Prove that the straight lines whose direction cosines are given by the relationsIf the projections of $OA$ and $OB$ on the plane $z=0$ make angles $phi_1$ and $phi_2$,respectively,with the $x-$axisIn space, four points, $A$, $B$, $C$, and $D$ are given such that $AB = AC$ and $DB = DC$. Prove that the lines $AD$ and $BC$ are perpendicular.Find the equation of straight lines through the point $(dfrac 1sqrt 3, 1)$ whose perpendicular distance from the origin is unity.Find the equations of two straight lines each of which is parallel toShowing the direction cosines of line perpendicular to two lines direction cosinesFind the angle between the two straight lines whose direction cosines are given by $l+m+n=0,l^2+m^2-n^2=0$Angle between lines whose direction cosines are related by given equations“Given three parallel straight lines. Construct a square three of whose vertices belong to these lines.” Are all three lines required?Condition when angle between two lines is $piover 3$How many lines are represented when only two direction cosines are given
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Prove that the straight lines whose direction cosines are given by the relations
If the projections of $OA$ and $OB$ on the plane $z=0$ make angles $phi_1$ and $phi_2$,respectively,with the $x-$axisIn space, four points, $A$, $B$, $C$, and $D$ are given such that $AB = AC$ and $DB = DC$. Prove that the lines $AD$ and $BC$ are perpendicular.Find the equation of straight lines through the point $(dfrac 1sqrt 3, 1)$ whose perpendicular distance from the origin is unity.Find the equations of two straight lines each of which is parallel toShowing the direction cosines of line perpendicular to two lines direction cosinesFind the angle between the two straight lines whose direction cosines are given by $l+m+n=0,l^2+m^2-n^2=0$Angle between lines whose direction cosines are related by given equations“Given three parallel straight lines. Construct a square three of whose vertices belong to these lines.” Are all three lines required?Condition when angle between two lines is $piover 3$How many lines are represented when only two direction cosines are given
$begingroup$
Prove that the straight lines whose direction cosines are given by the relations $al+bm+cn=0$ and $fmn+gnl+hlm=0$ are perpendicular if $dfrac fa +dfrac gb + dfrac hc=0$ and parallel if $sqrt af pm sqrt bg pm sqrt ch=0$
My Attempt:
Given relations are
$$al+bm+cn=0$$
$$fmn+gnl+hlm=0$$
Eliminating $n$ between these two equations, we get
$$fm (-dfrac al+bmc) + g(-dfrac al+bmc)l + hlm=0$$
$$agl^2+(af+bg-ch)lm+bfm^2=0$$
$$ag(dfrac lm)^2 + (ch-af+bg) (dfrac lm) + bf=0$$
geometry analytic-geometry 3d
asked Mar 15 at 3:33
blue_eyed_...blue_eyed_...
3,29521753
$endgroup$
add a comment |
$begingroup$
Prove that the straight lines whose direction cosines are given by the relations $al+bm+cn=0$ and $fmn+gnl+hlm=0$ are perpendicular if $dfrac fa +dfrac gb + dfrac hc=0$ and parallel if $sqrt af pm sqrt bg pm sqrt ch=0$
My Attempt:
Given relations are
$$al+bm+cn=0$$
$$fmn+gnl+hlm=0$$
Eliminating $n$ between these two equations, we get
$$fm (-dfrac al+bmc) + g(-dfrac al+bmc)l + hlm=0$$
$$agl^2+(af+bg-ch)lm+bfm^2=0$$
$$ag(dfrac lm)^2 + (ch-af+bg) (dfrac lm) + bf=0$$
geometry analytic-geometry 3d
asked Mar 15 at 3:33
blue_eyed_...blue_eyed_...
3,29521753
$endgroup$
$begingroup$
Could you elaborate more on the question? There are a lot of letters but no indication which of them are related to the direction cosines of the straight lines.
$endgroup$
– user
Mar 15 at 21:51
$begingroup$
@user I think the direction cosines are represented by $l,m,n$.
$endgroup$
– Shubham Johri
Mar 16 at 4:43
$begingroup$
@ShubhamJohris But there should be at least two sets of the direction cosines. If two lines have the same direction cosines $(l,m,n) $ they are parallel.
$endgroup$
– user
Mar 16 at 7:32
$begingroup$
@user The two sets are given by the solutions of the two equations
$endgroup$
– Shubham Johri
yesterday
$begingroup$
@ShubhamJohri This does not look convincing taking into account the possibility of "parallel" solutions.
$endgroup$
– user
yesterday
add a comment |
$begingroup$
Prove that the straight lines whose direction cosines are given by the relations $al+bm+cn=0$ and $fmn+gnl+hlm=0$ are perpendicular if $dfrac fa +dfrac gb + dfrac hc=0$ and parallel if $sqrt af pm sqrt bg pm sqrt ch=0$
My Attempt:
Given relations are
$$al+bm+cn=0$$
$$fmn+gnl+hlm=0$$
Eliminating $n$ between these two equations, we get
$$fm (-dfrac al+bmc) + g(-dfrac al+bmc)l + hlm=0$$
$$agl^2+(af+bg-ch)lm+bfm^2=0$$
$$ag(dfrac lm)^2 + (ch-af+bg) (dfrac lm) + bf=0$$
geometry analytic-geometry 3d
asked Mar 15 at 3:33
blue_eyed_...blue_eyed_...
3,29521753
$endgroup$
Prove that the straight lines whose direction cosines are given by the relations $al+bm+cn=0$ and $fmn+gnl+hlm=0$ are perpendicular if $dfrac fa +dfrac gb + dfrac hc=0$ and parallel if $sqrt af pm sqrt bg pm sqrt ch=0$
My Attempt:
Given relations are
$$al+bm+cn=0$$
$$fmn+gnl+hlm=0$$
Eliminating $n$ between these two equations, we get
$$fm (-dfrac al+bmc) + g(-dfrac al+bmc)l + hlm=0$$
$$agl^2+(af+bg-ch)lm+bfm^2=0$$
$$ag(dfrac lm)^2 + (ch-af+bg) (dfrac lm) + bf=0$$
geometry analytic-geometry 3d
geometry analytic-geometry 3d
asked Mar 15 at 3:33
blue_eyed_...blue_eyed_...
3,29521753
asked Mar 15 at 3:33
blue_eyed_...blue_eyed_...
3,29521753
asked Mar 15 at 3:33
blue_eyed_...blue_eyed_...
3,29521753
asked Mar 15 at 3:33
blue_eyed_...blue_eyed_...
3,29521753
asked Mar 15 at 3:33
blue_eyed_...blue_eyed_...
3,29521753
3,29521753
$begingroup$
Could you elaborate more on the question? There are a lot of letters but no indication which of them are related to the direction cosines of the straight lines.
$endgroup$
– user
Mar 15 at 21:51
$begingroup$
@user I think the direction cosines are represented by $l,m,n$.
$endgroup$
– Shubham Johri
Mar 16 at 4:43
$begingroup$
@ShubhamJohris But there should be at least two sets of the direction cosines. If two lines have the same direction cosines $(l,m,n) $ they are parallel.
$endgroup$
– user
Mar 16 at 7:32
$begingroup$
@user The two sets are given by the solutions of the two equations
$endgroup$
– Shubham Johri
yesterday
$begingroup$
@ShubhamJohri This does not look convincing taking into account the possibility of "parallel" solutions.
$endgroup$
– user
yesterday
add a comment |
$begingroup$
Could you elaborate more on the question? There are a lot of letters but no indication which of them are related to the direction cosines of the straight lines.
$endgroup$
– user
Mar 15 at 21:51
$begingroup$
@user I think the direction cosines are represented by $l,m,n$.
$endgroup$
– Shubham Johri
Mar 16 at 4:43
$begingroup$
@ShubhamJohris But there should be at least two sets of the direction cosines. If two lines have the same direction cosines $(l,m,n) $ they are parallel.
$endgroup$
– user
Mar 16 at 7:32
$begingroup$
@user The two sets are given by the solutions of the two equations
$endgroup$
– Shubham Johri
yesterday
$begingroup$
@ShubhamJohri This does not look convincing taking into account the possibility of "parallel" solutions.
$endgroup$
– user
yesterday
$begingroup$
Could you elaborate more on the question? There are a lot of letters but no indication which of them are related to the direction cosines of the straight lines.
$endgroup$
– user
Mar 15 at 21:51
$begingroup$
Could you elaborate more on the question? There are a lot of letters but no indication which of them are related to the direction cosines of the straight lines.
$endgroup$
– user
Mar 15 at 21:51
$begingroup$
@user I think the direction cosines are represented by $l,m,n$.
$endgroup$
– Shubham Johri
Mar 16 at 4:43
$begingroup$
@user I think the direction cosines are represented by $l,m,n$.
$endgroup$
– Shubham Johri
Mar 16 at 4:43
$begingroup$
@ShubhamJohris But there should be at least two sets of the direction cosines. If two lines have the same direction cosines $(l,m,n) $ they are parallel.
$endgroup$
– user
Mar 16 at 7:32
$begingroup$
@ShubhamJohris But there should be at least two sets of the direction cosines. If two lines have the same direction cosines $(l,m,n) $ they are parallel.
$endgroup$
– user
Mar 16 at 7:32
$begingroup$
@user The two sets are given by the solutions of the two equations
$endgroup$
– Shubham Johri
yesterday
$begingroup$
@user The two sets are given by the solutions of the two equations
$endgroup$
– Shubham Johri
yesterday
$begingroup$
@ShubhamJohri This does not look convincing taking into account the possibility of "parallel" solutions.
$endgroup$
– user
yesterday
$begingroup$
@ShubhamJohri This does not look convincing taking into account the possibility of "parallel" solutions.
$endgroup$
– user
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is sufficient to work with direction ratios to test perpendicularity and parallelism. Note that $lne0$, because $l=0implies m=n=l=0$ assuming $fne0$. Divide the first equation by $l$ and the second by $l^2$,$$a+bfrac ml+cfrac nl=0\ffrac mlfrac nl+gfrac nl+hfrac ml=0$$Substitute $frac ml=x,frac nl=y$ and solve the resulting system of equations. The required direction ratios will then be $(1,x_1,y_1),(1,x_2,y_2)$ for which the conditions for perpendicularity $(1+x_1x_2+y_1y_2=0)$ and parallelism $(fracx_1x_2=fracy_1y_2=1)$ can be easily derived.
The formulae for sum and product of roots of a quadratic will help in getting the conditions without having to solve the quadratic.
answered Mar 15 at 4:36
Shubham JohriShubham Johri
5,382818
$endgroup$
$begingroup$
I didn't understand why $l=0$ and $l=m=n=0$? Why assuming $fneq 0$?
$endgroup$
– blue_eyed_...
Mar 15 at 6:37
$begingroup$
When $l=0,fmn+gnl+hlm=fmn=0implies fne0wedge(m=0vee n=0)vee f=0$. When $fne0wedge m=0$, using the first equation, $al+bm+cn=cn=0implies n=0$. Similarly when $fne0wedge n=0$.
$endgroup$
– Shubham Johri
Mar 16 at 4:39
add a comment |
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1 Answer
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1 Answer
1
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oldest
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active
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active
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votes
$begingroup$
It is sufficient to work with direction ratios to test perpendicularity and parallelism. Note that $lne0$, because $l=0implies m=n=l=0$ assuming $fne0$. Divide the first equation by $l$ and the second by $l^2$,$$a+bfrac ml+cfrac nl=0\ffrac mlfrac nl+gfrac nl+hfrac ml=0$$Substitute $frac ml=x,frac nl=y$ and solve the resulting system of equations. The required direction ratios will then be $(1,x_1,y_1),(1,x_2,y_2)$ for which the conditions for perpendicularity $(1+x_1x_2+y_1y_2=0)$ and parallelism $(fracx_1x_2=fracy_1y_2=1)$ can be easily derived.
The formulae for sum and product of roots of a quadratic will help in getting the conditions without having to solve the quadratic.
answered Mar 15 at 4:36
Shubham JohriShubham Johri
5,382818
$endgroup$
$begingroup$
I didn't understand why $l=0$ and $l=m=n=0$? Why assuming $fneq 0$?
$endgroup$
– blue_eyed_...
Mar 15 at 6:37
$begingroup$
When $l=0,fmn+gnl+hlm=fmn=0implies fne0wedge(m=0vee n=0)vee f=0$. When $fne0wedge m=0$, using the first equation, $al+bm+cn=cn=0implies n=0$. Similarly when $fne0wedge n=0$.
$endgroup$
– Shubham Johri
Mar 16 at 4:39
add a comment |
$begingroup$
It is sufficient to work with direction ratios to test perpendicularity and parallelism. Note that $lne0$, because $l=0implies m=n=l=0$ assuming $fne0$. Divide the first equation by $l$ and the second by $l^2$,$$a+bfrac ml+cfrac nl=0\ffrac mlfrac nl+gfrac nl+hfrac ml=0$$Substitute $frac ml=x,frac nl=y$ and solve the resulting system of equations. The required direction ratios will then be $(1,x_1,y_1),(1,x_2,y_2)$ for which the conditions for perpendicularity $(1+x_1x_2+y_1y_2=0)$ and parallelism $(fracx_1x_2=fracy_1y_2=1)$ can be easily derived.
The formulae for sum and product of roots of a quadratic will help in getting the conditions without having to solve the quadratic.
answered Mar 15 at 4:36
Shubham JohriShubham Johri
5,382818
$endgroup$
$begingroup$
I didn't understand why $l=0$ and $l=m=n=0$? Why assuming $fneq 0$?
$endgroup$
– blue_eyed_...
Mar 15 at 6:37
$begingroup$
When $l=0,fmn+gnl+hlm=fmn=0implies fne0wedge(m=0vee n=0)vee f=0$. When $fne0wedge m=0$, using the first equation, $al+bm+cn=cn=0implies n=0$. Similarly when $fne0wedge n=0$.
$endgroup$
– Shubham Johri
Mar 16 at 4:39
add a comment |
$begingroup$
It is sufficient to work with direction ratios to test perpendicularity and parallelism. Note that $lne0$, because $l=0implies m=n=l=0$ assuming $fne0$. Divide the first equation by $l$ and the second by $l^2$,$$a+bfrac ml+cfrac nl=0\ffrac mlfrac nl+gfrac nl+hfrac ml=0$$Substitute $frac ml=x,frac nl=y$ and solve the resulting system of equations. The required direction ratios will then be $(1,x_1,y_1),(1,x_2,y_2)$ for which the conditions for perpendicularity $(1+x_1x_2+y_1y_2=0)$ and parallelism $(fracx_1x_2=fracy_1y_2=1)$ can be easily derived.
The formulae for sum and product of roots of a quadratic will help in getting the conditions without having to solve the quadratic.
answered Mar 15 at 4:36
Shubham JohriShubham Johri
5,382818
$endgroup$
It is sufficient to work with direction ratios to test perpendicularity and parallelism. Note that $lne0$, because $l=0implies m=n=l=0$ assuming $fne0$. Divide the first equation by $l$ and the second by $l^2$,$$a+bfrac ml+cfrac nl=0\ffrac mlfrac nl+gfrac nl+hfrac ml=0$$Substitute $frac ml=x,frac nl=y$ and solve the resulting system of equations. The required direction ratios will then be $(1,x_1,y_1),(1,x_2,y_2)$ for which the conditions for perpendicularity $(1+x_1x_2+y_1y_2=0)$ and parallelism $(fracx_1x_2=fracy_1y_2=1)$ can be easily derived.
The formulae for sum and product of roots of a quadratic will help in getting the conditions without having to solve the quadratic.
answered Mar 15 at 4:36
Shubham JohriShubham Johri
5,382818
edited Mar 16 at 4:39
answered Mar 15 at 4:36
Shubham JohriShubham Johri
5,382818
answered Mar 15 at 4:36
Shubham JohriShubham Johri
5,382818
answered Mar 15 at 4:36
Shubham JohriShubham Johri
5,382818
5,382818
$begingroup$
I didn't understand why $l=0$ and $l=m=n=0$? Why assuming $fneq 0$?
$endgroup$
– blue_eyed_...
Mar 15 at 6:37
$begingroup$
When $l=0,fmn+gnl+hlm=fmn=0implies fne0wedge(m=0vee n=0)vee f=0$. When $fne0wedge m=0$, using the first equation, $al+bm+cn=cn=0implies n=0$. Similarly when $fne0wedge n=0$.
$endgroup$
– Shubham Johri
Mar 16 at 4:39
add a comment |
$begingroup$
I didn't understand why $l=0$ and $l=m=n=0$? Why assuming $fneq 0$?
$endgroup$
– blue_eyed_...
Mar 15 at 6:37
$begingroup$
When $l=0,fmn+gnl+hlm=fmn=0implies fne0wedge(m=0vee n=0)vee f=0$. When $fne0wedge m=0$, using the first equation, $al+bm+cn=cn=0implies n=0$. Similarly when $fne0wedge n=0$.
$endgroup$
– Shubham Johri
Mar 16 at 4:39
$begingroup$
I didn't understand why $l=0$ and $l=m=n=0$? Why assuming $fneq 0$?
$endgroup$
– blue_eyed_...
Mar 15 at 6:37
$begingroup$
I didn't understand why $l=0$ and $l=m=n=0$? Why assuming $fneq 0$?
$endgroup$
– blue_eyed_...
Mar 15 at 6:37
$begingroup$
When $l=0,fmn+gnl+hlm=fmn=0implies fne0wedge(m=0vee n=0)vee f=0$. When $fne0wedge m=0$, using the first equation, $al+bm+cn=cn=0implies n=0$. Similarly when $fne0wedge n=0$.
$endgroup$
– Shubham Johri
Mar 16 at 4:39
$begingroup$
When $l=0,fmn+gnl+hlm=fmn=0implies fne0wedge(m=0vee n=0)vee f=0$. When $fne0wedge m=0$, using the first equation, $al+bm+cn=cn=0implies n=0$. Similarly when $fne0wedge n=0$.
$endgroup$
– Shubham Johri
Mar 16 at 4:39
add a comment |
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$begingroup$
Could you elaborate more on the question? There are a lot of letters but no indication which of them are related to the direction cosines of the straight lines.
$endgroup$
– user
Mar 15 at 21:51
$begingroup$
@user I think the direction cosines are represented by $l,m,n$.
$endgroup$
– Shubham Johri
Mar 16 at 4:43
$begingroup$
@ShubhamJohris But there should be at least two sets of the direction cosines. If two lines have the same direction cosines $(l,m,n) $ they are parallel.
$endgroup$
– user
Mar 16 at 7:32
$begingroup$
@user The two sets are given by the solutions of the two equations
$endgroup$
– Shubham Johri
yesterday
$begingroup$
@ShubhamJohri This does not look convincing taking into account the possibility of "parallel" solutions.
$endgroup$
– user
yesterday