Determining a vector orthogonal to $q_1=(1,1,1)$ and $q_3=(1,1,-2)$, why I'm wrong with my calculations?
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Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that ${q_1,q_2,q_3}$ is a arthogonal basis for $mathbb{R}^3$.
My problem is the following: I did take $v=(1,0,0)$ and I did verify that ${q_1,q_3,v}$ is a basis for $mathbb{R}^3$. Then I did take $$q_2=v-langle v|q_1rangle q_1-langle v|q_3rangle q_3=(-1,-2,1)$$
And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?
linear-algebra vectors orthogonality
edited Nov 22 at 18:50
quid♦
36.8k95093
asked Nov 22 at 16:51
Gödel
1,393319
add a comment |
up vote
1
down vote
favorite
Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that ${q_1,q_2,q_3}$ is a arthogonal basis for $mathbb{R}^3$.
My problem is the following: I did take $v=(1,0,0)$ and I did verify that ${q_1,q_3,v}$ is a basis for $mathbb{R}^3$. Then I did take $$q_2=v-langle v|q_1rangle q_1-langle v|q_3rangle q_3=(-1,-2,1)$$
And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?
linear-algebra vectors orthogonality
edited Nov 22 at 18:50
quid♦
36.8k95093
asked Nov 22 at 16:51
Gödel
1,393319
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that ${q_1,q_2,q_3}$ is a arthogonal basis for $mathbb{R}^3$.
My problem is the following: I did take $v=(1,0,0)$ and I did verify that ${q_1,q_3,v}$ is a basis for $mathbb{R}^3$. Then I did take $$q_2=v-langle v|q_1rangle q_1-langle v|q_3rangle q_3=(-1,-2,1)$$
And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?
linear-algebra vectors orthogonality
edited Nov 22 at 18:50
quid♦
36.8k95093
asked Nov 22 at 16:51
Gödel
1,393319
Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that ${q_1,q_2,q_3}$ is a arthogonal basis for $mathbb{R}^3$.
My problem is the following: I did take $v=(1,0,0)$ and I did verify that ${q_1,q_3,v}$ is a basis for $mathbb{R}^3$. Then I did take $$q_2=v-langle v|q_1rangle q_1-langle v|q_3rangle q_3=(-1,-2,1)$$
And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?
linear-algebra vectors orthogonality
linear-algebra vectors orthogonality
edited Nov 22 at 18:50
quid♦
36.8k95093
asked Nov 22 at 16:51
Gödel
1,393319
edited Nov 22 at 18:50
quid♦
36.8k95093
asked Nov 22 at 16:51
Gödel
1,393319
edited Nov 22 at 18:50
quid♦
36.8k95093
edited Nov 22 at 18:50
quid♦
36.8k95093
edited Nov 22 at 18:50
quid♦
36.8k95093
36.8k95093
asked Nov 22 at 16:51
Gödel
1,393319
asked Nov 22 at 16:51
Gödel
1,393319
asked Nov 22 at 16:51
Gödel
1,393319
1,393319
add a comment |
add a comment |
3 Answers
3
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2
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HINT
Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by
$$q_2=q_1times q_3$$
As an alternative by GS we have
$$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$
answered Nov 22 at 16:54
gimusi
92.7k94495
I don't know why I don't use cross product before, It is a simple way to solve the problem.
– Gödel
Nov 22 at 17:10
@Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
– gimusi
Nov 22 at 17:18
add a comment |
up vote
4
down vote
None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
$$ q_2=v-{langle v|q_1rangle over langle q_1|q_1rangle} q_1-{langle v|q_3rangle over langle q_3|q_3rangle} q_3 = ({1 over 2},-{1 over 2},0)$$.
answered Nov 22 at 16:56
SinTan1729
2,399622
add a comment |
up vote
2
down vote
$$q_2=v-frac{langle v|q_1rangle}{langle q_1|q_1rangle} q_1-frac{langle v|q_3rangle}{langle q_3|q_3rangle} q_3$$
answered Nov 22 at 16:54
Siong Thye Goh
97.7k1463116
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
HINT
Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by
$$q_2=q_1times q_3$$
As an alternative by GS we have
$$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$
answered Nov 22 at 16:54
gimusi
92.7k94495
I don't know why I don't use cross product before, It is a simple way to solve the problem.
– Gödel
Nov 22 at 17:10
@Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
– gimusi
Nov 22 at 17:18
add a comment |
up vote
2
down vote
accepted
HINT
Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by
$$q_2=q_1times q_3$$
As an alternative by GS we have
$$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$
answered Nov 22 at 16:54
gimusi
92.7k94495
I don't know why I don't use cross product before, It is a simple way to solve the problem.
– Gödel
Nov 22 at 17:10
@Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
– gimusi
Nov 22 at 17:18
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
HINT
Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by
$$q_2=q_1times q_3$$
As an alternative by GS we have
$$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$
answered Nov 22 at 16:54
gimusi
92.7k94495
HINT
Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by
$$q_2=q_1times q_3$$
As an alternative by GS we have
$$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$
answered Nov 22 at 16:54
gimusi
92.7k94495
edited Nov 22 at 17:06
answered Nov 22 at 16:54
gimusi
92.7k94495
answered Nov 22 at 16:54
gimusi
92.7k94495
answered Nov 22 at 16:54
gimusi
92.7k94495
92.7k94495
I don't know why I don't use cross product before, It is a simple way to solve the problem.
– Gödel
Nov 22 at 17:10
@Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
– gimusi
Nov 22 at 17:18
add a comment |
I don't know why I don't use cross product before, It is a simple way to solve the problem.
– Gödel
Nov 22 at 17:10
@Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
– gimusi
Nov 22 at 17:18
I don't know why I don't use cross product before, It is a simple way to solve the problem.
– Gödel
Nov 22 at 17:10
I don't know why I don't use cross product before, It is a simple way to solve the problem.
– Gödel
Nov 22 at 17:10
@Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
– gimusi
Nov 22 at 17:18
@Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
– gimusi
Nov 22 at 17:18
add a comment |
up vote
4
down vote
None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
$$ q_2=v-{langle v|q_1rangle over langle q_1|q_1rangle} q_1-{langle v|q_3rangle over langle q_3|q_3rangle} q_3 = ({1 over 2},-{1 over 2},0)$$.
answered Nov 22 at 16:56
SinTan1729
2,399622
add a comment |
up vote
4
down vote
None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
$$ q_2=v-{langle v|q_1rangle over langle q_1|q_1rangle} q_1-{langle v|q_3rangle over langle q_3|q_3rangle} q_3 = ({1 over 2},-{1 over 2},0)$$.
answered Nov 22 at 16:56
SinTan1729
2,399622
add a comment |
up vote
4
down vote
up vote
4
down vote
None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
$$ q_2=v-{langle v|q_1rangle over langle q_1|q_1rangle} q_1-{langle v|q_3rangle over langle q_3|q_3rangle} q_3 = ({1 over 2},-{1 over 2},0)$$.
answered Nov 22 at 16:56
SinTan1729
2,399622
None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
$$ q_2=v-{langle v|q_1rangle over langle q_1|q_1rangle} q_1-{langle v|q_3rangle over langle q_3|q_3rangle} q_3 = ({1 over 2},-{1 over 2},0)$$.
answered Nov 22 at 16:56
SinTan1729
2,399622
edited Nov 22 at 17:03
answered Nov 22 at 16:56
SinTan1729
2,399622
answered Nov 22 at 16:56
SinTan1729
2,399622
answered Nov 22 at 16:56
SinTan1729
2,399622
2,399622
add a comment |
add a comment |
up vote
2
down vote
$$q_2=v-frac{langle v|q_1rangle}{langle q_1|q_1rangle} q_1-frac{langle v|q_3rangle}{langle q_3|q_3rangle} q_3$$
answered Nov 22 at 16:54
Siong Thye Goh
97.7k1463116
add a comment |
up vote
2
down vote
$$q_2=v-frac{langle v|q_1rangle}{langle q_1|q_1rangle} q_1-frac{langle v|q_3rangle}{langle q_3|q_3rangle} q_3$$
answered Nov 22 at 16:54
Siong Thye Goh
97.7k1463116
add a comment |
up vote
2
down vote
up vote
2
down vote
$$q_2=v-frac{langle v|q_1rangle}{langle q_1|q_1rangle} q_1-frac{langle v|q_3rangle}{langle q_3|q_3rangle} q_3$$
answered Nov 22 at 16:54
Siong Thye Goh
97.7k1463116
$$q_2=v-frac{langle v|q_1rangle}{langle q_1|q_1rangle} q_1-frac{langle v|q_3rangle}{langle q_3|q_3rangle} q_3$$
answered Nov 22 at 16:54
Siong Thye Goh
97.7k1463116
answered Nov 22 at 16:54
Siong Thye Goh
97.7k1463116
answered Nov 22 at 16:54
Siong Thye Goh
97.7k1463116
answered Nov 22 at 16:54
Siong Thye Goh
97.7k1463116
97.7k1463116
add a comment |
add a comment |
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