Equivalent definition of plane as a locus
$begingroup$
I recently read definition of a plane as "given an equation in 3-variables, the locus of the equation will be a plane if every point of the line joining any two points on the locus also lie on the locus". While the definition intuitively makes sense, my question is how do I mathematically solve the issue i.e. what if I have to find a equation which satisfies the given definition?
Alternatively I want to do this -:
I have to find a equation satisfying $X(x) + Y(y) + Z(z) = 0$ such that
$X(x1) + Y(y1) + Z(z1) = 0$ - (i)
$X(x2) + Y(y2) + Z(z2) = 0$ - (ii)
where (x1, y1, z1) and (x2, y2, z2) are two points on a line satisfying the equation and then
$X(x1 + kx2/1+k) + Y(y1 + ky2/1 + k) + Z(z1 + kz2/1 + k) = 0$ - (iii)
where k is an arbitrary constant.
geometry
asked Dec 10 '18 at 6:19
Vishal GoyalVishal Goyal
11
$endgroup$
add a comment |
$begingroup$
I recently read definition of a plane as "given an equation in 3-variables, the locus of the equation will be a plane if every point of the line joining any two points on the locus also lie on the locus". While the definition intuitively makes sense, my question is how do I mathematically solve the issue i.e. what if I have to find a equation which satisfies the given definition?
Alternatively I want to do this -:
I have to find a equation satisfying $X(x) + Y(y) + Z(z) = 0$ such that
$X(x1) + Y(y1) + Z(z1) = 0$ - (i)
$X(x2) + Y(y2) + Z(z2) = 0$ - (ii)
where (x1, y1, z1) and (x2, y2, z2) are two points on a line satisfying the equation and then
$X(x1 + kx2/1+k) + Y(y1 + ky2/1 + k) + Z(z1 + kz2/1 + k) = 0$ - (iii)
where k is an arbitrary constant.
geometry
asked Dec 10 '18 at 6:19
Vishal GoyalVishal Goyal
11
$endgroup$
add a comment |
$begingroup$
I recently read definition of a plane as "given an equation in 3-variables, the locus of the equation will be a plane if every point of the line joining any two points on the locus also lie on the locus". While the definition intuitively makes sense, my question is how do I mathematically solve the issue i.e. what if I have to find a equation which satisfies the given definition?
Alternatively I want to do this -:
I have to find a equation satisfying $X(x) + Y(y) + Z(z) = 0$ such that
$X(x1) + Y(y1) + Z(z1) = 0$ - (i)
$X(x2) + Y(y2) + Z(z2) = 0$ - (ii)
where (x1, y1, z1) and (x2, y2, z2) are two points on a line satisfying the equation and then
$X(x1 + kx2/1+k) + Y(y1 + ky2/1 + k) + Z(z1 + kz2/1 + k) = 0$ - (iii)
where k is an arbitrary constant.
geometry
asked Dec 10 '18 at 6:19
Vishal GoyalVishal Goyal
11
$endgroup$
I recently read definition of a plane as "given an equation in 3-variables, the locus of the equation will be a plane if every point of the line joining any two points on the locus also lie on the locus". While the definition intuitively makes sense, my question is how do I mathematically solve the issue i.e. what if I have to find a equation which satisfies the given definition?
Alternatively I want to do this -:
I have to find a equation satisfying $X(x) + Y(y) + Z(z) = 0$ such that
$X(x1) + Y(y1) + Z(z1) = 0$ - (i)
$X(x2) + Y(y2) + Z(z2) = 0$ - (ii)
where (x1, y1, z1) and (x2, y2, z2) are two points on a line satisfying the equation and then
$X(x1 + kx2/1+k) + Y(y1 + ky2/1 + k) + Z(z1 + kz2/1 + k) = 0$ - (iii)
where k is an arbitrary constant.
geometry
geometry
asked Dec 10 '18 at 6:19
Vishal GoyalVishal Goyal
11
asked Dec 10 '18 at 6:19
Vishal GoyalVishal Goyal
11
asked Dec 10 '18 at 6:19
Vishal GoyalVishal Goyal
11
asked Dec 10 '18 at 6:19
Vishal GoyalVishal Goyal
11
asked Dec 10 '18 at 6:19
Vishal GoyalVishal Goyal
11
11
add a comment |
add a comment |
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