Mixing
Find the distance from $(1, 1, 1)$ to the subspace span ${(1, 1, 0), (0, 1, 1)}$ of $mathbb R^3$.
$begingroup$
Consider $mathbb R
^3$ with the usual inner product. If d is the distance from $Q=(1, 1, 1)$ to the subspace $P$
span ${(1, 1, 0), (0, 1, 1)}$of $mathbb R
^3$
, then $ 3d
^2 $ = ?
Can anyone give me a hint? I can not understand what to do with the inner product. I know that I will get an orthogonal vector of $(1, 1, 1)$ in the subspace span ${(1, 1, 0), (0, 1, 1)}$.
linear-algebra vector-spaces inner-product-space
asked Jan 25 at 6:58
cmicmi
1,136312
$endgroup$
add a comment |
$begingroup$
Consider $mathbb R
^3$ with the usual inner product. If d is the distance from $Q=(1, 1, 1)$ to the subspace $P$
span ${(1, 1, 0), (0, 1, 1)}$of $mathbb R
^3$
, then $ 3d
^2 $ = ?
Can anyone give me a hint? I can not understand what to do with the inner product. I know that I will get an orthogonal vector of $(1, 1, 1)$ in the subspace span ${(1, 1, 0), (0, 1, 1)}$.
linear-algebra vector-spaces inner-product-space
asked Jan 25 at 6:58
cmicmi
1,136312
$endgroup$
2
$begingroup$
Your subspace is a plane inside the $mathbb{R}^3$. Actually you want to calculate the distance between a point and a plane. Does this help?
$endgroup$
– Sqyuli
Jan 25 at 7:07
1
$begingroup$
This (math.stackexchange.com/questions/2101140/…) might help!
$endgroup$
– Chinnapparaj R
Jan 25 at 7:13
$begingroup$
@Sqyuli Then I have to find the vector in that subspace which is orthogonal to (1,1,1)? Is not it?
$endgroup$
– cmi
Jan 25 at 7:18
add a comment |
$begingroup$
Consider $mathbb R
^3$ with the usual inner product. If d is the distance from $Q=(1, 1, 1)$ to the subspace $P$
span ${(1, 1, 0), (0, 1, 1)}$of $mathbb R
^3$
, then $ 3d
^2 $ = ?
Can anyone give me a hint? I can not understand what to do with the inner product. I know that I will get an orthogonal vector of $(1, 1, 1)$ in the subspace span ${(1, 1, 0), (0, 1, 1)}$.
linear-algebra vector-spaces inner-product-space
asked Jan 25 at 6:58
cmicmi
1,136312
$endgroup$
Consider $mathbb R
^3$ with the usual inner product. If d is the distance from $Q=(1, 1, 1)$ to the subspace $P$
span ${(1, 1, 0), (0, 1, 1)}$of $mathbb R
^3$
, then $ 3d
^2 $ = ?
Can anyone give me a hint? I can not understand what to do with the inner product. I know that I will get an orthogonal vector of $(1, 1, 1)$ in the subspace span ${(1, 1, 0), (0, 1, 1)}$.
linear-algebra vector-spaces inner-product-space
linear-algebra vector-spaces inner-product-space
asked Jan 25 at 6:58
cmicmi
1,136312
asked Jan 25 at 6:58
cmicmi
1,136312
edited Jan 25 at 13:45
cmi
asked Jan 25 at 6:58
cmicmi
1,136312
asked Jan 25 at 6:58
cmicmi
1,136312
asked Jan 25 at 6:58
cmicmi
1,136312
1,136312
2
$begingroup$
Your subspace is a plane inside the $mathbb{R}^3$. Actually you want to calculate the distance between a point and a plane. Does this help?
$endgroup$
– Sqyuli
Jan 25 at 7:07
1
$begingroup$
This (math.stackexchange.com/questions/2101140/…) might help!
$endgroup$
– Chinnapparaj R
Jan 25 at 7:13
$begingroup$
@Sqyuli Then I have to find the vector in that subspace which is orthogonal to (1,1,1)? Is not it?
$endgroup$
– cmi
Jan 25 at 7:18
add a comment |
2
$begingroup$
Your subspace is a plane inside the $mathbb{R}^3$. Actually you want to calculate the distance between a point and a plane. Does this help?
$endgroup$
– Sqyuli
Jan 25 at 7:07
1
$begingroup$
This (math.stackexchange.com/questions/2101140/…) might help!
$endgroup$
– Chinnapparaj R
Jan 25 at 7:13
$begingroup$
@Sqyuli Then I have to find the vector in that subspace which is orthogonal to (1,1,1)? Is not it?
$endgroup$
– cmi
Jan 25 at 7:18
2
2
$begingroup$
Your subspace is a plane inside the $mathbb{R}^3$. Actually you want to calculate the distance between a point and a plane. Does this help?
$endgroup$
– Sqyuli
Jan 25 at 7:07
$begingroup$
Your subspace is a plane inside the $mathbb{R}^3$. Actually you want to calculate the distance between a point and a plane. Does this help?
$endgroup$
– Sqyuli
Jan 25 at 7:07
1
1
$begingroup$
This (math.stackexchange.com/questions/2101140/…) might help!
$endgroup$
– Chinnapparaj R
Jan 25 at 7:13
$begingroup$
This (math.stackexchange.com/questions/2101140/…) might help!
$endgroup$
– Chinnapparaj R
Jan 25 at 7:13
$begingroup$
@Sqyuli Then I have to find the vector in that subspace which is orthogonal to (1,1,1)? Is not it?
$endgroup$
– cmi
Jan 25 at 7:18
$begingroup$
@Sqyuli Then I have to find the vector in that subspace which is orthogonal to (1,1,1)? Is not it?
$endgroup$
– cmi
Jan 25 at 7:18
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that the vector $(1, -1, 1)$ is orthogonal to both of these vectors. Now
$$
(1, 1, 1) = frac{2}{3}(1, 1, 0) + frac{2}{3}(0, 1, 1) + frac{1}{3}(1, -1, 1).
$$
Hence the orthogonal projection is $frac{1}{3}(1, -1, 1)$ and so $d$ is the length of this vector. Then $d^2 = 1/3$ and $3d^2 = 1$.
answered Jan 25 at 7:28
hunterhunter
15.3k32640
$endgroup$
add a comment |
$begingroup$
Project the vector onto the subspace. Subtract the result from $(1,1,1)$. Then compute the norm.
If you're familiar with Gram-Schmidt, it's basically the same.
First I need to get an orthogonal basis for $P$: $(1,1,0)-frac12(0,1,1)=(1,frac12,-frac12)$.
So, we get $(1,1,1)-(0,1,1)-frac23(1,frac12, -frac12) =(frac13,-frac13,frac13) $.
Now the norm is $d=sqrt{frac13}$.
So $3d^2=1 $.
answered Jan 25 at 7:46
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Note that the vector $(1, -1, 1)$ is orthogonal to both of these vectors. Now
$$
(1, 1, 1) = frac{2}{3}(1, 1, 0) + frac{2}{3}(0, 1, 1) + frac{1}{3}(1, -1, 1).
$$
Hence the orthogonal projection is $frac{1}{3}(1, -1, 1)$ and so $d$ is the length of this vector. Then $d^2 = 1/3$ and $3d^2 = 1$.
answered Jan 25 at 7:28
hunterhunter
15.3k32640
$endgroup$
add a comment |
$begingroup$
Note that the vector $(1, -1, 1)$ is orthogonal to both of these vectors. Now
$$
(1, 1, 1) = frac{2}{3}(1, 1, 0) + frac{2}{3}(0, 1, 1) + frac{1}{3}(1, -1, 1).
$$
Hence the orthogonal projection is $frac{1}{3}(1, -1, 1)$ and so $d$ is the length of this vector. Then $d^2 = 1/3$ and $3d^2 = 1$.
answered Jan 25 at 7:28
hunterhunter
15.3k32640
$endgroup$
add a comment |
$begingroup$
Note that the vector $(1, -1, 1)$ is orthogonal to both of these vectors. Now
$$
(1, 1, 1) = frac{2}{3}(1, 1, 0) + frac{2}{3}(0, 1, 1) + frac{1}{3}(1, -1, 1).
$$
Hence the orthogonal projection is $frac{1}{3}(1, -1, 1)$ and so $d$ is the length of this vector. Then $d^2 = 1/3$ and $3d^2 = 1$.
answered Jan 25 at 7:28
hunterhunter
15.3k32640
$endgroup$
Note that the vector $(1, -1, 1)$ is orthogonal to both of these vectors. Now
$$
(1, 1, 1) = frac{2}{3}(1, 1, 0) + frac{2}{3}(0, 1, 1) + frac{1}{3}(1, -1, 1).
$$
Hence the orthogonal projection is $frac{1}{3}(1, -1, 1)$ and so $d$ is the length of this vector. Then $d^2 = 1/3$ and $3d^2 = 1$.
answered Jan 25 at 7:28
hunterhunter
15.3k32640
answered Jan 25 at 7:28
hunterhunter
15.3k32640
answered Jan 25 at 7:28
hunterhunter
15.3k32640
answered Jan 25 at 7:28
hunterhunter
15.3k32640
15.3k32640
add a comment |
add a comment |
$begingroup$
Project the vector onto the subspace. Subtract the result from $(1,1,1)$. Then compute the norm.
If you're familiar with Gram-Schmidt, it's basically the same.
First I need to get an orthogonal basis for $P$: $(1,1,0)-frac12(0,1,1)=(1,frac12,-frac12)$.
So, we get $(1,1,1)-(0,1,1)-frac23(1,frac12, -frac12) =(frac13,-frac13,frac13) $.
Now the norm is $d=sqrt{frac13}$.
So $3d^2=1 $.
answered Jan 25 at 7:46
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Project the vector onto the subspace. Subtract the result from $(1,1,1)$. Then compute the norm.
If you're familiar with Gram-Schmidt, it's basically the same.
First I need to get an orthogonal basis for $P$: $(1,1,0)-frac12(0,1,1)=(1,frac12,-frac12)$.
So, we get $(1,1,1)-(0,1,1)-frac23(1,frac12, -frac12) =(frac13,-frac13,frac13) $.
Now the norm is $d=sqrt{frac13}$.
So $3d^2=1 $.
answered Jan 25 at 7:46
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Project the vector onto the subspace. Subtract the result from $(1,1,1)$. Then compute the norm.
If you're familiar with Gram-Schmidt, it's basically the same.
First I need to get an orthogonal basis for $P$: $(1,1,0)-frac12(0,1,1)=(1,frac12,-frac12)$.
So, we get $(1,1,1)-(0,1,1)-frac23(1,frac12, -frac12) =(frac13,-frac13,frac13) $.
Now the norm is $d=sqrt{frac13}$.
So $3d^2=1 $.
answered Jan 25 at 7:46
Chris CusterChris Custer
14.2k3827
$endgroup$
Project the vector onto the subspace. Subtract the result from $(1,1,1)$. Then compute the norm.
If you're familiar with Gram-Schmidt, it's basically the same.
First I need to get an orthogonal basis for $P$: $(1,1,0)-frac12(0,1,1)=(1,frac12,-frac12)$.
So, we get $(1,1,1)-(0,1,1)-frac23(1,frac12, -frac12) =(frac13,-frac13,frac13) $.
Now the norm is $d=sqrt{frac13}$.
So $3d^2=1 $.
answered Jan 25 at 7:46
Chris CusterChris Custer
14.2k3827
edited Jan 26 at 10:10
answered Jan 25 at 7:46
Chris CusterChris Custer
14.2k3827
answered Jan 25 at 7:46
Chris CusterChris Custer
14.2k3827
answered Jan 25 at 7:46
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
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2
$begingroup$
Your subspace is a plane inside the $mathbb{R}^3$. Actually you want to calculate the distance between a point and a plane. Does this help?
$endgroup$
– Sqyuli
Jan 25 at 7:07
1
$begingroup$
This (math.stackexchange.com/questions/2101140/…) might help!
$endgroup$
– Chinnapparaj R
Jan 25 at 7:13
$begingroup$
@Sqyuli Then I have to find the vector in that subspace which is orthogonal to (1,1,1)? Is not it?
$endgroup$
– cmi
Jan 25 at 7:18