Find the plane $P$ passing through the origin such that the three planes $P$, $P_1=(x+y+z=1)$ and $P_2=...
$begingroup$
What I did is find the equation of a line of the intersection like
$$(x+y+z=1)+t(x-y+z=2)=0$$
Since it passes through the origin, I substitute $x=0$, $y=0$, $z=0$ into the equation and get $t$. Then I sub $t$ into the equation to obtain the equation of plane.
linear-algebra
edited Sep 20 '18 at 20:23
Daniel Buck
2,8201725
asked Sep 20 '18 at 19:59
Kelvin CheungKelvin Cheung
12
$endgroup$
add a comment |
$begingroup$
What I did is find the equation of a line of the intersection like
$$(x+y+z=1)+t(x-y+z=2)=0$$
Since it passes through the origin, I substitute $x=0$, $y=0$, $z=0$ into the equation and get $t$. Then I sub $t$ into the equation to obtain the equation of plane.
linear-algebra
edited Sep 20 '18 at 20:23
Daniel Buck
2,8201725
asked Sep 20 '18 at 19:59
Kelvin CheungKelvin Cheung
12
$endgroup$
$begingroup$
Welcome to Maths SX! What is the question?
$endgroup$
– Bernard
Sep 20 '18 at 20:07
$begingroup$
Find the plane P passing through the origin such that the three planes P, P1=(x+y+z=1） and P2 (x-y+z=2) meet along a line in R3.
$endgroup$
– Kelvin Cheung
Sep 20 '18 at 20:14
$begingroup$
I think you have the right idea, but you have to formulate it it in a correct way (what's a linear combination of equalities?)
$endgroup$
– Bernard
Sep 20 '18 at 20:17
$begingroup$
See this question and its solutions
$endgroup$
– user376343
Sep 20 '18 at 20:50
add a comment |
$begingroup$
What I did is find the equation of a line of the intersection like
$$(x+y+z=1)+t(x-y+z=2)=0$$
Since it passes through the origin, I substitute $x=0$, $y=0$, $z=0$ into the equation and get $t$. Then I sub $t$ into the equation to obtain the equation of plane.
linear-algebra
edited Sep 20 '18 at 20:23
Daniel Buck
2,8201725
asked Sep 20 '18 at 19:59
Kelvin CheungKelvin Cheung
12
$endgroup$
What I did is find the equation of a line of the intersection like
$$(x+y+z=1)+t(x-y+z=2)=0$$
Since it passes through the origin, I substitute $x=0$, $y=0$, $z=0$ into the equation and get $t$. Then I sub $t$ into the equation to obtain the equation of plane.
linear-algebra
linear-algebra
edited Sep 20 '18 at 20:23
Daniel Buck
2,8201725
asked Sep 20 '18 at 19:59
Kelvin CheungKelvin Cheung
12
edited Sep 20 '18 at 20:23
Daniel Buck
2,8201725
asked Sep 20 '18 at 19:59
Kelvin CheungKelvin Cheung
12
edited Sep 20 '18 at 20:23
Daniel Buck
2,8201725
edited Sep 20 '18 at 20:23
Daniel Buck
2,8201725
edited Sep 20 '18 at 20:23
Daniel Buck
2,8201725
2,8201725
asked Sep 20 '18 at 19:59
Kelvin CheungKelvin Cheung
12
asked Sep 20 '18 at 19:59
Kelvin CheungKelvin Cheung
12
asked Sep 20 '18 at 19:59
Kelvin CheungKelvin Cheung
12
12
$begingroup$
Welcome to Maths SX! What is the question?
$endgroup$
– Bernard
Sep 20 '18 at 20:07
$begingroup$
Find the plane P passing through the origin such that the three planes P, P1=(x+y+z=1） and P2 (x-y+z=2) meet along a line in R3.
$endgroup$
– Kelvin Cheung
Sep 20 '18 at 20:14
$begingroup$
I think you have the right idea, but you have to formulate it it in a correct way (what's a linear combination of equalities?)
$endgroup$
– Bernard
Sep 20 '18 at 20:17
$begingroup$
See this question and its solutions
$endgroup$
– user376343
Sep 20 '18 at 20:50
add a comment |
$begingroup$
Welcome to Maths SX! What is the question?
$endgroup$
– Bernard
Sep 20 '18 at 20:07
$begingroup$
Find the plane P passing through the origin such that the three planes P, P1=(x+y+z=1） and P2 (x-y+z=2) meet along a line in R3.
$endgroup$
– Kelvin Cheung
Sep 20 '18 at 20:14
$begingroup$
I think you have the right idea, but you have to formulate it it in a correct way (what's a linear combination of equalities?)
$endgroup$
– Bernard
Sep 20 '18 at 20:17
$begingroup$
See this question and its solutions
$endgroup$
– user376343
Sep 20 '18 at 20:50
$begingroup$
Welcome to Maths SX! What is the question?
$endgroup$
– Bernard
Sep 20 '18 at 20:07
$begingroup$
Welcome to Maths SX! What is the question?
$endgroup$
– Bernard
Sep 20 '18 at 20:07
$begingroup$
Find the plane P passing through the origin such that the three planes P, P1=(x+y+z=1） and P2 (x-y+z=2) meet along a line in R3.
$endgroup$
– Kelvin Cheung
Sep 20 '18 at 20:14
$begingroup$
Find the plane P passing through the origin such that the three planes P, P1=(x+y+z=1） and P2 (x-y+z=2) meet along a line in R3.
$endgroup$
– Kelvin Cheung
Sep 20 '18 at 20:14
$begingroup$
I think you have the right idea, but you have to formulate it it in a correct way (what's a linear combination of equalities?)
$endgroup$
– Bernard
Sep 20 '18 at 20:17
$begingroup$
I think you have the right idea, but you have to formulate it it in a correct way (what's a linear combination of equalities?)
$endgroup$
– Bernard
Sep 20 '18 at 20:17
$begingroup$
See this question and its solutions
$endgroup$
– user376343
Sep 20 '18 at 20:50
$begingroup$
See this question and its solutions
$endgroup$
– user376343
Sep 20 '18 at 20:50
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You’ve got the right idea, but there’s a detail that you might’ve overlooked. Every plane that passes through the intersection of $P_1$ and $P_2$ has an equation that’s a nontrivial linear combination of their equations: $$s(x+y+z-1)+t(x-y+z-2)=0.$$ What you’ve done is to set $s=1$ in this equation, which excludes $P_1$ from the set. For this problem, we know that $P_1$ doesn’t pass through the origin and therefore can’t be the solution, so it’s OK to do this, but you need to be careful that you don’t exclude a solution by doing this in other similar problems.
Continuing from the above linear combination, setting $x=y=z=0$ reduces the equation to $-s-2t=0$, or $s=-2t$. (In fact, since the constant term of a plane equation that passes through the origin is zero, we could’ve reached this constraint without bothering to compute any of the other terms.) Taking $t=-1$ produces the equation $x+3y+z=0$, which clearly passes through the origin as required.
answered Sep 20 '18 at 20:58
amdamd
29.7k21050
$endgroup$
add a comment |
$begingroup$
HINT
We can proceed as follows
- the general equation for $P$ is $ax+by+cz=0$ with normal $vec n=(a,b,c)$
- find the common line to $P_1$ and $P_2$; note that it suffices to find a common point $Q$ and the direction vector that is $vec v=vec n_1times vec n_2$
- then $vec n=vec {OQ}times vec v$
answered Sep 20 '18 at 20:42
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
Adding x + y + z = 1 and x - y + z = 2 gives 2x + 2z = 3 or z = 3/2- x. putting that into x+ y+ z= 1, x+ y+ 3/2- x= y+ 3/2= 1 so y= -3/2. Taking t= x as parameter, the line of intersection is x= t, y= -3/2, z= 3/2- t. Any plane that contains the origin can be written as z= Ax+ By. In order that it contains that line we must have 3/2- t= At- (3/2)B= 0, so (A+ 1)t=(5/2)B for all t. In order that this be true for all t, we must have A- 1= 0 and (5/2)B= 0. That is, z= x.
edited Sep 21 '18 at 23:53
amWhy
192k28225439
answered Sep 20 '18 at 20:52
user247327user247327
10.7k1515
$endgroup$
add a comment |
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3 Answers
3
active
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votes
3 Answers
3
active
oldest
votes
active
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votes
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$begingroup$
You’ve got the right idea, but there’s a detail that you might’ve overlooked. Every plane that passes through the intersection of $P_1$ and $P_2$ has an equation that’s a nontrivial linear combination of their equations: $$s(x+y+z-1)+t(x-y+z-2)=0.$$ What you’ve done is to set $s=1$ in this equation, which excludes $P_1$ from the set. For this problem, we know that $P_1$ doesn’t pass through the origin and therefore can’t be the solution, so it’s OK to do this, but you need to be careful that you don’t exclude a solution by doing this in other similar problems.
Continuing from the above linear combination, setting $x=y=z=0$ reduces the equation to $-s-2t=0$, or $s=-2t$. (In fact, since the constant term of a plane equation that passes through the origin is zero, we could’ve reached this constraint without bothering to compute any of the other terms.) Taking $t=-1$ produces the equation $x+3y+z=0$, which clearly passes through the origin as required.
answered Sep 20 '18 at 20:58
amdamd
29.7k21050
$endgroup$
add a comment |
$begingroup$
You’ve got the right idea, but there’s a detail that you might’ve overlooked. Every plane that passes through the intersection of $P_1$ and $P_2$ has an equation that’s a nontrivial linear combination of their equations: $$s(x+y+z-1)+t(x-y+z-2)=0.$$ What you’ve done is to set $s=1$ in this equation, which excludes $P_1$ from the set. For this problem, we know that $P_1$ doesn’t pass through the origin and therefore can’t be the solution, so it’s OK to do this, but you need to be careful that you don’t exclude a solution by doing this in other similar problems.
Continuing from the above linear combination, setting $x=y=z=0$ reduces the equation to $-s-2t=0$, or $s=-2t$. (In fact, since the constant term of a plane equation that passes through the origin is zero, we could’ve reached this constraint without bothering to compute any of the other terms.) Taking $t=-1$ produces the equation $x+3y+z=0$, which clearly passes through the origin as required.
answered Sep 20 '18 at 20:58
amdamd
29.7k21050
$endgroup$
add a comment |
$begingroup$
You’ve got the right idea, but there’s a detail that you might’ve overlooked. Every plane that passes through the intersection of $P_1$ and $P_2$ has an equation that’s a nontrivial linear combination of their equations: $$s(x+y+z-1)+t(x-y+z-2)=0.$$ What you’ve done is to set $s=1$ in this equation, which excludes $P_1$ from the set. For this problem, we know that $P_1$ doesn’t pass through the origin and therefore can’t be the solution, so it’s OK to do this, but you need to be careful that you don’t exclude a solution by doing this in other similar problems.
Continuing from the above linear combination, setting $x=y=z=0$ reduces the equation to $-s-2t=0$, or $s=-2t$. (In fact, since the constant term of a plane equation that passes through the origin is zero, we could’ve reached this constraint without bothering to compute any of the other terms.) Taking $t=-1$ produces the equation $x+3y+z=0$, which clearly passes through the origin as required.
answered Sep 20 '18 at 20:58
amdamd
29.7k21050
$endgroup$
You’ve got the right idea, but there’s a detail that you might’ve overlooked. Every plane that passes through the intersection of $P_1$ and $P_2$ has an equation that’s a nontrivial linear combination of their equations: $$s(x+y+z-1)+t(x-y+z-2)=0.$$ What you’ve done is to set $s=1$ in this equation, which excludes $P_1$ from the set. For this problem, we know that $P_1$ doesn’t pass through the origin and therefore can’t be the solution, so it’s OK to do this, but you need to be careful that you don’t exclude a solution by doing this in other similar problems.
Continuing from the above linear combination, setting $x=y=z=0$ reduces the equation to $-s-2t=0$, or $s=-2t$. (In fact, since the constant term of a plane equation that passes through the origin is zero, we could’ve reached this constraint without bothering to compute any of the other terms.) Taking $t=-1$ produces the equation $x+3y+z=0$, which clearly passes through the origin as required.
answered Sep 20 '18 at 20:58
amdamd
29.7k21050
edited Dec 6 '18 at 1:04
answered Sep 20 '18 at 20:58
amdamd
29.7k21050
answered Sep 20 '18 at 20:58
amdamd
29.7k21050
answered Sep 20 '18 at 20:58
amdamd
29.7k21050
29.7k21050
add a comment |
add a comment |
$begingroup$
HINT
We can proceed as follows
- the general equation for $P$ is $ax+by+cz=0$ with normal $vec n=(a,b,c)$
- find the common line to $P_1$ and $P_2$; note that it suffices to find a common point $Q$ and the direction vector that is $vec v=vec n_1times vec n_2$
- then $vec n=vec {OQ}times vec v$
answered Sep 20 '18 at 20:42
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
HINT
We can proceed as follows
- the general equation for $P$ is $ax+by+cz=0$ with normal $vec n=(a,b,c)$
- find the common line to $P_1$ and $P_2$; note that it suffices to find a common point $Q$ and the direction vector that is $vec v=vec n_1times vec n_2$
- then $vec n=vec {OQ}times vec v$
answered Sep 20 '18 at 20:42
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
HINT
We can proceed as follows
- the general equation for $P$ is $ax+by+cz=0$ with normal $vec n=(a,b,c)$
- find the common line to $P_1$ and $P_2$; note that it suffices to find a common point $Q$ and the direction vector that is $vec v=vec n_1times vec n_2$
- then $vec n=vec {OQ}times vec v$
answered Sep 20 '18 at 20:42
gimusigimusi
92.8k84494
$endgroup$
HINT
We can proceed as follows
- the general equation for $P$ is $ax+by+cz=0$ with normal $vec n=(a,b,c)$
- find the common line to $P_1$ and $P_2$; note that it suffices to find a common point $Q$ and the direction vector that is $vec v=vec n_1times vec n_2$
- then $vec n=vec {OQ}times vec v$
answered Sep 20 '18 at 20:42
gimusigimusi
92.8k84494
answered Sep 20 '18 at 20:42
gimusigimusi
92.8k84494
answered Sep 20 '18 at 20:42
gimusigimusi
92.8k84494
answered Sep 20 '18 at 20:42
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
$begingroup$
Adding x + y + z = 1 and x - y + z = 2 gives 2x + 2z = 3 or z = 3/2- x. putting that into x+ y+ z= 1, x+ y+ 3/2- x= y+ 3/2= 1 so y= -3/2. Taking t= x as parameter, the line of intersection is x= t, y= -3/2, z= 3/2- t. Any plane that contains the origin can be written as z= Ax+ By. In order that it contains that line we must have 3/2- t= At- (3/2)B= 0, so (A+ 1)t=(5/2)B for all t. In order that this be true for all t, we must have A- 1= 0 and (5/2)B= 0. That is, z= x.
edited Sep 21 '18 at 23:53
amWhy
192k28225439
answered Sep 20 '18 at 20:52
user247327user247327
10.7k1515
$endgroup$
add a comment |
$begingroup$
Adding x + y + z = 1 and x - y + z = 2 gives 2x + 2z = 3 or z = 3/2- x. putting that into x+ y+ z= 1, x+ y+ 3/2- x= y+ 3/2= 1 so y= -3/2. Taking t= x as parameter, the line of intersection is x= t, y= -3/2, z= 3/2- t. Any plane that contains the origin can be written as z= Ax+ By. In order that it contains that line we must have 3/2- t= At- (3/2)B= 0, so (A+ 1)t=(5/2)B for all t. In order that this be true for all t, we must have A- 1= 0 and (5/2)B= 0. That is, z= x.
edited Sep 21 '18 at 23:53
amWhy
192k28225439
answered Sep 20 '18 at 20:52
user247327user247327
10.7k1515
$endgroup$
add a comment |
$begingroup$
Adding x + y + z = 1 and x - y + z = 2 gives 2x + 2z = 3 or z = 3/2- x. putting that into x+ y+ z= 1, x+ y+ 3/2- x= y+ 3/2= 1 so y= -3/2. Taking t= x as parameter, the line of intersection is x= t, y= -3/2, z= 3/2- t. Any plane that contains the origin can be written as z= Ax+ By. In order that it contains that line we must have 3/2- t= At- (3/2)B= 0, so (A+ 1)t=(5/2)B for all t. In order that this be true for all t, we must have A- 1= 0 and (5/2)B= 0. That is, z= x.
edited Sep 21 '18 at 23:53
amWhy
192k28225439
answered Sep 20 '18 at 20:52
user247327user247327
10.7k1515
$endgroup$
Adding x + y + z = 1 and x - y + z = 2 gives 2x + 2z = 3 or z = 3/2- x. putting that into x+ y+ z= 1, x+ y+ 3/2- x= y+ 3/2= 1 so y= -3/2. Taking t= x as parameter, the line of intersection is x= t, y= -3/2, z= 3/2- t. Any plane that contains the origin can be written as z= Ax+ By. In order that it contains that line we must have 3/2- t= At- (3/2)B= 0, so (A+ 1)t=(5/2)B for all t. In order that this be true for all t, we must have A- 1= 0 and (5/2)B= 0. That is, z= x.
edited Sep 21 '18 at 23:53
amWhy
192k28225439
answered Sep 20 '18 at 20:52
user247327user247327
10.7k1515
edited Sep 21 '18 at 23:53
amWhy
192k28225439
edited Sep 21 '18 at 23:53
amWhy
192k28225439
edited Sep 21 '18 at 23:53
amWhy
192k28225439
192k28225439
answered Sep 20 '18 at 20:52
user247327user247327
10.7k1515
answered Sep 20 '18 at 20:52
user247327user247327
10.7k1515
answered Sep 20 '18 at 20:52
user247327user247327
10.7k1515
10.7k1515
add a comment |
add a comment |
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$begingroup$
Welcome to Maths SX! What is the question?
$endgroup$
– Bernard
Sep 20 '18 at 20:07
$begingroup$
Find the plane P passing through the origin such that the three planes P, P1=(x+y+z=1） and P2 (x-y+z=2) meet along a line in R3.
$endgroup$
– Kelvin Cheung
Sep 20 '18 at 20:14
$begingroup$
I think you have the right idea, but you have to formulate it it in a correct way (what's a linear combination of equalities?)
$endgroup$
– Bernard
Sep 20 '18 at 20:17
$begingroup$
See this question and its solutions
$endgroup$
– user376343
Sep 20 '18 at 20:50