Mixing
Show that two lines are perpendicular in the complex plane.
Clash Royale CLAN TAG#URR8PPP
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1
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I want to show that
$overlineABperpoverlineCD$ if and only if $bf Redfraca-bc-d=0$, where $a, b, c, d$ are the complex numbers corresponding to the points $A$, $B$, $C$ and $D$.
I am yet to find a proof for this. Could someone give me some hints?
complex-analysis complex-numbers proof-writing
edited Jul 29 at 18:14
Ted Shifrin
59.4k44386
asked Jul 28 at 18:40
tapsomilian
304
add a comment |Â
up vote
1
down vote
favorite
I want to show that
$overlineABperpoverlineCD$ if and only if $bf Redfraca-bc-d=0$, where $a, b, c, d$ are the complex numbers corresponding to the points $A$, $B$, $C$ and $D$.
I am yet to find a proof for this. Could someone give me some hints?
complex-analysis complex-numbers proof-writing
edited Jul 29 at 18:14
Ted Shifrin
59.4k44386
asked Jul 28 at 18:40
tapsomilian
304
4
Hint: $argbig((a-b)/(c-d)big)$ is the angle between $AB$ and $CD$.
– dxiv
Jul 28 at 18:42
1
After working out, a very long way is to assign complex numbers $a$=$a_1+a_2i$ etc for$A,B,C$ and $D$ and work out $Refraca-bc-d$ and multiply with its conjugate. Compare that expression to the expression by setting up slopes (rise over run) and using the fact that the product of two perpendicular lines is $-1$. It isn't the prettiest way to show, there has to be a quicker and more elegant way, perhaps the suggestion of dxiv
– imranfat
Jul 28 at 18:51
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to show that
$overlineABperpoverlineCD$ if and only if $bf Redfraca-bc-d=0$, where $a, b, c, d$ are the complex numbers corresponding to the points $A$, $B$, $C$ and $D$.
I am yet to find a proof for this. Could someone give me some hints?
complex-analysis complex-numbers proof-writing
edited Jul 29 at 18:14
Ted Shifrin
59.4k44386
asked Jul 28 at 18:40
tapsomilian
304
I want to show that
$overlineABperpoverlineCD$ if and only if $bf Redfraca-bc-d=0$, where $a, b, c, d$ are the complex numbers corresponding to the points $A$, $B$, $C$ and $D$.
I am yet to find a proof for this. Could someone give me some hints?
complex-analysis complex-numbers proof-writing
edited Jul 29 at 18:14
Ted Shifrin
59.4k44386
asked Jul 28 at 18:40
tapsomilian
304
edited Jul 29 at 18:14
Ted Shifrin
59.4k44386
edited Jul 29 at 18:14
Ted Shifrin
59.4k44386
edited Jul 29 at 18:14
Ted Shifrin
59.4k44386
59.4k44386
asked Jul 28 at 18:40
tapsomilian
304
asked Jul 28 at 18:40
tapsomilian
304
asked Jul 28 at 18:40
tapsomilian
304
304
4
Hint: $argbig((a-b)/(c-d)big)$ is the angle between $AB$ and $CD$.
– dxiv
Jul 28 at 18:42
1
After working out, a very long way is to assign complex numbers $a$=$a_1+a_2i$ etc for$A,B,C$ and $D$ and work out $Refraca-bc-d$ and multiply with its conjugate. Compare that expression to the expression by setting up slopes (rise over run) and using the fact that the product of two perpendicular lines is $-1$. It isn't the prettiest way to show, there has to be a quicker and more elegant way, perhaps the suggestion of dxiv
– imranfat
Jul 28 at 18:51
add a comment |Â
4
Hint: $argbig((a-b)/(c-d)big)$ is the angle between $AB$ and $CD$.
– dxiv
Jul 28 at 18:42
1
After working out, a very long way is to assign complex numbers $a$=$a_1+a_2i$ etc for$A,B,C$ and $D$ and work out $Refraca-bc-d$ and multiply with its conjugate. Compare that expression to the expression by setting up slopes (rise over run) and using the fact that the product of two perpendicular lines is $-1$. It isn't the prettiest way to show, there has to be a quicker and more elegant way, perhaps the suggestion of dxiv
– imranfat
Jul 28 at 18:51
4
4
Hint: $argbig((a-b)/(c-d)big)$ is the angle between $AB$ and $CD$.
– dxiv
Jul 28 at 18:42
Hint: $argbig((a-b)/(c-d)big)$ is the angle between $AB$ and $CD$.
– dxiv
Jul 28 at 18:42
1
1
After working out, a very long way is to assign complex numbers $a$=$a_1+a_2i$ etc for$A,B,C$ and $D$ and work out $Refraca-bc-d$ and multiply with its conjugate. Compare that expression to the expression by setting up slopes (rise over run) and using the fact that the product of two perpendicular lines is $-1$. It isn't the prettiest way to show, there has to be a quicker and more elegant way, perhaps the suggestion of dxiv
– imranfat
Jul 28 at 18:51
After working out, a very long way is to assign complex numbers $a$=$a_1+a_2i$ etc for$A,B,C$ and $D$ and work out $Refraca-bc-d$ and multiply with its conjugate. Compare that expression to the expression by setting up slopes (rise over run) and using the fact that the product of two perpendicular lines is $-1$. It isn't the prettiest way to show, there has to be a quicker and more elegant way, perhaps the suggestion of dxiv
– imranfat
Jul 28 at 18:51
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
We know the angle of vector $a-b$ with $x$-axis is $arg(a-b)$ and also the angle of vector $c-d$ with $x$-axis is $arg(c-d)$, then they are perpendicular if
$$arg(a-b)-arg(c-d)=dfracpi2$$
or
$$argdfraca-bc-d=dfracpi2$$
this shows that the number $dfraca-bc-d$ is purely imaginary, therfore $bf Redfraca-bc-d=0$.
answered Jul 29 at 1:09
user 108128
19k41544
add a comment |Â
up vote
0
down vote
The lines have direction vectors
$$
AB = b-a quad CD = d-c
$$
We have
$$
DeclareMathOperatorReRe
Re fracb-ad-c
= Re fracr_1 e^i phi_1r_2 e^i phi_2
= Re fracr_1r_2 e^i(phi_1 - phi_2)
= fracr_1r_2 cos(phi_1 - phi_2)
$$
where we switched to polar coordinates.
answered Jul 28 at 21:14
mvw
30.2k22250
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We know the angle of vector $a-b$ with $x$-axis is $arg(a-b)$ and also the angle of vector $c-d$ with $x$-axis is $arg(c-d)$, then they are perpendicular if
$$arg(a-b)-arg(c-d)=dfracpi2$$
or
$$argdfraca-bc-d=dfracpi2$$
this shows that the number $dfraca-bc-d$ is purely imaginary, therfore $bf Redfraca-bc-d=0$.
answered Jul 29 at 1:09
user 108128
19k41544
add a comment |Â
up vote
1
down vote
accepted
We know the angle of vector $a-b$ with $x$-axis is $arg(a-b)$ and also the angle of vector $c-d$ with $x$-axis is $arg(c-d)$, then they are perpendicular if
$$arg(a-b)-arg(c-d)=dfracpi2$$
or
$$argdfraca-bc-d=dfracpi2$$
this shows that the number $dfraca-bc-d$ is purely imaginary, therfore $bf Redfraca-bc-d=0$.
answered Jul 29 at 1:09
user 108128
19k41544
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We know the angle of vector $a-b$ with $x$-axis is $arg(a-b)$ and also the angle of vector $c-d$ with $x$-axis is $arg(c-d)$, then they are perpendicular if
$$arg(a-b)-arg(c-d)=dfracpi2$$
or
$$argdfraca-bc-d=dfracpi2$$
this shows that the number $dfraca-bc-d$ is purely imaginary, therfore $bf Redfraca-bc-d=0$.
answered Jul 29 at 1:09
user 108128
19k41544
We know the angle of vector $a-b$ with $x$-axis is $arg(a-b)$ and also the angle of vector $c-d$ with $x$-axis is $arg(c-d)$, then they are perpendicular if
$$arg(a-b)-arg(c-d)=dfracpi2$$
or
$$argdfraca-bc-d=dfracpi2$$
this shows that the number $dfraca-bc-d$ is purely imaginary, therfore $bf Redfraca-bc-d=0$.
answered Jul 29 at 1:09
user 108128
19k41544
answered Jul 29 at 1:09
user 108128
19k41544
answered Jul 29 at 1:09
user 108128
19k41544
answered Jul 29 at 1:09
user 108128
19k41544
19k41544
add a comment |Â
add a comment |Â
up vote
0
down vote
The lines have direction vectors
$$
AB = b-a quad CD = d-c
$$
We have
$$
DeclareMathOperatorReRe
Re fracb-ad-c
= Re fracr_1 e^i phi_1r_2 e^i phi_2
= Re fracr_1r_2 e^i(phi_1 - phi_2)
= fracr_1r_2 cos(phi_1 - phi_2)
$$
where we switched to polar coordinates.
answered Jul 28 at 21:14
mvw
30.2k22250
add a comment |Â
up vote
0
down vote
The lines have direction vectors
$$
AB = b-a quad CD = d-c
$$
We have
$$
DeclareMathOperatorReRe
Re fracb-ad-c
= Re fracr_1 e^i phi_1r_2 e^i phi_2
= Re fracr_1r_2 e^i(phi_1 - phi_2)
= fracr_1r_2 cos(phi_1 - phi_2)
$$
where we switched to polar coordinates.
answered Jul 28 at 21:14
mvw
30.2k22250
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The lines have direction vectors
$$
AB = b-a quad CD = d-c
$$
We have
$$
DeclareMathOperatorReRe
Re fracb-ad-c
= Re fracr_1 e^i phi_1r_2 e^i phi_2
= Re fracr_1r_2 e^i(phi_1 - phi_2)
= fracr_1r_2 cos(phi_1 - phi_2)
$$
where we switched to polar coordinates.
answered Jul 28 at 21:14
mvw
30.2k22250
The lines have direction vectors
$$
AB = b-a quad CD = d-c
$$
We have
$$
DeclareMathOperatorReRe
Re fracb-ad-c
= Re fracr_1 e^i phi_1r_2 e^i phi_2
= Re fracr_1r_2 e^i(phi_1 - phi_2)
= fracr_1r_2 cos(phi_1 - phi_2)
$$
where we switched to polar coordinates.
answered Jul 28 at 21:14
mvw
30.2k22250
edited Jul 28 at 21:46
answered Jul 28 at 21:14
mvw
30.2k22250
answered Jul 28 at 21:14
mvw
30.2k22250
answered Jul 28 at 21:14
mvw
30.2k22250
30.2k22250
add a comment |Â
add a comment |Â
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4
Hint: $argbig((a-b)/(c-d)big)$ is the angle between $AB$ and $CD$.
– dxiv
Jul 28 at 18:42
1
After working out, a very long way is to assign complex numbers $a$=$a_1+a_2i$ etc for$A,B,C$ and $D$ and work out $Refraca-bc-d$ and multiply with its conjugate. Compare that expression to the expression by setting up slopes (rise over run) and using the fact that the product of two perpendicular lines is $-1$. It isn't the prettiest way to show, there has to be a quicker and more elegant way, perhaps the suggestion of dxiv
– imranfat
Jul 28 at 18:51