Mixing
Showing the direction cosines of line perpendicular to two lines direction cosines
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The question is :- if $l_1$, $m_1$, $n_1$ and $l_2$, $m_2$, $n_2$
are the direction cosines of two mutually perpendicular
lines, show that the direction cosines of the line perpendicular to both of these
are ( $m_1n2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1$).
I know that for two mutually perpendicular lines
$$l_1l_2+m_1m_2+n_1n_2=0.$$
But I don't know the further what to do
Please can anyone guide me further?
Thank you.
linear-algebra vectors 3d
edited Aug 3 at 5:21
TRUSKI
5091422
asked Aug 3 at 4:35
Ritik
4310
add a comment |Â
up vote
0
down vote
favorite
The question is :- if $l_1$, $m_1$, $n_1$ and $l_2$, $m_2$, $n_2$
are the direction cosines of two mutually perpendicular
lines, show that the direction cosines of the line perpendicular to both of these
are ( $m_1n2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1$).
I know that for two mutually perpendicular lines
$$l_1l_2+m_1m_2+n_1n_2=0.$$
But I don't know the further what to do
Please can anyone guide me further?
Thank you.
linear-algebra vectors 3d
edited Aug 3 at 5:21
TRUSKI
5091422
asked Aug 3 at 4:35
Ritik
4310
To do a complete proof you have to show that those quantities are not all zero -- one or two may be zero -- using the fact that l1, m1, n1 are not all zero, l2, m2, n2 are not all zero, and that the lines are perpendicular. Once you have done that, use the same test you already know to compare that third line to each of the other two.
– EGoodman
Aug 3 at 4:40
You may find this helpful, as they're essentially attempting to get you to reinvent the cross product: en.wikipedia.org/wiki/Cross_product#Geometric_meaning
– EGoodman
Aug 3 at 4:40
Thank you very much sir I got the answer by your way .
– Ritik
Aug 3 at 4:46
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The question is :- if $l_1$, $m_1$, $n_1$ and $l_2$, $m_2$, $n_2$
are the direction cosines of two mutually perpendicular
lines, show that the direction cosines of the line perpendicular to both of these
are ( $m_1n2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1$).
I know that for two mutually perpendicular lines
$$l_1l_2+m_1m_2+n_1n_2=0.$$
But I don't know the further what to do
Please can anyone guide me further?
Thank you.
linear-algebra vectors 3d
edited Aug 3 at 5:21
TRUSKI
5091422
asked Aug 3 at 4:35
Ritik
4310
The question is :- if $l_1$, $m_1$, $n_1$ and $l_2$, $m_2$, $n_2$
are the direction cosines of two mutually perpendicular
lines, show that the direction cosines of the line perpendicular to both of these
are ( $m_1n2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1$).
I know that for two mutually perpendicular lines
$$l_1l_2+m_1m_2+n_1n_2=0.$$
But I don't know the further what to do
Please can anyone guide me further?
Thank you.
linear-algebra vectors 3d
edited Aug 3 at 5:21
TRUSKI
5091422
asked Aug 3 at 4:35
Ritik
4310
edited Aug 3 at 5:21
TRUSKI
5091422
edited Aug 3 at 5:21
TRUSKI
5091422
edited Aug 3 at 5:21
TRUSKI
5091422
5091422
asked Aug 3 at 4:35
Ritik
4310
asked Aug 3 at 4:35
Ritik
4310
asked Aug 3 at 4:35
Ritik
4310
4310
To do a complete proof you have to show that those quantities are not all zero -- one or two may be zero -- using the fact that l1, m1, n1 are not all zero, l2, m2, n2 are not all zero, and that the lines are perpendicular. Once you have done that, use the same test you already know to compare that third line to each of the other two.
– EGoodman
Aug 3 at 4:40
You may find this helpful, as they're essentially attempting to get you to reinvent the cross product: en.wikipedia.org/wiki/Cross_product#Geometric_meaning
– EGoodman
Aug 3 at 4:40
Thank you very much sir I got the answer by your way .
– Ritik
Aug 3 at 4:46
add a comment |Â
To do a complete proof you have to show that those quantities are not all zero -- one or two may be zero -- using the fact that l1, m1, n1 are not all zero, l2, m2, n2 are not all zero, and that the lines are perpendicular. Once you have done that, use the same test you already know to compare that third line to each of the other two.
– EGoodman
Aug 3 at 4:40
You may find this helpful, as they're essentially attempting to get you to reinvent the cross product: en.wikipedia.org/wiki/Cross_product#Geometric_meaning
– EGoodman
Aug 3 at 4:40
Thank you very much sir I got the answer by your way .
– Ritik
Aug 3 at 4:46
To do a complete proof you have to show that those quantities are not all zero -- one or two may be zero -- using the fact that l1, m1, n1 are not all zero, l2, m2, n2 are not all zero, and that the lines are perpendicular. Once you have done that, use the same test you already know to compare that third line to each of the other two.
– EGoodman
Aug 3 at 4:40
To do a complete proof you have to show that those quantities are not all zero -- one or two may be zero -- using the fact that l1, m1, n1 are not all zero, l2, m2, n2 are not all zero, and that the lines are perpendicular. Once you have done that, use the same test you already know to compare that third line to each of the other two.
– EGoodman
Aug 3 at 4:40
You may find this helpful, as they're essentially attempting to get you to reinvent the cross product: en.wikipedia.org/wiki/Cross_product#Geometric_meaning
– EGoodman
Aug 3 at 4:40
You may find this helpful, as they're essentially attempting to get you to reinvent the cross product: en.wikipedia.org/wiki/Cross_product#Geometric_meaning
– EGoodman
Aug 3 at 4:40
Thank you very much sir I got the answer by your way .
– Ritik
Aug 3 at 4:46
Thank you very much sir I got the answer by your way .
– Ritik
Aug 3 at 4:46
add a comment |Â
1 Answer
1
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oldest
votes
up vote
0
down vote
accepted
Let us take $u_1=(l_1,m_1,n_1)$ as the unit vector along one line.
$u_2=(l_2,m_2,n_2)$ as the unit vector along another line.
Since $u_1times u_2$ is a unit vector perpendicular to both $u_1$ and $u_2$, we can calculate $u_1times u_2$
$$u_1times u_2=beginvmatrix veci & vecj & veck \ l_1 & m_1 & n_1 \ l_2 & m_2 & n_2 endvmatrix$$
$$=veci(m_1n_2-m_2n_1)-vecj(n_2l_1-n_1l_2)+veck(l_1m_2-l_2m_1)$$
$$=veci(m_1n_2-m_2n_1)+vecj(n_1l_2-n_2l_1)+veck(l_1m_2-l_2m_1)$$
Therefore, the required direction cosines are $(m_1n_2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1)$
answered Aug 3 at 4:47
Key Flex
3,673422
Thanks I got the answer
– Ritik
Aug 3 at 4:48
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Let us take $u_1=(l_1,m_1,n_1)$ as the unit vector along one line.
$u_2=(l_2,m_2,n_2)$ as the unit vector along another line.
Since $u_1times u_2$ is a unit vector perpendicular to both $u_1$ and $u_2$, we can calculate $u_1times u_2$
$$u_1times u_2=beginvmatrix veci & vecj & veck \ l_1 & m_1 & n_1 \ l_2 & m_2 & n_2 endvmatrix$$
$$=veci(m_1n_2-m_2n_1)-vecj(n_2l_1-n_1l_2)+veck(l_1m_2-l_2m_1)$$
$$=veci(m_1n_2-m_2n_1)+vecj(n_1l_2-n_2l_1)+veck(l_1m_2-l_2m_1)$$
Therefore, the required direction cosines are $(m_1n_2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1)$
answered Aug 3 at 4:47
Key Flex
3,673422
Thanks I got the answer
– Ritik
Aug 3 at 4:48
add a comment |Â
up vote
0
down vote
accepted
Let us take $u_1=(l_1,m_1,n_1)$ as the unit vector along one line.
$u_2=(l_2,m_2,n_2)$ as the unit vector along another line.
Since $u_1times u_2$ is a unit vector perpendicular to both $u_1$ and $u_2$, we can calculate $u_1times u_2$
$$u_1times u_2=beginvmatrix veci & vecj & veck \ l_1 & m_1 & n_1 \ l_2 & m_2 & n_2 endvmatrix$$
$$=veci(m_1n_2-m_2n_1)-vecj(n_2l_1-n_1l_2)+veck(l_1m_2-l_2m_1)$$
$$=veci(m_1n_2-m_2n_1)+vecj(n_1l_2-n_2l_1)+veck(l_1m_2-l_2m_1)$$
Therefore, the required direction cosines are $(m_1n_2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1)$
answered Aug 3 at 4:47
Key Flex
3,673422
Thanks I got the answer
– Ritik
Aug 3 at 4:48
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let us take $u_1=(l_1,m_1,n_1)$ as the unit vector along one line.
$u_2=(l_2,m_2,n_2)$ as the unit vector along another line.
Since $u_1times u_2$ is a unit vector perpendicular to both $u_1$ and $u_2$, we can calculate $u_1times u_2$
$$u_1times u_2=beginvmatrix veci & vecj & veck \ l_1 & m_1 & n_1 \ l_2 & m_2 & n_2 endvmatrix$$
$$=veci(m_1n_2-m_2n_1)-vecj(n_2l_1-n_1l_2)+veck(l_1m_2-l_2m_1)$$
$$=veci(m_1n_2-m_2n_1)+vecj(n_1l_2-n_2l_1)+veck(l_1m_2-l_2m_1)$$
Therefore, the required direction cosines are $(m_1n_2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1)$
answered Aug 3 at 4:47
Key Flex
3,673422
Let us take $u_1=(l_1,m_1,n_1)$ as the unit vector along one line.
$u_2=(l_2,m_2,n_2)$ as the unit vector along another line.
Since $u_1times u_2$ is a unit vector perpendicular to both $u_1$ and $u_2$, we can calculate $u_1times u_2$
$$u_1times u_2=beginvmatrix veci & vecj & veck \ l_1 & m_1 & n_1 \ l_2 & m_2 & n_2 endvmatrix$$
$$=veci(m_1n_2-m_2n_1)-vecj(n_2l_1-n_1l_2)+veck(l_1m_2-l_2m_1)$$
$$=veci(m_1n_2-m_2n_1)+vecj(n_1l_2-n_2l_1)+veck(l_1m_2-l_2m_1)$$
Therefore, the required direction cosines are $(m_1n_2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1)$
answered Aug 3 at 4:47
Key Flex
3,673422
answered Aug 3 at 4:47
Key Flex
3,673422
answered Aug 3 at 4:47
Key Flex
3,673422
answered Aug 3 at 4:47
Key Flex
3,673422
3,673422
Thanks I got the answer
– Ritik
Aug 3 at 4:48
add a comment |Â
Thanks I got the answer
– Ritik
Aug 3 at 4:48
Thanks I got the answer
– Ritik
Aug 3 at 4:48
Thanks I got the answer
– Ritik
Aug 3 at 4:48
add a comment |Â
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To do a complete proof you have to show that those quantities are not all zero -- one or two may be zero -- using the fact that l1, m1, n1 are not all zero, l2, m2, n2 are not all zero, and that the lines are perpendicular. Once you have done that, use the same test you already know to compare that third line to each of the other two.
– EGoodman
Aug 3 at 4:40
You may find this helpful, as they're essentially attempting to get you to reinvent the cross product: en.wikipedia.org/wiki/Cross_product#Geometric_meaning
– EGoodman
Aug 3 at 4:40
Thank you very much sir I got the answer by your way .
– Ritik
Aug 3 at 4:46