Uniqueness of line through a point, perpendicular with an affine space
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Consider the affine space $D=q+D_0$, with $pin mathbb{R}^nbackslash D$ and $L$ the perpendicular line through $p$ on $D$. This line $L$ is unique, i.e. $L$ is the unique line through $p$, perpendicular with $D$, such that $L cap Dne emptyset $.
Proof: Consider an arbitrary line $M$ through $p$, perpendicular with $D$, with $M cap D ne emptyset$. Assume $M cap D = {z}$. Then $underline{M subseteq p+D_0^perp}$, and therefore $underline{z in (p+D_0^perp)cap D}$. But then we have $underline{D=z+D_0} $ and $underline{p+D_0^perp=z+D_0^perp}$, which imply: $(p+D_0^perp)cap D=(z+D_0^perp)cap(z+D_0) = z+(D_0^perpcap D_0) = {z}.$
We can conclude that $z$ is independent of the choice of $M$, and therefore $M$ is the unique line through $p$ and $z$.
This is the only proof in my book that I don't understand completely. I'm taking an exam tomorrow, so I would be infinitely grateful if someone could explain to me where the underlined parts come from.
linear-algebra
asked Jan 13 at 20:23
ZacharyZachary
1559
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$begingroup$
Consider the affine space $D=q+D_0$, with $pin mathbb{R}^nbackslash D$ and $L$ the perpendicular line through $p$ on $D$. This line $L$ is unique, i.e. $L$ is the unique line through $p$, perpendicular with $D$, such that $L cap Dne emptyset $.
Proof: Consider an arbitrary line $M$ through $p$, perpendicular with $D$, with $M cap D ne emptyset$. Assume $M cap D = {z}$. Then $underline{M subseteq p+D_0^perp}$, and therefore $underline{z in (p+D_0^perp)cap D}$. But then we have $underline{D=z+D_0} $ and $underline{p+D_0^perp=z+D_0^perp}$, which imply: $(p+D_0^perp)cap D=(z+D_0^perp)cap(z+D_0) = z+(D_0^perpcap D_0) = {z}.$
We can conclude that $z$ is independent of the choice of $M$, and therefore $M$ is the unique line through $p$ and $z$.
This is the only proof in my book that I don't understand completely. I'm taking an exam tomorrow, so I would be infinitely grateful if someone could explain to me where the underlined parts come from.
linear-algebra
asked Jan 13 at 20:23
ZacharyZachary
1559
$endgroup$
add a comment |
$begingroup$
Consider the affine space $D=q+D_0$, with $pin mathbb{R}^nbackslash D$ and $L$ the perpendicular line through $p$ on $D$. This line $L$ is unique, i.e. $L$ is the unique line through $p$, perpendicular with $D$, such that $L cap Dne emptyset $.
Proof: Consider an arbitrary line $M$ through $p$, perpendicular with $D$, with $M cap D ne emptyset$. Assume $M cap D = {z}$. Then $underline{M subseteq p+D_0^perp}$, and therefore $underline{z in (p+D_0^perp)cap D}$. But then we have $underline{D=z+D_0} $ and $underline{p+D_0^perp=z+D_0^perp}$, which imply: $(p+D_0^perp)cap D=(z+D_0^perp)cap(z+D_0) = z+(D_0^perpcap D_0) = {z}.$
We can conclude that $z$ is independent of the choice of $M$, and therefore $M$ is the unique line through $p$ and $z$.
This is the only proof in my book that I don't understand completely. I'm taking an exam tomorrow, so I would be infinitely grateful if someone could explain to me where the underlined parts come from.
linear-algebra
asked Jan 13 at 20:23
ZacharyZachary
1559
$endgroup$
Consider the affine space $D=q+D_0$, with $pin mathbb{R}^nbackslash D$ and $L$ the perpendicular line through $p$ on $D$. This line $L$ is unique, i.e. $L$ is the unique line through $p$, perpendicular with $D$, such that $L cap Dne emptyset $.
Proof: Consider an arbitrary line $M$ through $p$, perpendicular with $D$, with $M cap D ne emptyset$. Assume $M cap D = {z}$. Then $underline{M subseteq p+D_0^perp}$, and therefore $underline{z in (p+D_0^perp)cap D}$. But then we have $underline{D=z+D_0} $ and $underline{p+D_0^perp=z+D_0^perp}$, which imply: $(p+D_0^perp)cap D=(z+D_0^perp)cap(z+D_0) = z+(D_0^perpcap D_0) = {z}.$
We can conclude that $z$ is independent of the choice of $M$, and therefore $M$ is the unique line through $p$ and $z$.
This is the only proof in my book that I don't understand completely. I'm taking an exam tomorrow, so I would be infinitely grateful if someone could explain to me where the underlined parts come from.
linear-algebra
linear-algebra
asked Jan 13 at 20:23
ZacharyZachary
1559
asked Jan 13 at 20:23
ZacharyZachary
1559
asked Jan 13 at 20:23
ZacharyZachary
1559
asked Jan 13 at 20:23
ZacharyZachary
1559
asked Jan 13 at 20:23
ZacharyZachary
1559
1559
add a comment |
add a comment |
1 Answer
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The first underline simply expresses that $M$ is perpendicular to $D$, from which the second underline follows immediately ($Asubseteq B land ain Aimplies ain B$). But then, since $ain Acap Bimplies ain Aland ain B$, you also have that $zin p+D_0^perp$, so just as with $D=z+D_0$, this affine space can also be expressed as $z+D_0^perp$.
answered Jan 22 at 1:25
amdamd
29.7k21050
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$begingroup$
The first underline simply expresses that $M$ is perpendicular to $D$, from which the second underline follows immediately ($Asubseteq B land ain Aimplies ain B$). But then, since $ain Acap Bimplies ain Aland ain B$, you also have that $zin p+D_0^perp$, so just as with $D=z+D_0$, this affine space can also be expressed as $z+D_0^perp$.
answered Jan 22 at 1:25
amdamd
29.7k21050
$endgroup$
add a comment |
$begingroup$
The first underline simply expresses that $M$ is perpendicular to $D$, from which the second underline follows immediately ($Asubseteq B land ain Aimplies ain B$). But then, since $ain Acap Bimplies ain Aland ain B$, you also have that $zin p+D_0^perp$, so just as with $D=z+D_0$, this affine space can also be expressed as $z+D_0^perp$.
answered Jan 22 at 1:25
amdamd
29.7k21050
$endgroup$
add a comment |
$begingroup$
The first underline simply expresses that $M$ is perpendicular to $D$, from which the second underline follows immediately ($Asubseteq B land ain Aimplies ain B$). But then, since $ain Acap Bimplies ain Aland ain B$, you also have that $zin p+D_0^perp$, so just as with $D=z+D_0$, this affine space can also be expressed as $z+D_0^perp$.
answered Jan 22 at 1:25
amdamd
29.7k21050
$endgroup$
The first underline simply expresses that $M$ is perpendicular to $D$, from which the second underline follows immediately ($Asubseteq B land ain Aimplies ain B$). But then, since $ain Acap Bimplies ain Aland ain B$, you also have that $zin p+D_0^perp$, so just as with $D=z+D_0$, this affine space can also be expressed as $z+D_0^perp$.
answered Jan 22 at 1:25
amdamd
29.7k21050
answered Jan 22 at 1:25
amdamd
29.7k21050
answered Jan 22 at 1:25
amdamd
29.7k21050
answered Jan 22 at 1:25
amdamd
29.7k21050
29.7k21050
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