Show that $ab + bc +ac = 2abc +1 $ for given planes
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If the three planes with the equation below have a line in common, show that $ab + bc +ac = 2abc +1 $
$$pi_1 : x+by+cz = 0 \
pi_2: ax+y+cz = 0 \
pi_3: ax +by +z = 0$$
I am not sure how to solve this problem. I am not sure if this is solvable using matrix.
$$begin{bmatrix}1 & b & c\a & 1 & c \ a & b & 1end{bmatrix}$$
matrices vector-spaces
asked Nov 22 at 23:08
didgocks
678823
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up vote
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favorite
If the three planes with the equation below have a line in common, show that $ab + bc +ac = 2abc +1 $
$$pi_1 : x+by+cz = 0 \
pi_2: ax+y+cz = 0 \
pi_3: ax +by +z = 0$$
I am not sure how to solve this problem. I am not sure if this is solvable using matrix.
$$begin{bmatrix}1 & b & c\a & 1 & c \ a & b & 1end{bmatrix}$$
matrices vector-spaces
asked Nov 22 at 23:08
didgocks
678823
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If the three planes with the equation below have a line in common, show that $ab + bc +ac = 2abc +1 $
$$pi_1 : x+by+cz = 0 \
pi_2: ax+y+cz = 0 \
pi_3: ax +by +z = 0$$
I am not sure how to solve this problem. I am not sure if this is solvable using matrix.
$$begin{bmatrix}1 & b & c\a & 1 & c \ a & b & 1end{bmatrix}$$
matrices vector-spaces
asked Nov 22 at 23:08
didgocks
678823
If the three planes with the equation below have a line in common, show that $ab + bc +ac = 2abc +1 $
$$pi_1 : x+by+cz = 0 \
pi_2: ax+y+cz = 0 \
pi_3: ax +by +z = 0$$
I am not sure how to solve this problem. I am not sure if this is solvable using matrix.
$$begin{bmatrix}1 & b & c\a & 1 & c \ a & b & 1end{bmatrix}$$
matrices vector-spaces
matrices vector-spaces
asked Nov 22 at 23:08
didgocks
678823
asked Nov 22 at 23:08
didgocks
678823
edited Nov 22 at 23:38
asked Nov 22 at 23:08
didgocks
678823
asked Nov 22 at 23:08
didgocks
678823
asked Nov 22 at 23:08
didgocks
678823
678823
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4 Answers
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If the planes have a common line, then the system has infinite number of solutions, and consequently is the determinant of the matrix zero. Just compute the determinant.
answered Nov 23 at 0:31
user376343
2,7012822
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0
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Hint: Call your matrix $M$ and set $,v=(x,y,z)^T$, then we are looking for a nonzero $v$ (representing the direction vector of the common line) such that $Mv=0$, in other words the assumption is that the kernel of $M$ is not trivial.
answered Nov 22 at 23:13
Berci
59.3k23672
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Subtract the second equations from the first one; the third one from the second one and the first one from the third. Then you get three new equations which gives as a result: $(1-a)x=(1-b)y=(1-c)z$. Plug everything in terms of x in the first (second and third will also do) equation. The you get $x(1+frac{1-a}{1-b}b+frac{1-a}{1-c}c)=0$. For $xneq 0$ we can devide this equation by zero. We obtain after working a bit out: $frac{(1-b)(1-c)+(1-a)((1-c)b+(1-b)c)}{(1-b)(1-c)}=0$. Since this fraction is equal to zero, we can ignore the denominator. Just write out all the brackets in the numerator and you will find your desired result.
answered Nov 23 at 0:30
esmo
406
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0
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Since line is a one-dimensional subspace, therefore the kernel of the matrix must be a one-dimensional subspace i.e. the rank of the matrix must be 2 (the determinant must be zero) and no row be a multiple of any other row: $$begin{vmatrix}1 & b & c\a & 1 & c \ a & b & 1end{vmatrix}=(1-bc)-b(a-ac)+c(ab-a)=2abc+1-ab-ac-bc=0$$which results what we sought. Also the condition that no two rows are multiples of each other holds automatically.
answered Nov 24 at 15:13
Mostafa Ayaz
13.5k3836
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If the planes have a common line, then the system has infinite number of solutions, and consequently is the determinant of the matrix zero. Just compute the determinant.
answered Nov 23 at 0:31
user376343
2,7012822
add a comment |
up vote
1
down vote
If the planes have a common line, then the system has infinite number of solutions, and consequently is the determinant of the matrix zero. Just compute the determinant.
answered Nov 23 at 0:31
user376343
2,7012822
add a comment |
up vote
1
down vote
up vote
1
down vote
If the planes have a common line, then the system has infinite number of solutions, and consequently is the determinant of the matrix zero. Just compute the determinant.
answered Nov 23 at 0:31
user376343
2,7012822
If the planes have a common line, then the system has infinite number of solutions, and consequently is the determinant of the matrix zero. Just compute the determinant.
answered Nov 23 at 0:31
user376343
2,7012822
answered Nov 23 at 0:31
user376343
2,7012822
answered Nov 23 at 0:31
user376343
2,7012822
answered Nov 23 at 0:31
user376343
2,7012822
2,7012822
add a comment |
add a comment |
up vote
0
down vote
Hint: Call your matrix $M$ and set $,v=(x,y,z)^T$, then we are looking for a nonzero $v$ (representing the direction vector of the common line) such that $Mv=0$, in other words the assumption is that the kernel of $M$ is not trivial.
answered Nov 22 at 23:13
Berci
59.3k23672
add a comment |
up vote
0
down vote
Hint: Call your matrix $M$ and set $,v=(x,y,z)^T$, then we are looking for a nonzero $v$ (representing the direction vector of the common line) such that $Mv=0$, in other words the assumption is that the kernel of $M$ is not trivial.
answered Nov 22 at 23:13
Berci
59.3k23672
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: Call your matrix $M$ and set $,v=(x,y,z)^T$, then we are looking for a nonzero $v$ (representing the direction vector of the common line) such that $Mv=0$, in other words the assumption is that the kernel of $M$ is not trivial.
answered Nov 22 at 23:13
Berci
59.3k23672
Hint: Call your matrix $M$ and set $,v=(x,y,z)^T$, then we are looking for a nonzero $v$ (representing the direction vector of the common line) such that $Mv=0$, in other words the assumption is that the kernel of $M$ is not trivial.
answered Nov 22 at 23:13
Berci
59.3k23672
answered Nov 22 at 23:13
Berci
59.3k23672
answered Nov 22 at 23:13
Berci
59.3k23672
answered Nov 22 at 23:13
Berci
59.3k23672
59.3k23672
add a comment |
add a comment |
up vote
0
down vote
Subtract the second equations from the first one; the third one from the second one and the first one from the third. Then you get three new equations which gives as a result: $(1-a)x=(1-b)y=(1-c)z$. Plug everything in terms of x in the first (second and third will also do) equation. The you get $x(1+frac{1-a}{1-b}b+frac{1-a}{1-c}c)=0$. For $xneq 0$ we can devide this equation by zero. We obtain after working a bit out: $frac{(1-b)(1-c)+(1-a)((1-c)b+(1-b)c)}{(1-b)(1-c)}=0$. Since this fraction is equal to zero, we can ignore the denominator. Just write out all the brackets in the numerator and you will find your desired result.
answered Nov 23 at 0:30
esmo
406
add a comment |
up vote
0
down vote
Subtract the second equations from the first one; the third one from the second one and the first one from the third. Then you get three new equations which gives as a result: $(1-a)x=(1-b)y=(1-c)z$. Plug everything in terms of x in the first (second and third will also do) equation. The you get $x(1+frac{1-a}{1-b}b+frac{1-a}{1-c}c)=0$. For $xneq 0$ we can devide this equation by zero. We obtain after working a bit out: $frac{(1-b)(1-c)+(1-a)((1-c)b+(1-b)c)}{(1-b)(1-c)}=0$. Since this fraction is equal to zero, we can ignore the denominator. Just write out all the brackets in the numerator and you will find your desired result.
answered Nov 23 at 0:30
esmo
406
add a comment |
up vote
0
down vote
up vote
0
down vote
Subtract the second equations from the first one; the third one from the second one and the first one from the third. Then you get three new equations which gives as a result: $(1-a)x=(1-b)y=(1-c)z$. Plug everything in terms of x in the first (second and third will also do) equation. The you get $x(1+frac{1-a}{1-b}b+frac{1-a}{1-c}c)=0$. For $xneq 0$ we can devide this equation by zero. We obtain after working a bit out: $frac{(1-b)(1-c)+(1-a)((1-c)b+(1-b)c)}{(1-b)(1-c)}=0$. Since this fraction is equal to zero, we can ignore the denominator. Just write out all the brackets in the numerator and you will find your desired result.
answered Nov 23 at 0:30
esmo
406
Subtract the second equations from the first one; the third one from the second one and the first one from the third. Then you get three new equations which gives as a result: $(1-a)x=(1-b)y=(1-c)z$. Plug everything in terms of x in the first (second and third will also do) equation. The you get $x(1+frac{1-a}{1-b}b+frac{1-a}{1-c}c)=0$. For $xneq 0$ we can devide this equation by zero. We obtain after working a bit out: $frac{(1-b)(1-c)+(1-a)((1-c)b+(1-b)c)}{(1-b)(1-c)}=0$. Since this fraction is equal to zero, we can ignore the denominator. Just write out all the brackets in the numerator and you will find your desired result.
answered Nov 23 at 0:30
esmo
406
answered Nov 23 at 0:30
esmo
406
answered Nov 23 at 0:30
esmo
406
answered Nov 23 at 0:30
esmo
406
406
add a comment |
add a comment |
up vote
0
down vote
Since line is a one-dimensional subspace, therefore the kernel of the matrix must be a one-dimensional subspace i.e. the rank of the matrix must be 2 (the determinant must be zero) and no row be a multiple of any other row: $$begin{vmatrix}1 & b & c\a & 1 & c \ a & b & 1end{vmatrix}=(1-bc)-b(a-ac)+c(ab-a)=2abc+1-ab-ac-bc=0$$which results what we sought. Also the condition that no two rows are multiples of each other holds automatically.
answered Nov 24 at 15:13
Mostafa Ayaz
13.5k3836
add a comment |
up vote
0
down vote
Since line is a one-dimensional subspace, therefore the kernel of the matrix must be a one-dimensional subspace i.e. the rank of the matrix must be 2 (the determinant must be zero) and no row be a multiple of any other row: $$begin{vmatrix}1 & b & c\a & 1 & c \ a & b & 1end{vmatrix}=(1-bc)-b(a-ac)+c(ab-a)=2abc+1-ab-ac-bc=0$$which results what we sought. Also the condition that no two rows are multiples of each other holds automatically.
answered Nov 24 at 15:13
Mostafa Ayaz
13.5k3836
add a comment |
up vote
0
down vote
up vote
0
down vote
Since line is a one-dimensional subspace, therefore the kernel of the matrix must be a one-dimensional subspace i.e. the rank of the matrix must be 2 (the determinant must be zero) and no row be a multiple of any other row: $$begin{vmatrix}1 & b & c\a & 1 & c \ a & b & 1end{vmatrix}=(1-bc)-b(a-ac)+c(ab-a)=2abc+1-ab-ac-bc=0$$which results what we sought. Also the condition that no two rows are multiples of each other holds automatically.
answered Nov 24 at 15:13
Mostafa Ayaz
13.5k3836
Since line is a one-dimensional subspace, therefore the kernel of the matrix must be a one-dimensional subspace i.e. the rank of the matrix must be 2 (the determinant must be zero) and no row be a multiple of any other row: $$begin{vmatrix}1 & b & c\a & 1 & c \ a & b & 1end{vmatrix}=(1-bc)-b(a-ac)+c(ab-a)=2abc+1-ab-ac-bc=0$$which results what we sought. Also the condition that no two rows are multiples of each other holds automatically.
answered Nov 24 at 15:13
Mostafa Ayaz
13.5k3836
answered Nov 24 at 15:13
Mostafa Ayaz
13.5k3836
answered Nov 24 at 15:13
Mostafa Ayaz
13.5k3836
answered Nov 24 at 15:13
Mostafa Ayaz
13.5k3836
13.5k3836
add a comment |
add a comment |
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