How to determine if any point is in the acute or the obtuse angle between 2 planes
$begingroup$
Show that the point $(3, 2, -1)$ lies inside the acute angle formed by the planes
$5x-y+z+3=0$ and $4x-3y+2z+5=0$.
I have tried this by calculating the angles between the plane, passing through the line of intersection of these two planes, and these planes. But the angles can't be calculated without calculator. So I need to get the solution in an easier way without calculator.
I also tried to get a method similar as the case of origin. To get if a origin is in the acute or the obtuse angle b/w 2 planes we check the sign of $a_1a_2+b_1b_2+c_1c_2$ if equations of the planes are $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2$ provided both $d_1$ and $d_2$ are positive.
I want to get this type of solution for this case also.
geometry analytic-geometry
edited Dec 7 '18 at 13:45
DRF
4,497926
asked Dec 4 '18 at 7:46
Srijan GhoshSrijan Ghosh
314
$endgroup$
add a comment |
$begingroup$
Show that the point $(3, 2, -1)$ lies inside the acute angle formed by the planes
$5x-y+z+3=0$ and $4x-3y+2z+5=0$.
I have tried this by calculating the angles between the plane, passing through the line of intersection of these two planes, and these planes. But the angles can't be calculated without calculator. So I need to get the solution in an easier way without calculator.
I also tried to get a method similar as the case of origin. To get if a origin is in the acute or the obtuse angle b/w 2 planes we check the sign of $a_1a_2+b_1b_2+c_1c_2$ if equations of the planes are $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2$ provided both $d_1$ and $d_2$ are positive.
I want to get this type of solution for this case also.
geometry analytic-geometry
edited Dec 7 '18 at 13:45
DRF
4,497926
asked Dec 4 '18 at 7:46
Srijan GhoshSrijan Ghosh
314
$endgroup$
$begingroup$
The kind of solution you asked for is outlined in one of the answers below. But it seems to me that the premise of the question is wrong: I do not believe $(3,2,-1)$ is in the acute angle between the planes.
$endgroup$
– David K
Dec 8 '18 at 2:39
add a comment |
$begingroup$
Show that the point $(3, 2, -1)$ lies inside the acute angle formed by the planes
$5x-y+z+3=0$ and $4x-3y+2z+5=0$.
I have tried this by calculating the angles between the plane, passing through the line of intersection of these two planes, and these planes. But the angles can't be calculated without calculator. So I need to get the solution in an easier way without calculator.
I also tried to get a method similar as the case of origin. To get if a origin is in the acute or the obtuse angle b/w 2 planes we check the sign of $a_1a_2+b_1b_2+c_1c_2$ if equations of the planes are $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2$ provided both $d_1$ and $d_2$ are positive.
I want to get this type of solution for this case also.
geometry analytic-geometry
edited Dec 7 '18 at 13:45
DRF
4,497926
asked Dec 4 '18 at 7:46
Srijan GhoshSrijan Ghosh
314
$endgroup$
Show that the point $(3, 2, -1)$ lies inside the acute angle formed by the planes
$5x-y+z+3=0$ and $4x-3y+2z+5=0$.
I have tried this by calculating the angles between the plane, passing through the line of intersection of these two planes, and these planes. But the angles can't be calculated without calculator. So I need to get the solution in an easier way without calculator.
I also tried to get a method similar as the case of origin. To get if a origin is in the acute or the obtuse angle b/w 2 planes we check the sign of $a_1a_2+b_1b_2+c_1c_2$ if equations of the planes are $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2$ provided both $d_1$ and $d_2$ are positive.
I want to get this type of solution for this case also.
geometry analytic-geometry
geometry analytic-geometry
edited Dec 7 '18 at 13:45
DRF
4,497926
asked Dec 4 '18 at 7:46
Srijan GhoshSrijan Ghosh
314
edited Dec 7 '18 at 13:45
DRF
4,497926
asked Dec 4 '18 at 7:46
Srijan GhoshSrijan Ghosh
314
edited Dec 7 '18 at 13:45
DRF
4,497926
edited Dec 7 '18 at 13:45
DRF
4,497926
edited Dec 7 '18 at 13:45
DRF
4,497926
4,497926
asked Dec 4 '18 at 7:46
Srijan GhoshSrijan Ghosh
314
asked Dec 4 '18 at 7:46
Srijan GhoshSrijan Ghosh
314
asked Dec 4 '18 at 7:46
Srijan GhoshSrijan Ghosh
314
314
$begingroup$
The kind of solution you asked for is outlined in one of the answers below. But it seems to me that the premise of the question is wrong: I do not believe $(3,2,-1)$ is in the acute angle between the planes.
$endgroup$
– David K
Dec 8 '18 at 2:39
add a comment |
$begingroup$
The kind of solution you asked for is outlined in one of the answers below. But it seems to me that the premise of the question is wrong: I do not believe $(3,2,-1)$ is in the acute angle between the planes.
$endgroup$
– David K
Dec 8 '18 at 2:39
$begingroup$
The kind of solution you asked for is outlined in one of the answers below. But it seems to me that the premise of the question is wrong: I do not believe $(3,2,-1)$ is in the acute angle between the planes.
$endgroup$
– David K
Dec 8 '18 at 2:39
$begingroup$
The kind of solution you asked for is outlined in one of the answers below. But it seems to me that the premise of the question is wrong: I do not believe $(3,2,-1)$ is in the acute angle between the planes.
$endgroup$
– David K
Dec 8 '18 at 2:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here is a systematic method to solve all such questions. In general we have two planes $P,Q$ and a point $A$, and we wish to know whether $A$ is within the acute 'angle' between $P$ and $Q$. Let $B$ be the point on $P$ such that $AB ⊥ P$. Let $C$ be the intersection of $AB$ and $Q$. Then we simply have to check whether $A$ is between $B$ and $C$ or not. If $A$ is rational and $P,Q$ are given by rational equations, $B$ and $C$ will also be rational and easy to find.
Another way (pointed out by DRF in chat) is that you can translate the point and the planes by the same vector so that the point goes to the origin, and then apply the test you already have for that case. Translation is rigid, so the answer before and after the translation is the same.
answered Dec 7 '18 at 15:21
user21820user21820
38.8k543152
$endgroup$
add a comment |
$begingroup$
General plane passing through line of intersection of two planes $S_1=0$ and $S_2=0$ is $S_1+k*S_2=0$
In your case, the general plane is $5x-y+z+3+k(4x-3y+2z+5)=0$.
Since it passes through the given point $(3,2,-1)$, value of $k$ can be found.
Now add the two angles between new plane and two original planes to get the final answer.
edited Dec 4 '18 at 8:35
Math Girl
631318
answered Dec 4 '18 at 8:24
Sai TejaSai Teja
286
$endgroup$
$begingroup$
i have tried this, but the angle can't be determined without calculator. I want to get a solution without using calculator.
$endgroup$
– Srijan Ghosh
Dec 7 '18 at 13:01
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a systematic method to solve all such questions. In general we have two planes $P,Q$ and a point $A$, and we wish to know whether $A$ is within the acute 'angle' between $P$ and $Q$. Let $B$ be the point on $P$ such that $AB ⊥ P$. Let $C$ be the intersection of $AB$ and $Q$. Then we simply have to check whether $A$ is between $B$ and $C$ or not. If $A$ is rational and $P,Q$ are given by rational equations, $B$ and $C$ will also be rational and easy to find.
Another way (pointed out by DRF in chat) is that you can translate the point and the planes by the same vector so that the point goes to the origin, and then apply the test you already have for that case. Translation is rigid, so the answer before and after the translation is the same.
answered Dec 7 '18 at 15:21
user21820user21820
38.8k543152
$endgroup$
add a comment |
$begingroup$
Here is a systematic method to solve all such questions. In general we have two planes $P,Q$ and a point $A$, and we wish to know whether $A$ is within the acute 'angle' between $P$ and $Q$. Let $B$ be the point on $P$ such that $AB ⊥ P$. Let $C$ be the intersection of $AB$ and $Q$. Then we simply have to check whether $A$ is between $B$ and $C$ or not. If $A$ is rational and $P,Q$ are given by rational equations, $B$ and $C$ will also be rational and easy to find.
Another way (pointed out by DRF in chat) is that you can translate the point and the planes by the same vector so that the point goes to the origin, and then apply the test you already have for that case. Translation is rigid, so the answer before and after the translation is the same.
answered Dec 7 '18 at 15:21
user21820user21820
38.8k543152
$endgroup$
add a comment |
$begingroup$
Here is a systematic method to solve all such questions. In general we have two planes $P,Q$ and a point $A$, and we wish to know whether $A$ is within the acute 'angle' between $P$ and $Q$. Let $B$ be the point on $P$ such that $AB ⊥ P$. Let $C$ be the intersection of $AB$ and $Q$. Then we simply have to check whether $A$ is between $B$ and $C$ or not. If $A$ is rational and $P,Q$ are given by rational equations, $B$ and $C$ will also be rational and easy to find.
Another way (pointed out by DRF in chat) is that you can translate the point and the planes by the same vector so that the point goes to the origin, and then apply the test you already have for that case. Translation is rigid, so the answer before and after the translation is the same.
answered Dec 7 '18 at 15:21
user21820user21820
38.8k543152
$endgroup$
Here is a systematic method to solve all such questions. In general we have two planes $P,Q$ and a point $A$, and we wish to know whether $A$ is within the acute 'angle' between $P$ and $Q$. Let $B$ be the point on $P$ such that $AB ⊥ P$. Let $C$ be the intersection of $AB$ and $Q$. Then we simply have to check whether $A$ is between $B$ and $C$ or not. If $A$ is rational and $P,Q$ are given by rational equations, $B$ and $C$ will also be rational and easy to find.
Another way (pointed out by DRF in chat) is that you can translate the point and the planes by the same vector so that the point goes to the origin, and then apply the test you already have for that case. Translation is rigid, so the answer before and after the translation is the same.
answered Dec 7 '18 at 15:21
user21820user21820
38.8k543152
edited Dec 7 '18 at 15:41
answered Dec 7 '18 at 15:21
user21820user21820
38.8k543152
answered Dec 7 '18 at 15:21
user21820user21820
38.8k543152
answered Dec 7 '18 at 15:21
user21820user21820
38.8k543152
38.8k543152
add a comment |
add a comment |
$begingroup$
General plane passing through line of intersection of two planes $S_1=0$ and $S_2=0$ is $S_1+k*S_2=0$
In your case, the general plane is $5x-y+z+3+k(4x-3y+2z+5)=0$.
Since it passes through the given point $(3,2,-1)$, value of $k$ can be found.
Now add the two angles between new plane and two original planes to get the final answer.
edited Dec 4 '18 at 8:35
Math Girl
631318
answered Dec 4 '18 at 8:24
Sai TejaSai Teja
286
$endgroup$
$begingroup$
i have tried this, but the angle can't be determined without calculator. I want to get a solution without using calculator.
$endgroup$
– Srijan Ghosh
Dec 7 '18 at 13:01
add a comment |
$begingroup$
General plane passing through line of intersection of two planes $S_1=0$ and $S_2=0$ is $S_1+k*S_2=0$
In your case, the general plane is $5x-y+z+3+k(4x-3y+2z+5)=0$.
Since it passes through the given point $(3,2,-1)$, value of $k$ can be found.
Now add the two angles between new plane and two original planes to get the final answer.
edited Dec 4 '18 at 8:35
Math Girl
631318
answered Dec 4 '18 at 8:24
Sai TejaSai Teja
286
$endgroup$
$begingroup$
i have tried this, but the angle can't be determined without calculator. I want to get a solution without using calculator.
$endgroup$
– Srijan Ghosh
Dec 7 '18 at 13:01
add a comment |
$begingroup$
General plane passing through line of intersection of two planes $S_1=0$ and $S_2=0$ is $S_1+k*S_2=0$
In your case, the general plane is $5x-y+z+3+k(4x-3y+2z+5)=0$.
Since it passes through the given point $(3,2,-1)$, value of $k$ can be found.
Now add the two angles between new plane and two original planes to get the final answer.
edited Dec 4 '18 at 8:35
Math Girl
631318
answered Dec 4 '18 at 8:24
Sai TejaSai Teja
286
$endgroup$
General plane passing through line of intersection of two planes $S_1=0$ and $S_2=0$ is $S_1+k*S_2=0$
In your case, the general plane is $5x-y+z+3+k(4x-3y+2z+5)=0$.
Since it passes through the given point $(3,2,-1)$, value of $k$ can be found.
Now add the two angles between new plane and two original planes to get the final answer.
edited Dec 4 '18 at 8:35
Math Girl
631318
answered Dec 4 '18 at 8:24
Sai TejaSai Teja
286
edited Dec 4 '18 at 8:35
Math Girl
631318
edited Dec 4 '18 at 8:35
Math Girl
631318
edited Dec 4 '18 at 8:35
Math Girl
631318
631318
answered Dec 4 '18 at 8:24
Sai TejaSai Teja
286
answered Dec 4 '18 at 8:24
Sai TejaSai Teja
286
answered Dec 4 '18 at 8:24
Sai TejaSai Teja
286
286
$begingroup$
i have tried this, but the angle can't be determined without calculator. I want to get a solution without using calculator.
$endgroup$
– Srijan Ghosh
Dec 7 '18 at 13:01
add a comment |
$begingroup$
i have tried this, but the angle can't be determined without calculator. I want to get a solution without using calculator.
$endgroup$
– Srijan Ghosh
Dec 7 '18 at 13:01
$begingroup$
i have tried this, but the angle can't be determined without calculator. I want to get a solution without using calculator.
$endgroup$
– Srijan Ghosh
Dec 7 '18 at 13:01
$begingroup$
i have tried this, but the angle can't be determined without calculator. I want to get a solution without using calculator.
$endgroup$
– Srijan Ghosh
Dec 7 '18 at 13:01
add a comment |
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$begingroup$
The kind of solution you asked for is outlined in one of the answers below. But it seems to me that the premise of the question is wrong: I do not believe $(3,2,-1)$ is in the acute angle between the planes.
$endgroup$
– David K
Dec 8 '18 at 2:39