Given the angle between planes $pi_1$ and $pi_2$ is equal to the angle between…
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I'm not sure where I'm going wrong with this question but i keep coming to a hexic equation rather than a quartic equation.
the three planes:
$$pi_1: ax+2y+z=3$$
$$pi_2: x+ay+z=4$$
$$pi_3: x+y+az=5$$
Given the angle between planes $pi_1$ and $pi_2$ is equal to the angle between $pi_2$ and $pi_3$, show that $a$ must statisfy the quartic equation
$$5a^4+2a^3-2a^2-8a-3=0$$
plane-geometry
asked Sep 10 at 18:52
H.Linkhorn
358
add a comment |Â
up vote
0
down vote
favorite
I'm not sure where I'm going wrong with this question but i keep coming to a hexic equation rather than a quartic equation.
the three planes:
$$pi_1: ax+2y+z=3$$
$$pi_2: x+ay+z=4$$
$$pi_3: x+y+az=5$$
Given the angle between planes $pi_1$ and $pi_2$ is equal to the angle between $pi_2$ and $pi_3$, show that $a$ must statisfy the quartic equation
$$5a^4+2a^3-2a^2-8a-3=0$$
plane-geometry
asked Sep 10 at 18:52
H.Linkhorn
358
If you’d like someone to point out where you’re going wrong, then you should show your work.
– amd
Sep 10 at 20:15
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm not sure where I'm going wrong with this question but i keep coming to a hexic equation rather than a quartic equation.
the three planes:
$$pi_1: ax+2y+z=3$$
$$pi_2: x+ay+z=4$$
$$pi_3: x+y+az=5$$
Given the angle between planes $pi_1$ and $pi_2$ is equal to the angle between $pi_2$ and $pi_3$, show that $a$ must statisfy the quartic equation
$$5a^4+2a^3-2a^2-8a-3=0$$
plane-geometry
asked Sep 10 at 18:52
H.Linkhorn
358
I'm not sure where I'm going wrong with this question but i keep coming to a hexic equation rather than a quartic equation.
the three planes:
$$pi_1: ax+2y+z=3$$
$$pi_2: x+ay+z=4$$
$$pi_3: x+y+az=5$$
Given the angle between planes $pi_1$ and $pi_2$ is equal to the angle between $pi_2$ and $pi_3$, show that $a$ must statisfy the quartic equation
$$5a^4+2a^3-2a^2-8a-3=0$$
plane-geometry
plane-geometry
asked Sep 10 at 18:52
H.Linkhorn
358
asked Sep 10 at 18:52
H.Linkhorn
358
asked Sep 10 at 18:52
H.Linkhorn
358
asked Sep 10 at 18:52
H.Linkhorn
358
asked Sep 10 at 18:52
H.Linkhorn
358
358
If you’d like someone to point out where you’re going wrong, then you should show your work.
– amd
Sep 10 at 20:15
add a comment |Â
If you’d like someone to point out where you’re going wrong, then you should show your work.
– amd
Sep 10 at 20:15
If you’d like someone to point out where you’re going wrong, then you should show your work.
– amd
Sep 10 at 20:15
If you’d like someone to point out where you’re going wrong, then you should show your work.
– amd
Sep 10 at 20:15
add a comment |Â
1 Answer
1
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0
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Given the planes
$$
pi_k to left<p-p_k,vec n_kright> = 0
$$
with $p = (x,y,z)$
if
$$
fracleft< vec n_1,vec n_2right> ^2 = fracleft< vec n_2,vec n_3right> ^2
$$
then
$$
5a^4+2a^3-2a^2-8a-3 = 0
$$
here $left<cdot,cdotright>$ represents the scalar product of two vectors
answered Sep 10 at 19:23
Cesareo
6,2192413
what does the first line mean? I'm not very good with notation
– H.Linkhorn
Sep 10 at 19:30
@H.Linkhorn For the first plane we have $vec n_1 = (a,2,1)$ and $p_1 = (0,0,3)$
– Cesareo
Sep 10 at 20:14
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
Given the planes
$$
pi_k to left<p-p_k,vec n_kright> = 0
$$
with $p = (x,y,z)$
if
$$
fracleft< vec n_1,vec n_2right> ^2 = fracleft< vec n_2,vec n_3right> ^2
$$
then
$$
5a^4+2a^3-2a^2-8a-3 = 0
$$
here $left<cdot,cdotright>$ represents the scalar product of two vectors
answered Sep 10 at 19:23
Cesareo
6,2192413
what does the first line mean? I'm not very good with notation
– H.Linkhorn
Sep 10 at 19:30
@H.Linkhorn For the first plane we have $vec n_1 = (a,2,1)$ and $p_1 = (0,0,3)$
– Cesareo
Sep 10 at 20:14
add a comment |Â
up vote
0
down vote
Given the planes
$$
pi_k to left<p-p_k,vec n_kright> = 0
$$
with $p = (x,y,z)$
if
$$
fracleft< vec n_1,vec n_2right> ^2 = fracleft< vec n_2,vec n_3right> ^2
$$
then
$$
5a^4+2a^3-2a^2-8a-3 = 0
$$
here $left<cdot,cdotright>$ represents the scalar product of two vectors
answered Sep 10 at 19:23
Cesareo
6,2192413
what does the first line mean? I'm not very good with notation
– H.Linkhorn
Sep 10 at 19:30
@H.Linkhorn For the first plane we have $vec n_1 = (a,2,1)$ and $p_1 = (0,0,3)$
– Cesareo
Sep 10 at 20:14
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Given the planes
$$
pi_k to left<p-p_k,vec n_kright> = 0
$$
with $p = (x,y,z)$
if
$$
fracleft< vec n_1,vec n_2right> ^2 = fracleft< vec n_2,vec n_3right> ^2
$$
then
$$
5a^4+2a^3-2a^2-8a-3 = 0
$$
here $left<cdot,cdotright>$ represents the scalar product of two vectors
answered Sep 10 at 19:23
Cesareo
6,2192413
Given the planes
$$
pi_k to left<p-p_k,vec n_kright> = 0
$$
with $p = (x,y,z)$
if
$$
fracleft< vec n_1,vec n_2right> ^2 = fracleft< vec n_2,vec n_3right> ^2
$$
then
$$
5a^4+2a^3-2a^2-8a-3 = 0
$$
here $left<cdot,cdotright>$ represents the scalar product of two vectors
answered Sep 10 at 19:23
Cesareo
6,2192413
answered Sep 10 at 19:23
Cesareo
6,2192413
answered Sep 10 at 19:23
Cesareo
6,2192413
answered Sep 10 at 19:23
Cesareo
6,2192413
6,2192413
what does the first line mean? I'm not very good with notation
– H.Linkhorn
Sep 10 at 19:30
@H.Linkhorn For the first plane we have $vec n_1 = (a,2,1)$ and $p_1 = (0,0,3)$
– Cesareo
Sep 10 at 20:14
add a comment |Â
what does the first line mean? I'm not very good with notation
– H.Linkhorn
Sep 10 at 19:30
@H.Linkhorn For the first plane we have $vec n_1 = (a,2,1)$ and $p_1 = (0,0,3)$
– Cesareo
Sep 10 at 20:14
what does the first line mean? I'm not very good with notation
– H.Linkhorn
Sep 10 at 19:30
what does the first line mean? I'm not very good with notation
– H.Linkhorn
Sep 10 at 19:30
@H.Linkhorn For the first plane we have $vec n_1 = (a,2,1)$ and $p_1 = (0,0,3)$
– Cesareo
Sep 10 at 20:14
@H.Linkhorn For the first plane we have $vec n_1 = (a,2,1)$ and $p_1 = (0,0,3)$
– Cesareo
Sep 10 at 20:14
add a comment |Â
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If you’d like someone to point out where you’re going wrong, then you should show your work.
– amd
Sep 10 at 20:15