation :
$$text{for} v:=(x,y,z)^T, X_n v=left(begin{array}{rrr}0&-c&b\c&0&-a\-b&a&0end{array}right)left(begin{array}{r}x\y\zend{array}right)=underbrace{left(begin{array}{r}bz-cy\cx-az\ay-bxend{array}right)}_{n times v}$$
Remark : operator $X_{n}$ is classical. See for example http://www.blackmesapress.com/CrossProduct.htm.
answered Jan 8 at 15:40
Jean MarieJean Marie
29.4k42050
$endgroup$
$begingroup$
I have provided a second solution.
$endgroup$
– Jean Marie
Jan 10 at 18:51
add a comment |
$begingroup$
1st method : Working with coordinates, let $n=(a,b,c)^T$. One at least of its coordinates is non zero (otherwise it would be the null vector).
Let us assume it is $a neq 0$.
Let the generic vector orthogonal to $n$ be $u=(x,y,z)^T$ with $$ax+by+cz=0,$$
meaning that $x=-(b/a)y-(c/a)z$. Thus the general orthogonal vector has parametric representation :
$$begin{pmatrix}x\y\zend{pmatrix}
=begin{pmatrix}-(b/a)y-(c/a)z\y\zend{pmatrix}
=y begin{pmatrix}-b/a\1\0end{pmatrix}+ z begin{pmatrix}-c/a\0\1end{pmatrix}$$
for any real $y,z$, generating evidently a 2-dimensional space.
2nd method : if you are looking for a "closed form" formula, valid for any $n$, without distinguishing cases, here is one. Consider the skew-symmetric matrix :
$$X_n:=left(begin{array}{rrr}0&-c&b\c&0&-a\-b&a&0end{array}right) text{with} n=(a,b,c)^Ttag{1}$$
then, the normal plane to $n$ is the range space $Range(X_n)$.
Why that ? Because $X_n$ is the matrix naturally associated with the cross product with vector $n$, as proven by the following computation :
$$text{for} v:=(x,y,z)^T, X_n v=left(begin{array}{rrr}0&-c&b\c&0&-a\-b&a&0end{array}right)left(begin{array}{r}x\y\zend{array}right)=underbrace{left(begin{array}{r}bz-cy\cx-az\ay-bxend{array}right)}_{n times v}$$
Remark : operator $X_{n}$ is classical. See for example http://www.blackmesapress.com/CrossProduct.htm.
answered Jan 8 at 15:40
Jean MarieJean Marie
29.4k42050
$endgroup$
$begingroup$
I have provided a second solution.
$endgroup$
– Jean Marie
Jan 10 at 18:51
add a comment |
$begingroup$
1st method : Working with coordinates, let $n=(a,b,c)^T$. One at least of its coordinates is non zero (otherwise it would be the null vector).
Let us assume it is $a neq 0$.
Let the generic vector orthogonal to $n$ be $u=(x,y,z)^T$ with $$ax+by+cz=0,$$
meaning that $x=-(b/a)y-(c/a)z$. Thus the general orthogonal vector has parametric representation :
$$begin{pmatrix}x\y\zend{pmatrix}
=begin{pmatrix}-(b/a)y-(c/a)z\y\zend{pmatrix}
=y begin{pmatrix}-b/a\1\0end{pmatrix}+ z begin{pmatrix}-c/a\0\1end{pmatrix}$$
for any real $y,z$, generating evidently a 2-dimensional space.
2nd method : if you are looking for a "closed form" formula, valid for any $n$, without distinguishing cases, here is one. Consider the skew-symmetric matrix :
$$X_n:=left(begin{array}{rrr}0&-c&b\c&0&-a\-b&a&0end{array}right) text{with} n=(a,b,c)^Ttag{1}$$
then, the normal plane to $n$ is the range space $Range(X_n)$.
Why that ? Because $X_n$ is the matrix naturally associated with the cross product with vector $n$, as proven by the following computation :
$$text{for} v:=(x,y,z)^T, X_n v=left(begin{array}{rrr}0&-c&b\c&0&-a\-b&a&0end{array}right)left(begin{array}{r}x\y\zend{array}right)=underbrace{left(begin{array}{r}bz-cy\cx-az\ay-bxend{array}right)}_{n times v}$$
Remark : operator $X_{n}$ is classical. See for example http://www.blackmesapress.com/CrossProduct.htm.
answered Jan 8 at 15:40
Jean MarieJean Marie
29.4k42050
$endgroup$
1st method : Working with coordinates, let $n=(a,b,c)^T$. One at least of its coordinates is non zero (otherwise it would be the null vector).
Let us assume it is $a neq 0$.
Let the generic vector orthogonal to $n$ be $u=(x,y,z)^T$ with $$ax+by+cz=0,$$
meaning that $x=-(b/a)y-(c/a)z$. Thus the general orthogonal vector has parametric representation :
$$begin{pmatrix}x\y\zend{pmatrix}
=begin{pmatrix}-(b/a)y-(c/a)z\y\zend{pmatrix}
=y begin{pmatrix}-b/a\1\0end{pmatrix}+ z begin{pmatrix}-c/a\0\1end{pmatrix}$$
for any real $y,z$, generating evidently a 2-dimensional space.
2nd method : if you are looking for a "closed form" formula, valid for any $n$, without distinguishing cases, here is one. Consider the skew-symmetric matrix :
$$X_n:=left(begin{array}{rrr}0&-c&b\c&0&-a\-b&a&0end{array}right) text{with} n=(a,b,c)^Ttag{1}$$
then, the normal plane to $n$ is the range space $Range(X_n)$.
Why that ? Because $X_n$ is the matrix naturally associated with the cross product with vector $n$, as proven by the following computation :
$$text{for} v:=(x,y,z)^T, X_n v=left(begin{array}{rrr}0&-c&b\c&0&-a\-b&a&0end{array}right)left(begin{array}{r}x\y\zend{array}right)=underbrace{left(begin{array}{r}bz-cy\cx-az\ay-bxend{array}right)}_{n times v}$$
Remark : operator $X_{n}$ is classical. See for example http://www.blackmesapress.com/CrossProduct.htm.
answered Jan 8 at 15:40
Jean MarieJean Marie
29.4k42050
edited Jan 10 at 19:09
answered Jan 8 at 15:40
Jean MarieJean Marie
29.4k42050
answered Jan 8 at 15:40
Jean MarieJean Marie
29.4k42050
answered Jan 8 at 15:40
Jean MarieJean Marie
29.4k42050
29.4k42050
$begingroup$
I have provided a second solution.
$endgroup$
– Jean Marie
Jan 10 at 18:51
add a comment |
$begingroup$
I have provided a second solution.
$endgroup$
– Jean Marie
Jan 10 at 18:51
$begingroup$
I have provided a second solution.
$endgroup$
– Jean Marie
Jan 10 at 18:51
$begingroup$
I have provided a second solution.
$endgroup$
– Jean Marie
Jan 10 at 18:51
add a comment |
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$begingroup$
$n^Tx = 0$ is a linear equation in $x$, so use your preferred method for solving linear equations.
$endgroup$
– user3482749
Jan 8 at 15:23