Two vectors with the same normal surface projection and the same normal surface cross product, are equal?

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I have to vectors $mathbf a$ and $mathbf b$, that fulfill the following conditions:



$(mathbf a-mathbf b)cdot mathbf n=mathbf 0$



$(mathbf a-mathbf b)times mathbf n=mathbf 0$



being $mathbf n$ a unit surface normal.



My question is, is $mathbf a = mathbf b$ ?



I have confirmed this by doing the cross product in a reference frame for which its first direction is coincident with the surface normal. Since both the dot and cross products are invariant under reference frame transformations, the results should be confirmed for all coordinate systems. Is this reasoning ok?






geometry vectors cross-product







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  • A slick one-liner proof: $$ mathbfa=(mathbfacdotmathbfn)mathbfn-(mathbfatimesmathbfn)timesmathbfn=dots=mathbfb $$
    – user10354138
    20 mins ago















up vote
3
down vote

favorite












I have to vectors $mathbf a$ and $mathbf b$, that fulfill the following conditions:



$(mathbf a-mathbf b)cdot mathbf n=mathbf 0$



$(mathbf a-mathbf b)times mathbf n=mathbf 0$



being $mathbf n$ a unit surface normal.



My question is, is $mathbf a = mathbf b$ ?



I have confirmed this by doing the cross product in a reference frame for which its first direction is coincident with the surface normal. Since both the dot and cross products are invariant under reference frame transformations, the results should be confirmed for all coordinate systems. Is this reasoning ok?






geometry vectors cross-product







share|cite|improve this question





















  • A slick one-liner proof: $$ mathbfa=(mathbfacdotmathbfn)mathbfn-(mathbfatimesmathbfn)timesmathbfn=dots=mathbfb $$
    – user10354138
    20 mins ago













up vote
3
down vote

favorite









up vote
3
down vote

favorite











I have to vectors $mathbf a$ and $mathbf b$, that fulfill the following conditions:



$(mathbf a-mathbf b)cdot mathbf n=mathbf 0$



$(mathbf a-mathbf b)times mathbf n=mathbf 0$



being $mathbf n$ a unit surface normal.



My question is, is $mathbf a = mathbf b$ ?



I have confirmed this by doing the cross product in a reference frame for which its first direction is coincident with the surface normal. Since both the dot and cross products are invariant under reference frame transformations, the results should be confirmed for all coordinate systems. Is this reasoning ok?






geometry vectors cross-product







share|cite|improve this question













I have to vectors $mathbf a$ and $mathbf b$, that fulfill the following conditions:



$(mathbf a-mathbf b)cdot mathbf n=mathbf 0$



$(mathbf a-mathbf b)times mathbf n=mathbf 0$



being $mathbf n$ a unit surface normal.



My question is, is $mathbf a = mathbf b$ ?



I have confirmed this by doing the cross product in a reference frame for which its first direction is coincident with the surface normal. Since both the dot and cross products are invariant under reference frame transformations, the results should be confirmed for all coordinate systems. Is this reasoning ok?






geometry vectors cross-product




geometry vectors cross-product






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asked 56 mins ago









nodarkside

205




205











  • A slick one-liner proof: $$ mathbfa=(mathbfacdotmathbfn)mathbfn-(mathbfatimesmathbfn)timesmathbfn=dots=mathbfb $$
    – user10354138
    20 mins ago

















  • A slick one-liner proof: $$ mathbfa=(mathbfacdotmathbfn)mathbfn-(mathbfatimesmathbfn)timesmathbfn=dots=mathbfb $$
    – user10354138
    20 mins ago
















A slick one-liner proof: $$ mathbfa=(mathbfacdotmathbfn)mathbfn-(mathbfatimesmathbfn)timesmathbfn=dots=mathbfb $$
– user10354138
20 mins ago





A slick one-liner proof: $$ mathbfa=(mathbfacdotmathbfn)mathbfn-(mathbfatimesmathbfn)timesmathbfn=dots=mathbfb $$
– user10354138
20 mins ago











2 Answers
2






active

oldest

votes

















up vote
4
down vote



accepted










Yes it is correct, indeed we have that



  • $(vec a-vec b)times vec n=vec 0 implies vec a-vec b$ is a multiple of $vec n$ that is $vec a-vec b=kvec n$

  • $(vec a-vec b)cdot vec n=0 implies vec a-vec b$ is orthogonal to $vec n$ that is $kvec ncdot vec n=0 implies k=0$

therefore



$$vec a-vec b=vec 0 implies vec a=vec b$$






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  • Thanks very much @gimusi !
    – nodarkside
    45 mins ago










  • @nodarkside You are welcome! Bye
    – gimusi
    44 mins ago

















up vote
3
down vote













You could also do this using the fact that $vecv cdot vecw = vertvecvrvert lvert vecw rvert costheta$ and $vecv times vecw = vertvecvrvert lvert vecw rvert sintheta$. In your case this leads to
beginalign*
(veca - vecb) cdot vecn = lvertveca - vecbrvert lvertvecnrvert costheta &= 0 \
(veca - vecb) times vecn = lvertveca - vecbrvert lvertvecnrvert sintheta &= 0
endalign*
From here, it is safe to cancel the $lvertvecnrvert$, as it is unit normal. We cannot safely cancel out the $costheta$ or $sintheta$ because these may be $0$, but notice that if $sintheta$ = 0, then $costheta neq 0$ and vice versa. Thus, we are left to the conclusion that $lvert veca - vecb rvert = 0$, so $veca = vecb$.






share|cite|improve this answer










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Alerra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • Thanks very much @Alerra! This proof is also very useful.
    – nodarkside
    7 mins ago










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










Yes it is correct, indeed we have that



  • $(vec a-vec b)times vec n=vec 0 implies vec a-vec b$ is a multiple of $vec n$ that is $vec a-vec b=kvec n$

  • $(vec a-vec b)cdot vec n=0 implies vec a-vec b$ is orthogonal to $vec n$ that is $kvec ncdot vec n=0 implies k=0$

therefore



$$vec a-vec b=vec 0 implies vec a=vec b$$






share|cite|improve this answer




















  • Thanks very much @gimusi !
    – nodarkside
    45 mins ago










  • @nodarkside You are welcome! Bye
    – gimusi
    44 mins ago














up vote
4
down vote



accepted










Yes it is correct, indeed we have that



  • $(vec a-vec b)times vec n=vec 0 implies vec a-vec b$ is a multiple of $vec n$ that is $vec a-vec b=kvec n$

  • $(vec a-vec b)cdot vec n=0 implies vec a-vec b$ is orthogonal to $vec n$ that is $kvec ncdot vec n=0 implies k=0$

therefore



$$vec a-vec b=vec 0 implies vec a=vec b$$






share|cite|improve this answer




















  • Thanks very much @gimusi !
    – nodarkside
    45 mins ago










  • @nodarkside You are welcome! Bye
    – gimusi
    44 mins ago












up vote
4
down vote



accepted







up vote
4
down vote



accepted






Yes it is correct, indeed we have that



  • $(vec a-vec b)times vec n=vec 0 implies vec a-vec b$ is a multiple of $vec n$ that is $vec a-vec b=kvec n$

  • $(vec a-vec b)cdot vec n=0 implies vec a-vec b$ is orthogonal to $vec n$ that is $kvec ncdot vec n=0 implies k=0$

therefore



$$vec a-vec b=vec 0 implies vec a=vec b$$






share|cite|improve this answer












Yes it is correct, indeed we have that



  • $(vec a-vec b)times vec n=vec 0 implies vec a-vec b$ is a multiple of $vec n$ that is $vec a-vec b=kvec n$

  • $(vec a-vec b)cdot vec n=0 implies vec a-vec b$ is orthogonal to $vec n$ that is $kvec ncdot vec n=0 implies k=0$

therefore



$$vec a-vec b=vec 0 implies vec a=vec b$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 52 mins ago









gimusi

73.9k73889




73.9k73889











  • Thanks very much @gimusi !
    – nodarkside
    45 mins ago










  • @nodarkside You are welcome! Bye
    – gimusi
    44 mins ago
















  • Thanks very much @gimusi !
    – nodarkside
    45 mins ago










  • @nodarkside You are welcome! Bye
    – gimusi
    44 mins ago















Thanks very much @gimusi !
– nodarkside
45 mins ago




Thanks very much @gimusi !
– nodarkside
45 mins ago












@nodarkside You are welcome! Bye
– gimusi
44 mins ago




@nodarkside You are welcome! Bye
– gimusi
44 mins ago










up vote
3
down vote













You could also do this using the fact that $vecv cdot vecw = vertvecvrvert lvert vecw rvert costheta$ and $vecv times vecw = vertvecvrvert lvert vecw rvert sintheta$. In your case this leads to
beginalign*
(veca - vecb) cdot vecn = lvertveca - vecbrvert lvertvecnrvert costheta &= 0 \
(veca - vecb) times vecn = lvertveca - vecbrvert lvertvecnrvert sintheta &= 0
endalign*
From here, it is safe to cancel the $lvertvecnrvert$, as it is unit normal. We cannot safely cancel out the $costheta$ or $sintheta$ because these may be $0$, but notice that if $sintheta$ = 0, then $costheta neq 0$ and vice versa. Thus, we are left to the conclusion that $lvert veca - vecb rvert = 0$, so $veca = vecb$.






share|cite|improve this answer










New contributor




Alerra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















  • Thanks very much @Alerra! This proof is also very useful.
    – nodarkside
    7 mins ago














up vote
3
down vote













You could also do this using the fact that $vecv cdot vecw = vertvecvrvert lvert vecw rvert costheta$ and $vecv times vecw = vertvecvrvert lvert vecw rvert sintheta$. In your case this leads to
beginalign*
(veca - vecb) cdot vecn = lvertveca - vecbrvert lvertvecnrvert costheta &= 0 \
(veca - vecb) times vecn = lvertveca - vecbrvert lvertvecnrvert sintheta &= 0
endalign*
From here, it is safe to cancel the $lvertvecnrvert$, as it is unit normal. We cannot safely cancel out the $costheta$ or $sintheta$ because these may be $0$, but notice that if $sintheta$ = 0, then $costheta neq 0$ and vice versa. Thus, we are left to the conclusion that $lvert veca - vecb rvert = 0$, so $veca = vecb$.






share|cite|improve this answer










New contributor




Alerra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















  • Thanks very much @Alerra! This proof is also very useful.
    – nodarkside
    7 mins ago












up vote
3
down vote










up vote
3
down vote









You could also do this using the fact that $vecv cdot vecw = vertvecvrvert lvert vecw rvert costheta$ and $vecv times vecw = vertvecvrvert lvert vecw rvert sintheta$. In your case this leads to
beginalign*
(veca - vecb) cdot vecn = lvertveca - vecbrvert lvertvecnrvert costheta &= 0 \
(veca - vecb) times vecn = lvertveca - vecbrvert lvertvecnrvert sintheta &= 0
endalign*
From here, it is safe to cancel the $lvertvecnrvert$, as it is unit normal. We cannot safely cancel out the $costheta$ or $sintheta$ because these may be $0$, but notice that if $sintheta$ = 0, then $costheta neq 0$ and vice versa. Thus, we are left to the conclusion that $lvert veca - vecb rvert = 0$, so $veca = vecb$.






share|cite|improve this answer










New contributor




Alerra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









You could also do this using the fact that $vecv cdot vecw = vertvecvrvert lvert vecw rvert costheta$ and $vecv times vecw = vertvecvrvert lvert vecw rvert sintheta$. In your case this leads to
beginalign*
(veca - vecb) cdot vecn = lvertveca - vecbrvert lvertvecnrvert costheta &= 0 \
(veca - vecb) times vecn = lvertveca - vecbrvert lvertvecnrvert sintheta &= 0
endalign*
From here, it is safe to cancel the $lvertvecnrvert$, as it is unit normal. We cannot safely cancel out the $costheta$ or $sintheta$ because these may be $0$, but notice that if $sintheta$ = 0, then $costheta neq 0$ and vice versa. Thus, we are left to the conclusion that $lvert veca - vecb rvert = 0$, so $veca = vecb$.







share|cite|improve this answer










New contributor




Alerra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer








edited 29 mins ago





















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answered 42 mins ago









Alerra

1605




1605




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New contributor





Alerra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Alerra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • Thanks very much @Alerra! This proof is also very useful.
    – nodarkside
    7 mins ago
















  • Thanks very much @Alerra! This proof is also very useful.
    – nodarkside
    7 mins ago















Thanks very much @Alerra! This proof is also very useful.
– nodarkside
7 mins ago




Thanks very much @Alerra! This proof is also very useful.
– nodarkside
7 mins ago

















 

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