Vectors proof: Find all planes passing through intersection of two planes

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I would appreciate help on the vectors chapter of bostock and chandler core course. They give the following proof purportedly finding all planes passing through the intersection line of two nonparallel planes. The two planes are $mathbfrbullet hat n_1=d_1$ and $mathbfrbullet hat n_2=d_2$ where $bf hat n_1, hat n_2$ are the unit vectors perpendicular to the planes. The proof given is as follows:



Proof. Multiply the second equation by an arbitrary number $k$ to get $mathbfrbulletkmathbfhat n_2=kd_2$ and subtract this equation from the first one to get $mathbfrbullet (hat n_1-kmathbfhat n_2)=d_1-kd_2$. This is a general equation giving all planes passing through the intersection of the first two planes. $square$



The problem I have with this proof is that the conclusion seems to be more than the authors prove. It seems that they only prove that if there are two planes, then the last equation gives infinitely many planes passing through their intersection. However, I don't see how the proof shows that this gives all planes passing through their intersection. I attempted to prove this latter result as follows:



I claim that $mathbfrbullet (hat n_1-kmathbfhat n_2)=d_1-kd_2$ gives all such planes.



Proof: Let $mathbfrbullet hat n_3=d_3$ be an arbitrary such plane. Since $d_1-kd_2$ can represent any number, it suffices to show that $mathbfhat n_1-kmathbfhat n_2$ can represent $mathbfhat n_3$.



Proof of this: Note that $mathbfhat n_1,hat n_2$ are perpendicular to the line of intersection and hence they form (ie as base vectors) the plane perpendicular to it. This plane contains all perpendicular vectors of the planes passing through the line of intersection, hence it includes $mathbfhat n_3$. Thus $mathbfhat n_3=amathbfhat n_1+bmathbfhat n_2$ for some a,b, giving $frac1amathbfhat n_3=mathbfhat n_1-(-fracba)mathbfhat n_2$. It follows that $mathbfrbullet hat n_3=d_3$ can be written as $mathbfrbullet(hat n_1-(-fracba)mathbfhat n_2)=frac1ad_3$ and we are done. $square$



I am mainly looking for assurance that (1) my proof is correct, and (2) what I have done is not just an elaboration of something that can be "seen" trivially.



Thank you very much




algebra-precalculus proof-verification vectors




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    I would appreciate help on the vectors chapter of bostock and chandler core course. They give the following proof purportedly finding all planes passing through the intersection line of two nonparallel planes. The two planes are $mathbfrbullet hat n_1=d_1$ and $mathbfrbullet hat n_2=d_2$ where $bf hat n_1, hat n_2$ are the unit vectors perpendicular to the planes. The proof given is as follows:



    Proof. Multiply the second equation by an arbitrary number $k$ to get $mathbfrbulletkmathbfhat n_2=kd_2$ and subtract this equation from the first one to get $mathbfrbullet (hat n_1-kmathbfhat n_2)=d_1-kd_2$. This is a general equation giving all planes passing through the intersection of the first two planes. $square$



    The problem I have with this proof is that the conclusion seems to be more than the authors prove. It seems that they only prove that if there are two planes, then the last equation gives infinitely many planes passing through their intersection. However, I don't see how the proof shows that this gives all planes passing through their intersection. I attempted to prove this latter result as follows:



    I claim that $mathbfrbullet (hat n_1-kmathbfhat n_2)=d_1-kd_2$ gives all such planes.



    Proof: Let $mathbfrbullet hat n_3=d_3$ be an arbitrary such plane. Since $d_1-kd_2$ can represent any number, it suffices to show that $mathbfhat n_1-kmathbfhat n_2$ can represent $mathbfhat n_3$.



    Proof of this: Note that $mathbfhat n_1,hat n_2$ are perpendicular to the line of intersection and hence they form (ie as base vectors) the plane perpendicular to it. This plane contains all perpendicular vectors of the planes passing through the line of intersection, hence it includes $mathbfhat n_3$. Thus $mathbfhat n_3=amathbfhat n_1+bmathbfhat n_2$ for some a,b, giving $frac1amathbfhat n_3=mathbfhat n_1-(-fracba)mathbfhat n_2$. It follows that $mathbfrbullet hat n_3=d_3$ can be written as $mathbfrbullet(hat n_1-(-fracba)mathbfhat n_2)=frac1ad_3$ and we are done. $square$



    I am mainly looking for assurance that (1) my proof is correct, and (2) what I have done is not just an elaboration of something that can be "seen" trivially.



    Thank you very much




    algebra-precalculus proof-verification vectors




    share|cite|improve this question























      up vote
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      up vote
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      down vote

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      I would appreciate help on the vectors chapter of bostock and chandler core course. They give the following proof purportedly finding all planes passing through the intersection line of two nonparallel planes. The two planes are $mathbfrbullet hat n_1=d_1$ and $mathbfrbullet hat n_2=d_2$ where $bf hat n_1, hat n_2$ are the unit vectors perpendicular to the planes. The proof given is as follows:



      Proof. Multiply the second equation by an arbitrary number $k$ to get $mathbfrbulletkmathbfhat n_2=kd_2$ and subtract this equation from the first one to get $mathbfrbullet (hat n_1-kmathbfhat n_2)=d_1-kd_2$. This is a general equation giving all planes passing through the intersection of the first two planes. $square$



      The problem I have with this proof is that the conclusion seems to be more than the authors prove. It seems that they only prove that if there are two planes, then the last equation gives infinitely many planes passing through their intersection. However, I don't see how the proof shows that this gives all planes passing through their intersection. I attempted to prove this latter result as follows:



      I claim that $mathbfrbullet (hat n_1-kmathbfhat n_2)=d_1-kd_2$ gives all such planes.



      Proof: Let $mathbfrbullet hat n_3=d_3$ be an arbitrary such plane. Since $d_1-kd_2$ can represent any number, it suffices to show that $mathbfhat n_1-kmathbfhat n_2$ can represent $mathbfhat n_3$.



      Proof of this: Note that $mathbfhat n_1,hat n_2$ are perpendicular to the line of intersection and hence they form (ie as base vectors) the plane perpendicular to it. This plane contains all perpendicular vectors of the planes passing through the line of intersection, hence it includes $mathbfhat n_3$. Thus $mathbfhat n_3=amathbfhat n_1+bmathbfhat n_2$ for some a,b, giving $frac1amathbfhat n_3=mathbfhat n_1-(-fracba)mathbfhat n_2$. It follows that $mathbfrbullet hat n_3=d_3$ can be written as $mathbfrbullet(hat n_1-(-fracba)mathbfhat n_2)=frac1ad_3$ and we are done. $square$



      I am mainly looking for assurance that (1) my proof is correct, and (2) what I have done is not just an elaboration of something that can be "seen" trivially.



      Thank you very much




      algebra-precalculus proof-verification vectors




      share|cite|improve this question













      I would appreciate help on the vectors chapter of bostock and chandler core course. They give the following proof purportedly finding all planes passing through the intersection line of two nonparallel planes. The two planes are $mathbfrbullet hat n_1=d_1$ and $mathbfrbullet hat n_2=d_2$ where $bf hat n_1, hat n_2$ are the unit vectors perpendicular to the planes. The proof given is as follows:



      Proof. Multiply the second equation by an arbitrary number $k$ to get $mathbfrbulletkmathbfhat n_2=kd_2$ and subtract this equation from the first one to get $mathbfrbullet (hat n_1-kmathbfhat n_2)=d_1-kd_2$. This is a general equation giving all planes passing through the intersection of the first two planes. $square$



      The problem I have with this proof is that the conclusion seems to be more than the authors prove. It seems that they only prove that if there are two planes, then the last equation gives infinitely many planes passing through their intersection. However, I don't see how the proof shows that this gives all planes passing through their intersection. I attempted to prove this latter result as follows:



      I claim that $mathbfrbullet (hat n_1-kmathbfhat n_2)=d_1-kd_2$ gives all such planes.



      Proof: Let $mathbfrbullet hat n_3=d_3$ be an arbitrary such plane. Since $d_1-kd_2$ can represent any number, it suffices to show that $mathbfhat n_1-kmathbfhat n_2$ can represent $mathbfhat n_3$.



      Proof of this: Note that $mathbfhat n_1,hat n_2$ are perpendicular to the line of intersection and hence they form (ie as base vectors) the plane perpendicular to it. This plane contains all perpendicular vectors of the planes passing through the line of intersection, hence it includes $mathbfhat n_3$. Thus $mathbfhat n_3=amathbfhat n_1+bmathbfhat n_2$ for some a,b, giving $frac1amathbfhat n_3=mathbfhat n_1-(-fracba)mathbfhat n_2$. It follows that $mathbfrbullet hat n_3=d_3$ can be written as $mathbfrbullet(hat n_1-(-fracba)mathbfhat n_2)=frac1ad_3$ and we are done. $square$



      I am mainly looking for assurance that (1) my proof is correct, and (2) what I have done is not just an elaboration of something that can be "seen" trivially.



      Thank you very much





      algebra-precalculus proof-verification vectors





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