How to find the direction cosines of the line which bisects the angle positive degrees between positive...












0












$begingroup$


I have a text book solution like -



The line bisects angle between Y and Z axes. Therefore the line lies in the YZ plane. Hence the X-axis is perpendicular to the line.



Now how come X-axis is perpendicular to the line in question if the fig. for this problem is like below where X-axis clearly doesn't look perpendicular to the line in question.



enter image description here






linear-algebra geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Figures are generally not drawn to scale
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 16:50










  • $begingroup$
    agreed...but there should at least be some coherence to what is being claimed in textbook answer, because if the bisecting line has to be perpendicular to x-axis it cannot exist in yz plane but rather in xy plane..please tell me how i am wrong?
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 16:55
















0












$begingroup$


I have a text book solution like -



The line bisects angle between Y and Z axes. Therefore the line lies in the YZ plane. Hence the X-axis is perpendicular to the line.



Now how come X-axis is perpendicular to the line in question if the fig. for this problem is like below where X-axis clearly doesn't look perpendicular to the line in question.



enter image description here






linear-algebra geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Figures are generally not drawn to scale
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 16:50










  • $begingroup$
    agreed...but there should at least be some coherence to what is being claimed in textbook answer, because if the bisecting line has to be perpendicular to x-axis it cannot exist in yz plane but rather in xy plane..please tell me how i am wrong?
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 16:55














0












0








0





$begingroup$


I have a text book solution like -



The line bisects angle between Y and Z axes. Therefore the line lies in the YZ plane. Hence the X-axis is perpendicular to the line.



Now how come X-axis is perpendicular to the line in question if the fig. for this problem is like below where X-axis clearly doesn't look perpendicular to the line in question.



enter image description here






linear-algebra geometry







share|cite|improve this question









$endgroup$




I have a text book solution like -



The line bisects angle between Y and Z axes. Therefore the line lies in the YZ plane. Hence the X-axis is perpendicular to the line.



Now how come X-axis is perpendicular to the line in question if the fig. for this problem is like below where X-axis clearly doesn't look perpendicular to the line in question.



enter image description here






linear-algebra geometry




linear-algebra geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 16:40









Shaikh SakibShaikh Sakib

195




195












  • $begingroup$
    Figures are generally not drawn to scale
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 16:50










  • $begingroup$
    agreed...but there should at least be some coherence to what is being claimed in textbook answer, because if the bisecting line has to be perpendicular to x-axis it cannot exist in yz plane but rather in xy plane..please tell me how i am wrong?
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 16:55


















  • $begingroup$
    Figures are generally not drawn to scale
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 16:50










  • $begingroup$
    agreed...but there should at least be some coherence to what is being claimed in textbook answer, because if the bisecting line has to be perpendicular to x-axis it cannot exist in yz plane but rather in xy plane..please tell me how i am wrong?
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 16:55
















$begingroup$
Figures are generally not drawn to scale
$endgroup$
– Shubham Johri
Dec 11 '18 at 16:50




$begingroup$
Figures are generally not drawn to scale
$endgroup$
– Shubham Johri
Dec 11 '18 at 16:50












$begingroup$
agreed...but there should at least be some coherence to what is being claimed in textbook answer, because if the bisecting line has to be perpendicular to x-axis it cannot exist in yz plane but rather in xy plane..please tell me how i am wrong?
$endgroup$
– Shaikh Sakib
Dec 11 '18 at 16:55




$begingroup$
agreed...but there should at least be some coherence to what is being claimed in textbook answer, because if the bisecting line has to be perpendicular to x-axis it cannot exist in yz plane but rather in xy plane..please tell me how i am wrong?
$endgroup$
– Shaikh Sakib
Dec 11 '18 at 16:55










1 Answer
1






active

oldest

votes


















2












$begingroup$

Look at this photo of a book:



enter image description here
The two edges of the book meeting at a corner do not meet at 90 degrees in the photo.



What we learn from this simple example is the projection from 3 dimensions to 2 dimensions generally does not preserve angles.



On the other hand, the $x$-axis is perpendicular to both the $y$ and $z$ axes, and hence (by bi-linearity of dot product, if you like) is perpendicular to any linear combination of them, hence to their bisector.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok by that logic even if the bisecting line exists in xy or xz plane it will still be perpendicular to all the axes from a 3d perspective..like that of a cube
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 16:59










  • $begingroup$
    @ShaikhSakib: yes, the line perpendicular to the plane is perpendicular to any line in that plane.
    $endgroup$
    – Vasya
    Dec 11 '18 at 17:01










  • $begingroup$
    hmm..that explains it..have accepted your answer @JohnHughes. Thank you.
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 17:03











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Look at this photo of a book:



enter image description here
The two edges of the book meeting at a corner do not meet at 90 degrees in the photo.



What we learn from this simple example is the projection from 3 dimensions to 2 dimensions generally does not preserve angles.



On the other hand, the $x$-axis is perpendicular to both the $y$ and $z$ axes, and hence (by bi-linearity of dot product, if you like) is perpendicular to any linear combination of them, hence to their bisector.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok by that logic even if the bisecting line exists in xy or xz plane it will still be perpendicular to all the axes from a 3d perspective..like that of a cube
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 16:59










  • $begingroup$
    @ShaikhSakib: yes, the line perpendicular to the plane is perpendicular to any line in that plane.
    $endgroup$
    – Vasya
    Dec 11 '18 at 17:01










  • $begingroup$
    hmm..that explains it..have accepted your answer @JohnHughes. Thank you.
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 17:03
















2












$begingroup$

Look at this photo of a book:



enter image description here
The two edges of the book meeting at a corner do not meet at 90 degrees in the photo.



What we learn from this simple example is the projection from 3 dimensions to 2 dimensions generally does not preserve angles.



On the other hand, the $x$-axis is perpendicular to both the $y$ and $z$ axes, and hence (by bi-linearity of dot product, if you like) is perpendicular to any linear combination of them, hence to their bisector.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok by that logic even if the bisecting line exists in xy or xz plane it will still be perpendicular to all the axes from a 3d perspective..like that of a cube
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 16:59










  • $begingroup$
    @ShaikhSakib: yes, the line perpendicular to the plane is perpendicular to any line in that plane.
    $endgroup$
    – Vasya
    Dec 11 '18 at 17:01










  • $begingroup$
    hmm..that explains it..have accepted your answer @JohnHughes. Thank you.
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 17:03














2












2








2





$begingroup$

Look at this photo of a book:



enter image description here
The two edges of the book meeting at a corner do not meet at 90 degrees in the photo.



What we learn from this simple example is the projection from 3 dimensions to 2 dimensions generally does not preserve angles.



On the other hand, the $x$-axis is perpendicular to both the $y$ and $z$ axes, and hence (by bi-linearity of dot product, if you like) is perpendicular to any linear combination of them, hence to their bisector.






share|cite|improve this answer









$endgroup$



Look at this photo of a book:



enter image description here
The two edges of the book meeting at a corner do not meet at 90 degrees in the photo.



What we learn from this simple example is the projection from 3 dimensions to 2 dimensions generally does not preserve angles.



On the other hand, the $x$-axis is perpendicular to both the $y$ and $z$ axes, and hence (by bi-linearity of dot product, if you like) is perpendicular to any linear combination of them, hence to their bisector.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 11 '18 at 16:56









John HughesJohn Hughes

64.2k24191




64.2k24191












  • $begingroup$
    ok by that logic even if the bisecting line exists in xy or xz plane it will still be perpendicular to all the axes from a 3d perspective..like that of a cube
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 16:59










  • $begingroup$
    @ShaikhSakib: yes, the line perpendicular to the plane is perpendicular to any line in that plane.
    $endgroup$
    – Vasya
    Dec 11 '18 at 17:01










  • $begingroup$
    hmm..that explains it..have accepted your answer @JohnHughes. Thank you.
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 17:03


















  • $begingroup$
    ok by that logic even if the bisecting line exists in xy or xz plane it will still be perpendicular to all the axes from a 3d perspective..like that of a cube
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 16:59










  • $begingroup$
    @ShaikhSakib: yes, the line perpendicular to the plane is perpendicular to any line in that plane.
    $endgroup$
    – Vasya
    Dec 11 '18 at 17:01










  • $begingroup$
    hmm..that explains it..have accepted your answer @JohnHughes. Thank you.
    $endgroup$
    – Shaikh Sakib
    Dec 11 '18 at 17:03
















$begingroup$
ok by that logic even if the bisecting line exists in xy or xz plane it will still be perpendicular to all the axes from a 3d perspective..like that of a cube
$endgroup$
– Shaikh Sakib
Dec 11 '18 at 16:59




$begingroup$
ok by that logic even if the bisecting line exists in xy or xz plane it will still be perpendicular to all the axes from a 3d perspective..like that of a cube
$endgroup$
– Shaikh Sakib
Dec 11 '18 at 16:59












$begingroup$
@ShaikhSakib: yes, the line perpendicular to the plane is perpendicular to any line in that plane.
$endgroup$
– Vasya
Dec 11 '18 at 17:01




$begingroup$
@ShaikhSakib: yes, the line perpendicular to the plane is perpendicular to any line in that plane.
$endgroup$
– Vasya
Dec 11 '18 at 17:01












$begingroup$
hmm..that explains it..have accepted your answer @JohnHughes. Thank you.
$endgroup$
– Shaikh Sakib
Dec 11 '18 at 17:03




$begingroup$
hmm..that explains it..have accepted your answer @JohnHughes. Thank you.
$endgroup$
– Shaikh Sakib
Dec 11 '18 at 17:03


















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