Two statements regarding orthogonal unit vectors and orthogonal complements respectively
1
$begingroup$
I am new to linear algebra, and I was confused regarding the following question. I would really appreciate it, if anybody could give some feedback...
True or False?
$left(frac{1}{sqrt14},frac{-2}{sqrt14},frac{3}{sqrt14}right)$ is an orthogonal unit vector for $Sp{(1,-1,-1),(2,7,4)}$.
If ${U={(x,y,z) in R^3|x+2y+3z=0}}$, then $U^bot=Sp{(2,4,6)}$.
For 1, I tried the Gram Schmidt process, which did not lead me to $left(frac{1}{sqrt14},frac{-2}{sqrt14},frac{3}{sqrt14}right)$, so my interim conclusion is that it is wrong.
For 2, I reasoned that
$x=-2y-3z$, so we can rewrite the vector as $(-2y-3z,y,z)$, and indeed $(-2y-3z)·2+4y+6z=0$, so the answer is "true".
Thank you!
linear-algebra vectors orthogonality
asked Jan 18 at 8:35
daltadalta
1578
$endgroup$
add a comment |
1
$begingroup$
I am new to linear algebra, and I was confused regarding the following question. I would really appreciate it, if anybody could give some feedback...
True or False?
$left(frac{1}{sqrt14},frac{-2}{sqrt14},frac{3}{sqrt14}right)$ is an orthogonal unit vector for $Sp{(1,-1,-1),(2,7,4)}$.
If ${U={(x,y,z) in R^3|x+2y+3z=0}}$, then $U^bot=Sp{(2,4,6)}$.
For 1, I tried the Gram Schmidt process, which did not lead me to $left(frac{1}{sqrt14},frac{-2}{sqrt14},frac{3}{sqrt14}right)$, so my interim conclusion is that it is wrong.
For 2, I reasoned that
$x=-2y-3z$, so we can rewrite the vector as $(-2y-3z,y,z)$, and indeed $(-2y-3z)·2+4y+6z=0$, so the answer is "true".
Thank you!
linear-algebra vectors orthogonality
asked Jan 18 at 8:35
daltadalta
1578
$endgroup$
$begingroup$
gram schmidt only constructs A orthogonal unit vector, there might be more (in particular at least 2), So I would recommend you check orthogonality by scalar product and that the norm is $1$, that is a clearer argument, and checks literally the definition, For the second, I recommend you construct a basis of $U$, and then check the same way if $begin{pmatrix}2\4\6\ end{pmatrix}$ is orthogonal to it, (but it looks very much like it! as the orthogonal vector of your plane is already hidden in the equation.)
$endgroup$
– Enkidu
Jan 18 at 8:40
add a comment |
1
1
1
$begingroup$
I am new to linear algebra, and I was confused regarding the following question. I would really appreciate it, if anybody could give some feedback...
True or False?
$left(frac{1}{sqrt14},frac{-2}{sqrt14},frac{3}{sqrt14}right)$ is an orthogonal unit vector for $Sp{(1,-1,-1),(2,7,4)}$.
If ${U={(x,y,z) in R^3|x+2y+3z=0}}$, then $U^bot=Sp{(2,4,6)}$.
For 1, I tried the Gram Schmidt process, which did not lead me to $left(frac{1}{sqrt14},frac{-2}{sqrt14},frac{3}{sqrt14}right)$, so my interim conclusion is that it is wrong.
For 2, I reasoned that
$x=-2y-3z$, so we can rewrite the vector as $(-2y-3z,y,z)$, and indeed $(-2y-3z)·2+4y+6z=0$, so the answer is "true".
Thank you!
linear-algebra vectors orthogonality
asked Jan 18 at 8:35
daltadalta
1578
$endgroup$
I am new to linear algebra, and I was confused regarding the following question. I would really appreciate it, if anybody could give some feedback...
True or False?
$left(frac{1}{sqrt14},frac{-2}{sqrt14},frac{3}{sqrt14}right)$ is an orthogonal unit vector for $Sp{(1,-1,-1),(2,7,4)}$.
If ${U={(x,y,z) in R^3|x+2y+3z=0}}$, then $U^bot=Sp{(2,4,6)}$.
For 1, I tried the Gram Schmidt process, which did not lead me to $left(frac{1}{sqrt14},frac{-2}{sqrt14},frac{3}{sqrt14}right)$, so my interim conclusion is that it is wrong.
For 2, I reasoned that
$x=-2y-3z$, so we can rewrite the vector as $(-2y-3z,y,z)$, and indeed $(-2y-3z)·2+4y+6z=0$, so the answer is "true".
Thank you!
linear-algebra vectors orthogonality
linear-algebra vectors orthogonality
asked Jan 18 at 8:35
daltadalta
1578
asked Jan 18 at 8:35
daltadalta
1578
edited Jan 18 at 9:02
dalta
asked Jan 18 at 8:35
daltadalta
1578
asked Jan 18 at 8:35
daltadalta
1578
asked Jan 18 at 8:35
daltadalta
1578
1578
$begingroup$
gram schmidt only constructs A orthogonal unit vector, there might be more (in particular at least 2), So I would recommend you check orthogonality by scalar product and that the norm is $1$, that is a clearer argument, and checks literally the definition, For the second, I recommend you construct a basis of $U$, and then check the same way if $begin{pmatrix}2\4\6\ end{pmatrix}$ is orthogonal to it, (but it looks very much like it! as the orthogonal vector of your plane is already hidden in the equation.)
$endgroup$
– Enkidu
Jan 18 at 8:40
add a comment |
$begingroup$
gram schmidt only constructs A orthogonal unit vector, there might be more (in particular at least 2), So I would recommend you check orthogonality by scalar product and that the norm is $1$, that is a clearer argument, and checks literally the definition, For the second, I recommend you construct a basis of $U$, and then check the same way if $begin{pmatrix}2\4\6\ end{pmatrix}$ is orthogonal to it, (but it looks very much like it! as the orthogonal vector of your plane is already hidden in the equation.)
$endgroup$
– Enkidu
Jan 18 at 8:40
$begingroup$
gram schmidt only constructs A orthogonal unit vector, there might be more (in particular at least 2), So I would recommend you check orthogonality by scalar product and that the norm is $1$, that is a clearer argument, and checks literally the definition, For the second, I recommend you construct a basis of $U$, and then check the same way if $begin{pmatrix}2\4\6\ end{pmatrix}$ is orthogonal to it, (but it looks very much like it! as the orthogonal vector of your plane is already hidden in the equation.)
$endgroup$
– Enkidu
Jan 18 at 8:40
$begingroup$
gram schmidt only constructs A orthogonal unit vector, there might be more (in particular at least 2), So I would recommend you check orthogonality by scalar product and that the norm is $1$, that is a clearer argument, and checks literally the definition, For the second, I recommend you construct a basis of $U$, and then check the same way if $begin{pmatrix}2\4\6\ end{pmatrix}$ is orthogonal to it, (but it looks very much like it! as the orthogonal vector of your plane is already hidden in the equation.)
$endgroup$
– Enkidu
Jan 18 at 8:40
add a comment |
1 Answer
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1
$begingroup$
- The given vector is orthogonal to $(1,-1,-1)$ as well as $(2,7,4)$ and hence it is orthogonal to the span of these two. There is no need to use Gram Schmidt.
- Use some geometry. The equation $x+2y+3z=0$ represents a plane for which $(1,2,3)$ is a normal vector. Since the orthogonal complement of a plane is one dimensional $U^{perp}$ must be $span {(1,2,3)}=span {(2,4,6)}$. Hence 2) is true.
answered Jan 18 at 8:45
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
$endgroup$
$begingroup$
Thank you! The thing that confuses me is that $1·frac{1}{sqrt14} -1·frac{-2}{sqrt14} -1·frac{3}{sqrt14}=0$, and the same for (2,7,4), so why is it not orthogonal?
$endgroup$
– dalta
Jan 18 at 8:57
$begingroup$
@dalta Please check 1) carefully. There is a typo in it and my answer is based on what you have typed.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:02
$begingroup$
Oh wow, I am sorry! I just edited accordingly...
$endgroup$
– dalta
Jan 18 at 9:03
$begingroup$
so the answer to 1 is 'yes' then, right?
$endgroup$
– dalta
Jan 18 at 9:04
$begingroup$
@dalta I have edited the answer.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:12
add a comment |
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1
$begingroup$
- The given vector is orthogonal to $(1,-1,-1)$ as well as $(2,7,4)$ and hence it is orthogonal to the span of these two. There is no need to use Gram Schmidt.
- Use some geometry. The equation $x+2y+3z=0$ represents a plane for which $(1,2,3)$ is a normal vector. Since the orthogonal complement of a plane is one dimensional $U^{perp}$ must be $span {(1,2,3)}=span {(2,4,6)}$. Hence 2) is true.
answered Jan 18 at 8:45
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
$endgroup$
$begingroup$
Thank you! The thing that confuses me is that $1·frac{1}{sqrt14} -1·frac{-2}{sqrt14} -1·frac{3}{sqrt14}=0$, and the same for (2,7,4), so why is it not orthogonal?
$endgroup$
– dalta
Jan 18 at 8:57
$begingroup$
@dalta Please check 1) carefully. There is a typo in it and my answer is based on what you have typed.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:02
$begingroup$
Oh wow, I am sorry! I just edited accordingly...
$endgroup$
– dalta
Jan 18 at 9:03
$begingroup$
so the answer to 1 is 'yes' then, right?
$endgroup$
– dalta
Jan 18 at 9:04
$begingroup$
@dalta I have edited the answer.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:12
add a comment |
1
$begingroup$
- The given vector is orthogonal to $(1,-1,-1)$ as well as $(2,7,4)$ and hence it is orthogonal to the span of these two. There is no need to use Gram Schmidt.
- Use some geometry. The equation $x+2y+3z=0$ represents a plane for which $(1,2,3)$ is a normal vector. Since the orthogonal complement of a plane is one dimensional $U^{perp}$ must be $span {(1,2,3)}=span {(2,4,6)}$. Hence 2) is true.
answered Jan 18 at 8:45
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
$endgroup$
$begingroup$
Thank you! The thing that confuses me is that $1·frac{1}{sqrt14} -1·frac{-2}{sqrt14} -1·frac{3}{sqrt14}=0$, and the same for (2,7,4), so why is it not orthogonal?
$endgroup$
– dalta
Jan 18 at 8:57
$begingroup$
@dalta Please check 1) carefully. There is a typo in it and my answer is based on what you have typed.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:02
$begingroup$
Oh wow, I am sorry! I just edited accordingly...
$endgroup$
– dalta
Jan 18 at 9:03
$begingroup$
so the answer to 1 is 'yes' then, right?
$endgroup$
– dalta
Jan 18 at 9:04
$begingroup$
@dalta I have edited the answer.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:12
add a comment |
1
1
1
$begingroup$
- The given vector is orthogonal to $(1,-1,-1)$ as well as $(2,7,4)$ and hence it is orthogonal to the span of these two. There is no need to use Gram Schmidt.
- Use some geometry. The equation $x+2y+3z=0$ represents a plane for which $(1,2,3)$ is a normal vector. Since the orthogonal complement of a plane is one dimensional $U^{perp}$ must be $span {(1,2,3)}=span {(2,4,6)}$. Hence 2) is true.
answered Jan 18 at 8:45
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
$endgroup$
- The given vector is orthogonal to $(1,-1,-1)$ as well as $(2,7,4)$ and hence it is orthogonal to the span of these two. There is no need to use Gram Schmidt.
- Use some geometry. The equation $x+2y+3z=0$ represents a plane for which $(1,2,3)$ is a normal vector. Since the orthogonal complement of a plane is one dimensional $U^{perp}$ must be $span {(1,2,3)}=span {(2,4,6)}$. Hence 2) is true.
answered Jan 18 at 8:45
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
edited Jan 18 at 9:11
answered Jan 18 at 8:45
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
answered Jan 18 at 8:45
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
answered Jan 18 at 8:45
Kavi Rama MurthyKavi Rama Murthy
75.5k53270
75.5k53270
$begingroup$
Thank you! The thing that confuses me is that $1·frac{1}{sqrt14} -1·frac{-2}{sqrt14} -1·frac{3}{sqrt14}=0$, and the same for (2,7,4), so why is it not orthogonal?
$endgroup$
– dalta
Jan 18 at 8:57
$begingroup$
@dalta Please check 1) carefully. There is a typo in it and my answer is based on what you have typed.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:02
$begingroup$
Oh wow, I am sorry! I just edited accordingly...
$endgroup$
– dalta
Jan 18 at 9:03
$begingroup$
so the answer to 1 is 'yes' then, right?
$endgroup$
– dalta
Jan 18 at 9:04
$begingroup$
@dalta I have edited the answer.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:12
add a comment |
$begingroup$
Thank you! The thing that confuses me is that $1·frac{1}{sqrt14} -1·frac{-2}{sqrt14} -1·frac{3}{sqrt14}=0$, and the same for (2,7,4), so why is it not orthogonal?
$endgroup$
– dalta
Jan 18 at 8:57
$begingroup$
@dalta Please check 1) carefully. There is a typo in it and my answer is based on what you have typed.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:02
$begingroup$
Oh wow, I am sorry! I just edited accordingly...
$endgroup$
– dalta
Jan 18 at 9:03
$begingroup$
so the answer to 1 is 'yes' then, right?
$endgroup$
– dalta
Jan 18 at 9:04
$begingroup$
@dalta I have edited the answer.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:12
$begingroup$
Thank you! The thing that confuses me is that $1·frac{1}{sqrt14} -1·frac{-2}{sqrt14} -1·frac{3}{sqrt14}=0$, and the same for (2,7,4), so why is it not orthogonal?
$endgroup$
– dalta
Jan 18 at 8:57
$begingroup$
Thank you! The thing that confuses me is that $1·frac{1}{sqrt14} -1·frac{-2}{sqrt14} -1·frac{3}{sqrt14}=0$, and the same for (2,7,4), so why is it not orthogonal?
$endgroup$
– dalta
Jan 18 at 8:57
$begingroup$
@dalta Please check 1) carefully. There is a typo in it and my answer is based on what you have typed.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:02
$begingroup$
@dalta Please check 1) carefully. There is a typo in it and my answer is based on what you have typed.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:02
$begingroup$
Oh wow, I am sorry! I just edited accordingly...
$endgroup$
– dalta
Jan 18 at 9:03
$begingroup$
Oh wow, I am sorry! I just edited accordingly...
$endgroup$
– dalta
Jan 18 at 9:03
$begingroup$
so the answer to 1 is 'yes' then, right?
$endgroup$
– dalta
Jan 18 at 9:04
$begingroup$
so the answer to 1 is 'yes' then, right?
$endgroup$
– dalta
Jan 18 at 9:04
$begingroup$
@dalta I have edited the answer.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:12
$begingroup$
@dalta I have edited the answer.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:12
add a comment |
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$begingroup$
gram schmidt only constructs A orthogonal unit vector, there might be more (in particular at least 2), So I would recommend you check orthogonality by scalar product and that the norm is $1$, that is a clearer argument, and checks literally the definition, For the second, I recommend you construct a basis of $U$, and then check the same way if $begin{pmatrix}2\4\6\ end{pmatrix}$ is orthogonal to it, (but it looks very much like it! as the orthogonal vector of your plane is already hidden in the equation.)
$endgroup$
– Enkidu
Jan 18 at 8:40