Pythagoras theorem in oblique coordinates The Next CEO of Stack OverflowDot product in coordinatesApproximation of difference of distances to a distant pointFind the value of $a$ for which two lines are perpendicularProving that the dot product is distributive?Prove the distributive property of the dot product using its geometric definition?Components of a vector onto nonorthogonal, nonunit basis vectorsCan one show that $fracvecucdot vecv$ is always on the range of $cos theta $?Angular Difference between two rotation matrices on XZ planeClarification on product rule for vectors in non-Cartesian coordinatesAngle between two non-coplanar vectors

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Pythagoras theorem in oblique coordinates



The Next CEO of Stack OverflowDot product in coordinatesApproximation of difference of distances to a distant pointFind the value of $a$ for which two lines are perpendicularProving that the dot product is distributive?Prove the distributive property of the dot product using its geometric definition?Components of a vector onto nonorthogonal, nonunit basis vectorsCan one show that $fracvecucdot vecvvecu$ is always on the range of $cos theta $?Angular Difference between two rotation matrices on XZ planeClarification on product rule for vectors in non-Cartesian coordinatesAngle between two non-coplanar vectors










0












$begingroup$


consider two oblique oblique basis vectors of unit length $vecr_1, vecr_2$



then any vector $vecv = pvecr_1+qvecr_2$



define the dot product between two vectors a and b as $|b|$ (ie length of b) * (the length projection of(say) a on b made by an oblique line from a on b with angle same as that formed between $vecr_1, vecr_2$)



ie such that $vecr_1 cdot vecr_2 = 0$



this is in contrast to the project of a on b along a right angle as done with orthogonal basis where you can use the cosine.



It's easy to show that this definition of the dot product is indeed distributive.



from our definition of v



$vecv cdot vecv = p^2 + q^2 $



Then $|v|^2 = p^2 + q^2$



which geometrically isn't the case,because p and q are oblique components, not perpendicular ones, this should not give you the length of v.










share|cite|improve this question











$endgroup$











  • $begingroup$
    You just found out that in oblique coordinates $p^2+q^2$ is not the length. The Pythagoras' theorem only applies to right triangles.
    $endgroup$
    – Yves Daoust
    Mar 18 at 16:01










  • $begingroup$
    @YvesDaoust, well but v dot v has to be the square of the length of v by our definition (edited above)
    $endgroup$
    – Vrisk
    Mar 18 at 16:08










  • $begingroup$
    No, it doesn't have to. In oblique coordinates, this is false.
    $endgroup$
    – Yves Daoust
    Mar 18 at 16:24











  • $begingroup$
    @YvesDaoust we define the dot product as length of one vector * oblique projection of length of other vector on that vector. Now if both vectors are same, don't we have the square of the length?
    $endgroup$
    – Vrisk
    Mar 18 at 16:29










  • $begingroup$
    No, you get the scalar square of the vector.
    $endgroup$
    – Yves Daoust
    Mar 18 at 17:05















0












$begingroup$


consider two oblique oblique basis vectors of unit length $vecr_1, vecr_2$



then any vector $vecv = pvecr_1+qvecr_2$



define the dot product between two vectors a and b as $|b|$ (ie length of b) * (the length projection of(say) a on b made by an oblique line from a on b with angle same as that formed between $vecr_1, vecr_2$)



ie such that $vecr_1 cdot vecr_2 = 0$



this is in contrast to the project of a on b along a right angle as done with orthogonal basis where you can use the cosine.



It's easy to show that this definition of the dot product is indeed distributive.



from our definition of v



$vecv cdot vecv = p^2 + q^2 $



Then $|v|^2 = p^2 + q^2$



which geometrically isn't the case,because p and q are oblique components, not perpendicular ones, this should not give you the length of v.










share|cite|improve this question











$endgroup$











  • $begingroup$
    You just found out that in oblique coordinates $p^2+q^2$ is not the length. The Pythagoras' theorem only applies to right triangles.
    $endgroup$
    – Yves Daoust
    Mar 18 at 16:01










  • $begingroup$
    @YvesDaoust, well but v dot v has to be the square of the length of v by our definition (edited above)
    $endgroup$
    – Vrisk
    Mar 18 at 16:08










  • $begingroup$
    No, it doesn't have to. In oblique coordinates, this is false.
    $endgroup$
    – Yves Daoust
    Mar 18 at 16:24











  • $begingroup$
    @YvesDaoust we define the dot product as length of one vector * oblique projection of length of other vector on that vector. Now if both vectors are same, don't we have the square of the length?
    $endgroup$
    – Vrisk
    Mar 18 at 16:29










  • $begingroup$
    No, you get the scalar square of the vector.
    $endgroup$
    – Yves Daoust
    Mar 18 at 17:05













0












0








0


0



$begingroup$


consider two oblique oblique basis vectors of unit length $vecr_1, vecr_2$



then any vector $vecv = pvecr_1+qvecr_2$



define the dot product between two vectors a and b as $|b|$ (ie length of b) * (the length projection of(say) a on b made by an oblique line from a on b with angle same as that formed between $vecr_1, vecr_2$)



ie such that $vecr_1 cdot vecr_2 = 0$



this is in contrast to the project of a on b along a right angle as done with orthogonal basis where you can use the cosine.



It's easy to show that this definition of the dot product is indeed distributive.



from our definition of v



$vecv cdot vecv = p^2 + q^2 $



Then $|v|^2 = p^2 + q^2$



which geometrically isn't the case,because p and q are oblique components, not perpendicular ones, this should not give you the length of v.










share|cite|improve this question











$endgroup$




consider two oblique oblique basis vectors of unit length $vecr_1, vecr_2$



then any vector $vecv = pvecr_1+qvecr_2$



define the dot product between two vectors a and b as $|b|$ (ie length of b) * (the length projection of(say) a on b made by an oblique line from a on b with angle same as that formed between $vecr_1, vecr_2$)



ie such that $vecr_1 cdot vecr_2 = 0$



this is in contrast to the project of a on b along a right angle as done with orthogonal basis where you can use the cosine.



It's easy to show that this definition of the dot product is indeed distributive.



from our definition of v



$vecv cdot vecv = p^2 + q^2 $



Then $|v|^2 = p^2 + q^2$



which geometrically isn't the case,because p and q are oblique components, not perpendicular ones, this should not give you the length of v.







geometry vectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 18 at 21:39









Jean Marie

31k42255




31k42255










asked Mar 18 at 15:47









VriskVrisk

12012




12012











  • $begingroup$
    You just found out that in oblique coordinates $p^2+q^2$ is not the length. The Pythagoras' theorem only applies to right triangles.
    $endgroup$
    – Yves Daoust
    Mar 18 at 16:01










  • $begingroup$
    @YvesDaoust, well but v dot v has to be the square of the length of v by our definition (edited above)
    $endgroup$
    – Vrisk
    Mar 18 at 16:08










  • $begingroup$
    No, it doesn't have to. In oblique coordinates, this is false.
    $endgroup$
    – Yves Daoust
    Mar 18 at 16:24











  • $begingroup$
    @YvesDaoust we define the dot product as length of one vector * oblique projection of length of other vector on that vector. Now if both vectors are same, don't we have the square of the length?
    $endgroup$
    – Vrisk
    Mar 18 at 16:29










  • $begingroup$
    No, you get the scalar square of the vector.
    $endgroup$
    – Yves Daoust
    Mar 18 at 17:05
















  • $begingroup$
    You just found out that in oblique coordinates $p^2+q^2$ is not the length. The Pythagoras' theorem only applies to right triangles.
    $endgroup$
    – Yves Daoust
    Mar 18 at 16:01










  • $begingroup$
    @YvesDaoust, well but v dot v has to be the square of the length of v by our definition (edited above)
    $endgroup$
    – Vrisk
    Mar 18 at 16:08










  • $begingroup$
    No, it doesn't have to. In oblique coordinates, this is false.
    $endgroup$
    – Yves Daoust
    Mar 18 at 16:24











  • $begingroup$
    @YvesDaoust we define the dot product as length of one vector * oblique projection of length of other vector on that vector. Now if both vectors are same, don't we have the square of the length?
    $endgroup$
    – Vrisk
    Mar 18 at 16:29










  • $begingroup$
    No, you get the scalar square of the vector.
    $endgroup$
    – Yves Daoust
    Mar 18 at 17:05















$begingroup$
You just found out that in oblique coordinates $p^2+q^2$ is not the length. The Pythagoras' theorem only applies to right triangles.
$endgroup$
– Yves Daoust
Mar 18 at 16:01




$begingroup$
You just found out that in oblique coordinates $p^2+q^2$ is not the length. The Pythagoras' theorem only applies to right triangles.
$endgroup$
– Yves Daoust
Mar 18 at 16:01












$begingroup$
@YvesDaoust, well but v dot v has to be the square of the length of v by our definition (edited above)
$endgroup$
– Vrisk
Mar 18 at 16:08




$begingroup$
@YvesDaoust, well but v dot v has to be the square of the length of v by our definition (edited above)
$endgroup$
– Vrisk
Mar 18 at 16:08












$begingroup$
No, it doesn't have to. In oblique coordinates, this is false.
$endgroup$
– Yves Daoust
Mar 18 at 16:24





$begingroup$
No, it doesn't have to. In oblique coordinates, this is false.
$endgroup$
– Yves Daoust
Mar 18 at 16:24













$begingroup$
@YvesDaoust we define the dot product as length of one vector * oblique projection of length of other vector on that vector. Now if both vectors are same, don't we have the square of the length?
$endgroup$
– Vrisk
Mar 18 at 16:29




$begingroup$
@YvesDaoust we define the dot product as length of one vector * oblique projection of length of other vector on that vector. Now if both vectors are same, don't we have the square of the length?
$endgroup$
– Vrisk
Mar 18 at 16:29












$begingroup$
No, you get the scalar square of the vector.
$endgroup$
– Yves Daoust
Mar 18 at 17:05




$begingroup$
No, you get the scalar square of the vector.
$endgroup$
– Yves Daoust
Mar 18 at 17:05










1 Answer
1






active

oldest

votes


















0












$begingroup$

Let me expand on what Yves Daust is saying with a specific example.



Let $vec r_1 = (1,0)$ and $vec r_2 =left(frac sqrt 22, frac sqrt 22right)$ and suppose we consider the vector $vec v = vec r_1 + vec r_2$.



By the definition of your inner product, $|vec v|^2 = 1^2 + 1^2 = 2$.



Now if we express $vec v$ in terms of canonical coordinates, it is $$vec v = vec r_1 + vec r_2 = (1, 0) + left(frac sqrt 22, frac sqrt 22right) = left(1+frac sqrt 22, frac sqrt 22right)$$



And therefore its Euclidean norm is given by $$|vec v|^2 = left(1+frac sqrt 22right)^2 + left(fracsqrt 22right)^2 = left(1 + sqrt 2 + frac 12right) + frac12 = 2 + sqrt 2$$



So very clearly, $|vec v| ne |vec v|$.



$|vec v|$ is a norm. It satisfies all the properties necessary to be a norm. But the key word here is "a". There are many, many, norms that can be placed on $Bbb R^2$, so simply being a norm does not make it equal to that other more famous norm.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Okay, I see. Clearly something is off here, I'm not able to pinpoint where however. By my defination a.b = length of b*(the length projection of(say) a on b made by an oblique line from a on b ), then if v= a = b, then length of v = length of projection of v on v . therefore |v|^2 = square of length of v ?
    $endgroup$
    – Vrisk
    Mar 19 at 14:37











Your Answer





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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Let me expand on what Yves Daust is saying with a specific example.



Let $vec r_1 = (1,0)$ and $vec r_2 =left(frac sqrt 22, frac sqrt 22right)$ and suppose we consider the vector $vec v = vec r_1 + vec r_2$.



By the definition of your inner product, $|vec v|^2 = 1^2 + 1^2 = 2$.



Now if we express $vec v$ in terms of canonical coordinates, it is $$vec v = vec r_1 + vec r_2 = (1, 0) + left(frac sqrt 22, frac sqrt 22right) = left(1+frac sqrt 22, frac sqrt 22right)$$



And therefore its Euclidean norm is given by $$|vec v|^2 = left(1+frac sqrt 22right)^2 + left(fracsqrt 22right)^2 = left(1 + sqrt 2 + frac 12right) + frac12 = 2 + sqrt 2$$



So very clearly, $|vec v| ne |vec v|$.



$|vec v|$ is a norm. It satisfies all the properties necessary to be a norm. But the key word here is "a". There are many, many, norms that can be placed on $Bbb R^2$, so simply being a norm does not make it equal to that other more famous norm.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Okay, I see. Clearly something is off here, I'm not able to pinpoint where however. By my defination a.b = length of b*(the length projection of(say) a on b made by an oblique line from a on b ), then if v= a = b, then length of v = length of projection of v on v . therefore |v|^2 = square of length of v ?
    $endgroup$
    – Vrisk
    Mar 19 at 14:37















0












$begingroup$

Let me expand on what Yves Daust is saying with a specific example.



Let $vec r_1 = (1,0)$ and $vec r_2 =left(frac sqrt 22, frac sqrt 22right)$ and suppose we consider the vector $vec v = vec r_1 + vec r_2$.



By the definition of your inner product, $|vec v|^2 = 1^2 + 1^2 = 2$.



Now if we express $vec v$ in terms of canonical coordinates, it is $$vec v = vec r_1 + vec r_2 = (1, 0) + left(frac sqrt 22, frac sqrt 22right) = left(1+frac sqrt 22, frac sqrt 22right)$$



And therefore its Euclidean norm is given by $$|vec v|^2 = left(1+frac sqrt 22right)^2 + left(fracsqrt 22right)^2 = left(1 + sqrt 2 + frac 12right) + frac12 = 2 + sqrt 2$$



So very clearly, $|vec v| ne |vec v|$.



$|vec v|$ is a norm. It satisfies all the properties necessary to be a norm. But the key word here is "a". There are many, many, norms that can be placed on $Bbb R^2$, so simply being a norm does not make it equal to that other more famous norm.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Okay, I see. Clearly something is off here, I'm not able to pinpoint where however. By my defination a.b = length of b*(the length projection of(say) a on b made by an oblique line from a on b ), then if v= a = b, then length of v = length of projection of v on v . therefore |v|^2 = square of length of v ?
    $endgroup$
    – Vrisk
    Mar 19 at 14:37













0












0








0





$begingroup$

Let me expand on what Yves Daust is saying with a specific example.



Let $vec r_1 = (1,0)$ and $vec r_2 =left(frac sqrt 22, frac sqrt 22right)$ and suppose we consider the vector $vec v = vec r_1 + vec r_2$.



By the definition of your inner product, $|vec v|^2 = 1^2 + 1^2 = 2$.



Now if we express $vec v$ in terms of canonical coordinates, it is $$vec v = vec r_1 + vec r_2 = (1, 0) + left(frac sqrt 22, frac sqrt 22right) = left(1+frac sqrt 22, frac sqrt 22right)$$



And therefore its Euclidean norm is given by $$|vec v|^2 = left(1+frac sqrt 22right)^2 + left(fracsqrt 22right)^2 = left(1 + sqrt 2 + frac 12right) + frac12 = 2 + sqrt 2$$



So very clearly, $|vec v| ne |vec v|$.



$|vec v|$ is a norm. It satisfies all the properties necessary to be a norm. But the key word here is "a". There are many, many, norms that can be placed on $Bbb R^2$, so simply being a norm does not make it equal to that other more famous norm.






share|cite|improve this answer









$endgroup$



Let me expand on what Yves Daust is saying with a specific example.



Let $vec r_1 = (1,0)$ and $vec r_2 =left(frac sqrt 22, frac sqrt 22right)$ and suppose we consider the vector $vec v = vec r_1 + vec r_2$.



By the definition of your inner product, $|vec v|^2 = 1^2 + 1^2 = 2$.



Now if we express $vec v$ in terms of canonical coordinates, it is $$vec v = vec r_1 + vec r_2 = (1, 0) + left(frac sqrt 22, frac sqrt 22right) = left(1+frac sqrt 22, frac sqrt 22right)$$



And therefore its Euclidean norm is given by $$|vec v|^2 = left(1+frac sqrt 22right)^2 + left(fracsqrt 22right)^2 = left(1 + sqrt 2 + frac 12right) + frac12 = 2 + sqrt 2$$



So very clearly, $|vec v| ne |vec v|$.



$|vec v|$ is a norm. It satisfies all the properties necessary to be a norm. But the key word here is "a". There are many, many, norms that can be placed on $Bbb R^2$, so simply being a norm does not make it equal to that other more famous norm.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 18 at 23:23









Paul SinclairPaul Sinclair

20.7k21543




20.7k21543











  • $begingroup$
    Okay, I see. Clearly something is off here, I'm not able to pinpoint where however. By my defination a.b = length of b*(the length projection of(say) a on b made by an oblique line from a on b ), then if v= a = b, then length of v = length of projection of v on v . therefore |v|^2 = square of length of v ?
    $endgroup$
    – Vrisk
    Mar 19 at 14:37
















  • $begingroup$
    Okay, I see. Clearly something is off here, I'm not able to pinpoint where however. By my defination a.b = length of b*(the length projection of(say) a on b made by an oblique line from a on b ), then if v= a = b, then length of v = length of projection of v on v . therefore |v|^2 = square of length of v ?
    $endgroup$
    – Vrisk
    Mar 19 at 14:37















$begingroup$
Okay, I see. Clearly something is off here, I'm not able to pinpoint where however. By my defination a.b = length of b*(the length projection of(say) a on b made by an oblique line from a on b ), then if v= a = b, then length of v = length of projection of v on v . therefore |v|^2 = square of length of v ?
$endgroup$
– Vrisk
Mar 19 at 14:37




$begingroup$
Okay, I see. Clearly something is off here, I'm not able to pinpoint where however. By my defination a.b = length of b*(the length projection of(say) a on b made by an oblique line from a on b ), then if v= a = b, then length of v = length of projection of v on v . therefore |v|^2 = square of length of v ?
$endgroup$
– Vrisk
Mar 19 at 14:37

















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