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Find sphere equation
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How to find sphere equation with center in plane $pi =x-5y+z-2=0$ and tangent to plane $alpha = 2x+3y-4z-1=0 $ at $M(1,1,1)$
Where I've stopped:
Knowing the sphere equation is: $$(X-a)^2 + (Y-b)^2 + (Z-c)^2 = r^2$$
Center of sphere is $C(a,b,c)$ and can be found in plane $pi$
$r^2$ is radius and can be found doing distance from center of sphere to tangent point M ( $d(C,M)$ )
My problem is center point, how can I find this? My first thought was something using plane $pi$ but then plane $alpha$ was given in question and never used.
geometry 3d spheres solid-geometry
edited 16 hours ago
Michael Rozenberg
86.9k1575178
asked 17 hours ago
Mihai Preda
614
add a comment |Â
up vote
0
down vote
favorite
How to find sphere equation with center in plane $pi =x-5y+z-2=0$ and tangent to plane $alpha = 2x+3y-4z-1=0 $ at $M(1,1,1)$
Where I've stopped:
Knowing the sphere equation is: $$(X-a)^2 + (Y-b)^2 + (Z-c)^2 = r^2$$
Center of sphere is $C(a,b,c)$ and can be found in plane $pi$
$r^2$ is radius and can be found doing distance from center of sphere to tangent point M ( $d(C,M)$ )
My problem is center point, how can I find this? My first thought was something using plane $pi$ but then plane $alpha$ was given in question and never used.
geometry 3d spheres solid-geometry
edited 16 hours ago
Michael Rozenberg
86.9k1575178
asked 17 hours ago
Mihai Preda
614
The answer is not nice.
– Michael Rozenberg
16 hours ago
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to find sphere equation with center in plane $pi =x-5y+z-2=0$ and tangent to plane $alpha = 2x+3y-4z-1=0 $ at $M(1,1,1)$
Where I've stopped:
Knowing the sphere equation is: $$(X-a)^2 + (Y-b)^2 + (Z-c)^2 = r^2$$
Center of sphere is $C(a,b,c)$ and can be found in plane $pi$
$r^2$ is radius and can be found doing distance from center of sphere to tangent point M ( $d(C,M)$ )
My problem is center point, how can I find this? My first thought was something using plane $pi$ but then plane $alpha$ was given in question and never used.
geometry 3d spheres solid-geometry
edited 16 hours ago
Michael Rozenberg
86.9k1575178
asked 17 hours ago
Mihai Preda
614
How to find sphere equation with center in plane $pi =x-5y+z-2=0$ and tangent to plane $alpha = 2x+3y-4z-1=0 $ at $M(1,1,1)$
Where I've stopped:
Knowing the sphere equation is: $$(X-a)^2 + (Y-b)^2 + (Z-c)^2 = r^2$$
Center of sphere is $C(a,b,c)$ and can be found in plane $pi$
$r^2$ is radius and can be found doing distance from center of sphere to tangent point M ( $d(C,M)$ )
My problem is center point, how can I find this? My first thought was something using plane $pi$ but then plane $alpha$ was given in question and never used.
geometry 3d spheres solid-geometry
edited 16 hours ago
Michael Rozenberg
86.9k1575178
asked 17 hours ago
Mihai Preda
614
edited 16 hours ago
Michael Rozenberg
86.9k1575178
edited 16 hours ago
Michael Rozenberg
86.9k1575178
edited 16 hours ago
Michael Rozenberg
86.9k1575178
86.9k1575178
asked 17 hours ago
Mihai Preda
614
asked 17 hours ago
Mihai Preda
614
asked 17 hours ago
Mihai Preda
614
614
The answer is not nice.
– Michael Rozenberg
16 hours ago
add a comment |Â
The answer is not nice.
– Michael Rozenberg
16 hours ago
The answer is not nice.
– Michael Rozenberg
16 hours ago
The answer is not nice.
– Michael Rozenberg
16 hours ago
add a comment |Â
3 Answers
3
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oldest
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up vote
1
down vote
Hint: Radius and tangent are perpendicular. Where does the perpendicular to $alpha$ at $M$ hit $pi$?
A different way: Parametrize the plane $pi$, then find for which value of the parameters the vector $vecMP$ (with $Pinpi$) is orthogonal to $alpha$.
answered 17 hours ago
Arnaud Mortier
17.6k21757
In projection of M on $alpha$ $M'$ ?
– Mihai Preda
17 hours ago
@MihaiPreda Sorry, what? I suggest that you find the line that is orthogonal to $alpha$ at point $M$ and then where this line hits $pi$.
– Arnaud Mortier
17 hours ago
add a comment |Â
up vote
0
down vote
$$MC: (1,1,1)+t(2,3,-4).$$
Let $C(1+2t,1+3t,1-4t).$
Thus,
$$1+2t-5(1+3t)+1-4t-2=0,$$ which gives $t=-frac517,$
$$Cleft(frac717,frac217,frac3717right),$$
$$MC=sqrtfrac725289$$ and the answer is
$$left(x-frac717right)^2+left(y-frac217right)^2+left(z-frac3717right)^2=frac725289.$$
answered 16 hours ago
Michael Rozenberg
86.9k1575178
add a comment |Â
up vote
0
down vote
The center of the sphere is the intersection of the plane $pi $ with the line passing through $M$ and perpendicular to $alpha$
The equation of the line passing through M and perpendicular to $alpha$ is $$x=1+2t, y=1+3t, z=1-4t$$
The point of intersection with $pi $ is $$ C(7/17,2/17,37/17)$$which is the center of the sphere.
The radius is simply $CM$ the distance between $C$ and $M$.
answered 16 hours ago
Mohammad Riazi-Kermani
26.9k41849
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: Radius and tangent are perpendicular. Where does the perpendicular to $alpha$ at $M$ hit $pi$?
A different way: Parametrize the plane $pi$, then find for which value of the parameters the vector $vecMP$ (with $Pinpi$) is orthogonal to $alpha$.
answered 17 hours ago
Arnaud Mortier
17.6k21757
In projection of M on $alpha$ $M'$ ?
– Mihai Preda
17 hours ago
@MihaiPreda Sorry, what? I suggest that you find the line that is orthogonal to $alpha$ at point $M$ and then where this line hits $pi$.
– Arnaud Mortier
17 hours ago
add a comment |Â
up vote
1
down vote
Hint: Radius and tangent are perpendicular. Where does the perpendicular to $alpha$ at $M$ hit $pi$?
A different way: Parametrize the plane $pi$, then find for which value of the parameters the vector $vecMP$ (with $Pinpi$) is orthogonal to $alpha$.
answered 17 hours ago
Arnaud Mortier
17.6k21757
In projection of M on $alpha$ $M'$ ?
– Mihai Preda
17 hours ago
@MihaiPreda Sorry, what? I suggest that you find the line that is orthogonal to $alpha$ at point $M$ and then where this line hits $pi$.
– Arnaud Mortier
17 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: Radius and tangent are perpendicular. Where does the perpendicular to $alpha$ at $M$ hit $pi$?
A different way: Parametrize the plane $pi$, then find for which value of the parameters the vector $vecMP$ (with $Pinpi$) is orthogonal to $alpha$.
answered 17 hours ago
Arnaud Mortier
17.6k21757
Hint: Radius and tangent are perpendicular. Where does the perpendicular to $alpha$ at $M$ hit $pi$?
A different way: Parametrize the plane $pi$, then find for which value of the parameters the vector $vecMP$ (with $Pinpi$) is orthogonal to $alpha$.
answered 17 hours ago
Arnaud Mortier
17.6k21757
edited 17 hours ago
answered 17 hours ago
Arnaud Mortier
17.6k21757
answered 17 hours ago
Arnaud Mortier
17.6k21757
answered 17 hours ago
Arnaud Mortier
17.6k21757
17.6k21757
In projection of M on $alpha$ $M'$ ?
– Mihai Preda
17 hours ago
@MihaiPreda Sorry, what? I suggest that you find the line that is orthogonal to $alpha$ at point $M$ and then where this line hits $pi$.
– Arnaud Mortier
17 hours ago
add a comment |Â
In projection of M on $alpha$ $M'$ ?
– Mihai Preda
17 hours ago
@MihaiPreda Sorry, what? I suggest that you find the line that is orthogonal to $alpha$ at point $M$ and then where this line hits $pi$.
– Arnaud Mortier
17 hours ago
In projection of M on $alpha$ $M'$ ?
– Mihai Preda
17 hours ago
In projection of M on $alpha$ $M'$ ?
– Mihai Preda
17 hours ago
@MihaiPreda Sorry, what? I suggest that you find the line that is orthogonal to $alpha$ at point $M$ and then where this line hits $pi$.
– Arnaud Mortier
17 hours ago
@MihaiPreda Sorry, what? I suggest that you find the line that is orthogonal to $alpha$ at point $M$ and then where this line hits $pi$.
– Arnaud Mortier
17 hours ago
add a comment |Â
up vote
0
down vote
$$MC: (1,1,1)+t(2,3,-4).$$
Let $C(1+2t,1+3t,1-4t).$
Thus,
$$1+2t-5(1+3t)+1-4t-2=0,$$ which gives $t=-frac517,$
$$Cleft(frac717,frac217,frac3717right),$$
$$MC=sqrtfrac725289$$ and the answer is
$$left(x-frac717right)^2+left(y-frac217right)^2+left(z-frac3717right)^2=frac725289.$$
answered 16 hours ago
Michael Rozenberg
86.9k1575178
add a comment |Â
up vote
0
down vote
$$MC: (1,1,1)+t(2,3,-4).$$
Let $C(1+2t,1+3t,1-4t).$
Thus,
$$1+2t-5(1+3t)+1-4t-2=0,$$ which gives $t=-frac517,$
$$Cleft(frac717,frac217,frac3717right),$$
$$MC=sqrtfrac725289$$ and the answer is
$$left(x-frac717right)^2+left(y-frac217right)^2+left(z-frac3717right)^2=frac725289.$$
answered 16 hours ago
Michael Rozenberg
86.9k1575178
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$MC: (1,1,1)+t(2,3,-4).$$
Let $C(1+2t,1+3t,1-4t).$
Thus,
$$1+2t-5(1+3t)+1-4t-2=0,$$ which gives $t=-frac517,$
$$Cleft(frac717,frac217,frac3717right),$$
$$MC=sqrtfrac725289$$ and the answer is
$$left(x-frac717right)^2+left(y-frac217right)^2+left(z-frac3717right)^2=frac725289.$$
answered 16 hours ago
Michael Rozenberg
86.9k1575178
$$MC: (1,1,1)+t(2,3,-4).$$
Let $C(1+2t,1+3t,1-4t).$
Thus,
$$1+2t-5(1+3t)+1-4t-2=0,$$ which gives $t=-frac517,$
$$Cleft(frac717,frac217,frac3717right),$$
$$MC=sqrtfrac725289$$ and the answer is
$$left(x-frac717right)^2+left(y-frac217right)^2+left(z-frac3717right)^2=frac725289.$$
answered 16 hours ago
Michael Rozenberg
86.9k1575178
answered 16 hours ago
Michael Rozenberg
86.9k1575178
answered 16 hours ago
Michael Rozenberg
86.9k1575178
answered 16 hours ago
Michael Rozenberg
86.9k1575178
86.9k1575178
add a comment |Â
add a comment |Â
up vote
0
down vote
The center of the sphere is the intersection of the plane $pi $ with the line passing through $M$ and perpendicular to $alpha$
The equation of the line passing through M and perpendicular to $alpha$ is $$x=1+2t, y=1+3t, z=1-4t$$
The point of intersection with $pi $ is $$ C(7/17,2/17,37/17)$$which is the center of the sphere.
The radius is simply $CM$ the distance between $C$ and $M$.
answered 16 hours ago
Mohammad Riazi-Kermani
26.9k41849
add a comment |Â
up vote
0
down vote
The center of the sphere is the intersection of the plane $pi $ with the line passing through $M$ and perpendicular to $alpha$
The equation of the line passing through M and perpendicular to $alpha$ is $$x=1+2t, y=1+3t, z=1-4t$$
The point of intersection with $pi $ is $$ C(7/17,2/17,37/17)$$which is the center of the sphere.
The radius is simply $CM$ the distance between $C$ and $M$.
answered 16 hours ago
Mohammad Riazi-Kermani
26.9k41849
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The center of the sphere is the intersection of the plane $pi $ with the line passing through $M$ and perpendicular to $alpha$
The equation of the line passing through M and perpendicular to $alpha$ is $$x=1+2t, y=1+3t, z=1-4t$$
The point of intersection with $pi $ is $$ C(7/17,2/17,37/17)$$which is the center of the sphere.
The radius is simply $CM$ the distance between $C$ and $M$.
answered 16 hours ago
Mohammad Riazi-Kermani
26.9k41849
The center of the sphere is the intersection of the plane $pi $ with the line passing through $M$ and perpendicular to $alpha$
The equation of the line passing through M and perpendicular to $alpha$ is $$x=1+2t, y=1+3t, z=1-4t$$
The point of intersection with $pi $ is $$ C(7/17,2/17,37/17)$$which is the center of the sphere.
The radius is simply $CM$ the distance between $C$ and $M$.
answered 16 hours ago
Mohammad Riazi-Kermani
26.9k41849
answered 16 hours ago
Mohammad Riazi-Kermani
26.9k41849
answered 16 hours ago
Mohammad Riazi-Kermani
26.9k41849
answered 16 hours ago
Mohammad Riazi-Kermani
26.9k41849
26.9k41849
add a comment |Â
add a comment |Â
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The answer is not nice.
– Michael Rozenberg
16 hours ago