Find the planes that are parallel to a line and have a distance of 2 to the point P(1,1,1)
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I have to give the equations of the planes that go through the point $(3,4,5)$ that are parallel to the lines with direction $(0,2,1)$ and that are on a distance 2 from the point $(1,1,1)$.
So if you say that the other vector of this plane is given by $(a, b, c)$ then I came to the conclusion that the equations of these planes are given by
$alpha leftrightarrow (2c-b)x + ay - 2az = -6a-3b+6c$
but now I don't know how I can use the distance to calculate the specific equations because you can't really derive a relation between $a,b$ and $c$.
Is there something I'm missing or an easier approach to solve this problem?
euclidean-geometry
asked Jun 6 at 9:53
Mee98
19010
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up vote
1
down vote
favorite
I have to give the equations of the planes that go through the point $(3,4,5)$ that are parallel to the lines with direction $(0,2,1)$ and that are on a distance 2 from the point $(1,1,1)$.
So if you say that the other vector of this plane is given by $(a, b, c)$ then I came to the conclusion that the equations of these planes are given by
$alpha leftrightarrow (2c-b)x + ay - 2az = -6a-3b+6c$
but now I don't know how I can use the distance to calculate the specific equations because you can't really derive a relation between $a,b$ and $c$.
Is there something I'm missing or an easier approach to solve this problem?
euclidean-geometry
asked Jun 6 at 9:53
Mee98
19010
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have to give the equations of the planes that go through the point $(3,4,5)$ that are parallel to the lines with direction $(0,2,1)$ and that are on a distance 2 from the point $(1,1,1)$.
So if you say that the other vector of this plane is given by $(a, b, c)$ then I came to the conclusion that the equations of these planes are given by
$alpha leftrightarrow (2c-b)x + ay - 2az = -6a-3b+6c$
but now I don't know how I can use the distance to calculate the specific equations because you can't really derive a relation between $a,b$ and $c$.
Is there something I'm missing or an easier approach to solve this problem?
euclidean-geometry
asked Jun 6 at 9:53
Mee98
19010
I have to give the equations of the planes that go through the point $(3,4,5)$ that are parallel to the lines with direction $(0,2,1)$ and that are on a distance 2 from the point $(1,1,1)$.
So if you say that the other vector of this plane is given by $(a, b, c)$ then I came to the conclusion that the equations of these planes are given by
$alpha leftrightarrow (2c-b)x + ay - 2az = -6a-3b+6c$
but now I don't know how I can use the distance to calculate the specific equations because you can't really derive a relation between $a,b$ and $c$.
Is there something I'm missing or an easier approach to solve this problem?
euclidean-geometry
euclidean-geometry
asked Jun 6 at 9:53
Mee98
19010
asked Jun 6 at 9:53
Mee98
19010
asked Jun 6 at 9:53
Mee98
19010
asked Jun 6 at 9:53
Mee98
19010
asked Jun 6 at 9:53
Mee98
19010
19010
add a comment |
add a comment |
2 Answers
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HINT
Let consider the plane equation $ax+by+cz+1=0$ (recall indeed that we can assume wlog $d=1$) and the given conditions
$3a+4b+5c+1=0$
$0cdot a+2cdot b+1cdot c=0$
$frac{|1cdot a+1cdot b+1cdot c+1|}{sqrt{a^2+b^2+c^2}}=2$
from which we can find $a,b,c$.
answered Jun 6 at 9:58
gimusi
1
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Working this problem from a perspective-geometric point of view, the first two conditions define a pencil of planes that contain the points $[3:4:5:1]$ and $[0:2:1:0]$ (the point at infinity on the given set of parallel lines). These planes are null vectors of the matrix $$begin{bmatrix}3&4&5&1\0&2&1&0end{bmatrix}.$$ This space is spanned by $[1:0:0:-3]$ and $[2:1:-1:0]$, so the planes containing the two points are linear combinations of these basis vectors: $$mathbfpi(lambda,mu) = lambda[1:0:0:-3]+mu[2:1:-2:0] = [lambda+2mu:mu:-2mu:-3lambda],$$ i.e., every plane that satisfies the first two conditions has an equation of the form $$(lambda+2mu)x + mu y - 2mu z - 3lambda = 0$$ for $lambda$ and $mu$ not both zero. Using the usual formula for the distance between a point and plane and squaring to eliminate the square root will give you a homogeneous quadratic equation in $lambda$ and $mu$, which in this case easily factors to give two linearly independent solutions.
answered Jun 6 at 19:42
amd
29k21050
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
HINT
Let consider the plane equation $ax+by+cz+1=0$ (recall indeed that we can assume wlog $d=1$) and the given conditions
$3a+4b+5c+1=0$
$0cdot a+2cdot b+1cdot c=0$
$frac{|1cdot a+1cdot b+1cdot c+1|}{sqrt{a^2+b^2+c^2}}=2$
from which we can find $a,b,c$.
answered Jun 6 at 9:58
gimusi
1
add a comment |
up vote
1
down vote
accepted
HINT
Let consider the plane equation $ax+by+cz+1=0$ (recall indeed that we can assume wlog $d=1$) and the given conditions
$3a+4b+5c+1=0$
$0cdot a+2cdot b+1cdot c=0$
$frac{|1cdot a+1cdot b+1cdot c+1|}{sqrt{a^2+b^2+c^2}}=2$
from which we can find $a,b,c$.
answered Jun 6 at 9:58
gimusi
1
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
HINT
Let consider the plane equation $ax+by+cz+1=0$ (recall indeed that we can assume wlog $d=1$) and the given conditions
$3a+4b+5c+1=0$
$0cdot a+2cdot b+1cdot c=0$
$frac{|1cdot a+1cdot b+1cdot c+1|}{sqrt{a^2+b^2+c^2}}=2$
from which we can find $a,b,c$.
answered Jun 6 at 9:58
gimusi
1
HINT
Let consider the plane equation $ax+by+cz+1=0$ (recall indeed that we can assume wlog $d=1$) and the given conditions
$3a+4b+5c+1=0$
$0cdot a+2cdot b+1cdot c=0$
$frac{|1cdot a+1cdot b+1cdot c+1|}{sqrt{a^2+b^2+c^2}}=2$
from which we can find $a,b,c$.
answered Jun 6 at 9:58
gimusi
1
edited Jun 6 at 10:19
answered Jun 6 at 9:58
gimusi
1
answered Jun 6 at 9:58
gimusi
1
answered Jun 6 at 9:58
gimusi
1
1
add a comment |
add a comment |
up vote
0
down vote
Working this problem from a perspective-geometric point of view, the first two conditions define a pencil of planes that contain the points $[3:4:5:1]$ and $[0:2:1:0]$ (the point at infinity on the given set of parallel lines). These planes are null vectors of the matrix $$begin{bmatrix}3&4&5&1\0&2&1&0end{bmatrix}.$$ This space is spanned by $[1:0:0:-3]$ and $[2:1:-1:0]$, so the planes containing the two points are linear combinations of these basis vectors: $$mathbfpi(lambda,mu) = lambda[1:0:0:-3]+mu[2:1:-2:0] = [lambda+2mu:mu:-2mu:-3lambda],$$ i.e., every plane that satisfies the first two conditions has an equation of the form $$(lambda+2mu)x + mu y - 2mu z - 3lambda = 0$$ for $lambda$ and $mu$ not both zero. Using the usual formula for the distance between a point and plane and squaring to eliminate the square root will give you a homogeneous quadratic equation in $lambda$ and $mu$, which in this case easily factors to give two linearly independent solutions.
answered Jun 6 at 19:42
amd
29k21050
add a comment |
up vote
0
down vote
Working this problem from a perspective-geometric point of view, the first two conditions define a pencil of planes that contain the points $[3:4:5:1]$ and $[0:2:1:0]$ (the point at infinity on the given set of parallel lines). These planes are null vectors of the matrix $$begin{bmatrix}3&4&5&1\0&2&1&0end{bmatrix}.$$ This space is spanned by $[1:0:0:-3]$ and $[2:1:-1:0]$, so the planes containing the two points are linear combinations of these basis vectors: $$mathbfpi(lambda,mu) = lambda[1:0:0:-3]+mu[2:1:-2:0] = [lambda+2mu:mu:-2mu:-3lambda],$$ i.e., every plane that satisfies the first two conditions has an equation of the form $$(lambda+2mu)x + mu y - 2mu z - 3lambda = 0$$ for $lambda$ and $mu$ not both zero. Using the usual formula for the distance between a point and plane and squaring to eliminate the square root will give you a homogeneous quadratic equation in $lambda$ and $mu$, which in this case easily factors to give two linearly independent solutions.
answered Jun 6 at 19:42
amd
29k21050
add a comment |
up vote
0
down vote
up vote
0
down vote
Working this problem from a perspective-geometric point of view, the first two conditions define a pencil of planes that contain the points $[3:4:5:1]$ and $[0:2:1:0]$ (the point at infinity on the given set of parallel lines). These planes are null vectors of the matrix $$begin{bmatrix}3&4&5&1\0&2&1&0end{bmatrix}.$$ This space is spanned by $[1:0:0:-3]$ and $[2:1:-1:0]$, so the planes containing the two points are linear combinations of these basis vectors: $$mathbfpi(lambda,mu) = lambda[1:0:0:-3]+mu[2:1:-2:0] = [lambda+2mu:mu:-2mu:-3lambda],$$ i.e., every plane that satisfies the first two conditions has an equation of the form $$(lambda+2mu)x + mu y - 2mu z - 3lambda = 0$$ for $lambda$ and $mu$ not both zero. Using the usual formula for the distance between a point and plane and squaring to eliminate the square root will give you a homogeneous quadratic equation in $lambda$ and $mu$, which in this case easily factors to give two linearly independent solutions.
answered Jun 6 at 19:42
amd
29k21050
Working this problem from a perspective-geometric point of view, the first two conditions define a pencil of planes that contain the points $[3:4:5:1]$ and $[0:2:1:0]$ (the point at infinity on the given set of parallel lines). These planes are null vectors of the matrix $$begin{bmatrix}3&4&5&1\0&2&1&0end{bmatrix}.$$ This space is spanned by $[1:0:0:-3]$ and $[2:1:-1:0]$, so the planes containing the two points are linear combinations of these basis vectors: $$mathbfpi(lambda,mu) = lambda[1:0:0:-3]+mu[2:1:-2:0] = [lambda+2mu:mu:-2mu:-3lambda],$$ i.e., every plane that satisfies the first two conditions has an equation of the form $$(lambda+2mu)x + mu y - 2mu z - 3lambda = 0$$ for $lambda$ and $mu$ not both zero. Using the usual formula for the distance between a point and plane and squaring to eliminate the square root will give you a homogeneous quadratic equation in $lambda$ and $mu$, which in this case easily factors to give two linearly independent solutions.
answered Jun 6 at 19:42
amd
29k21050
edited Dec 6 at 1:24
answered Jun 6 at 19:42
amd
29k21050
answered Jun 6 at 19:42
amd
29k21050
answered Jun 6 at 19:42
amd
29k21050
29k21050
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