Find distance to the origin of the tangent plane of $x^2-y^2+2z^2=5$ in the point $(2,-1,1)$












2












$begingroup$


I'm asked to find the distance to the origin of the tangent plane of the surface $x^2-y^2+2z^2=5$ in the point $(2,-1,1)$. This seems to be an optimization problem with constraint, which I think can be solved with Lagrange multiplier.



The tangent plane is of the form $f_x(p_0)(x-x_0)+f_y(p_0)(y-y_0)+f_z(p_0)(z-z_0)$ where $f_x=2x, f_y=-2y, f_z=4z$. So in that point, we get $4(x-2)+2(y+1)+4(z-1)$. This will be our constraint $g(x,y,z)$



The function we want to optimize is the Euclidian distance squared $(x-0)^2+(y-0)^2+(z-0)^2:=f(x,y,z)$



So with lagrange :$(2x,2y,2z)=lambda(4,2,4)$ so $(x,y,z)=lambda(2,1,2)$.



So $lambda=frac{x}{2}=y=frac{z}{2}$

So $x=z=2y$





My questions are :

1) Is what I did until now correct ?

2) If yes, how am I supposed to find the given point(s) now ? I have everything expressed in term of one variable but how do I find the value of this variable ? I thought about plugging $x$ and $z$ in terms of $y$ into the original surface equation $x^2-y^2+2z^2=5$ but we're asked to find the distance of the tangent plane of that equation at a point, so the tangent plane of that point isn't necessarily member of that surface, or am I wrong ?

Thanks for your help !






real-analysis calculus analysis multivariable-calculus optimization







share|cite|improve this question









$endgroup$












  • $begingroup$
    You haven’t used your constraint, which is the equation of the tangent plane.
    $endgroup$
    – amd
    Jan 16 at 18:11






  • 1




    $begingroup$
    Note that the distance of a plane from the origin can be read directly from its point-normal equation, so casting this as a Lagrange multiplier problem is a bit overkill.
    $endgroup$
    – amd
    Jan 16 at 18:12










  • $begingroup$
    You’ve found that a critical point of the Lagrangian must have $x=z=2y$. Find a point on the tangent plane that satisfies this condition.
    $endgroup$
    – amd
    Jan 16 at 18:13










  • $begingroup$
    @amd Okay, but still. If we go it that way, what do you mean with " haven’t used your constraint" ? I used it as my constraint.
    $endgroup$
    – Poujh
    Jan 16 at 18:13








  • 1




    $begingroup$
    Sounds like the right approach to me.
    $endgroup$
    – amd
    Jan 16 at 18:25
















2












$begingroup$


I'm asked to find the distance to the origin of the tangent plane of the surface $x^2-y^2+2z^2=5$ in the point $(2,-1,1)$. This seems to be an optimization problem with constraint, which I think can be solved with Lagrange multiplier.



The tangent plane is of the form $f_x(p_0)(x-x_0)+f_y(p_0)(y-y_0)+f_z(p_0)(z-z_0)$ where $f_x=2x, f_y=-2y, f_z=4z$. So in that point, we get $4(x-2)+2(y+1)+4(z-1)$. This will be our constraint $g(x,y,z)$



The function we want to optimize is the Euclidian distance squared $(x-0)^2+(y-0)^2+(z-0)^2:=f(x,y,z)$



So with lagrange :$(2x,2y,2z)=lambda(4,2,4)$ so $(x,y,z)=lambda(2,1,2)$.



So $lambda=frac{x}{2}=y=frac{z}{2}$

So $x=z=2y$





My questions are :

1) Is what I did until now correct ?

2) If yes, how am I supposed to find the given point(s) now ? I have everything expressed in term of one variable but how do I find the value of this variable ? I thought about plugging $x$ and $z$ in terms of $y$ into the original surface equation $x^2-y^2+2z^2=5$ but we're asked to find the distance of the tangent plane of that equation at a point, so the tangent plane of that point isn't necessarily member of that surface, or am I wrong ?

Thanks for your help !






real-analysis calculus analysis multivariable-calculus optimization







share|cite|improve this question









$endgroup$












  • $begingroup$
    You haven’t used your constraint, which is the equation of the tangent plane.
    $endgroup$
    – amd
    Jan 16 at 18:11






  • 1




    $begingroup$
    Note that the distance of a plane from the origin can be read directly from its point-normal equation, so casting this as a Lagrange multiplier problem is a bit overkill.
    $endgroup$
    – amd
    Jan 16 at 18:12










  • $begingroup$
    You’ve found that a critical point of the Lagrangian must have $x=z=2y$. Find a point on the tangent plane that satisfies this condition.
    $endgroup$
    – amd
    Jan 16 at 18:13










  • $begingroup$
    @amd Okay, but still. If we go it that way, what do you mean with " haven’t used your constraint" ? I used it as my constraint.
    $endgroup$
    – Poujh
    Jan 16 at 18:13








  • 1




    $begingroup$
    Sounds like the right approach to me.
    $endgroup$
    – amd
    Jan 16 at 18:25














2












2








2





$begingroup$


I'm asked to find the distance to the origin of the tangent plane of the surface $x^2-y^2+2z^2=5$ in the point $(2,-1,1)$. This seems to be an optimization problem with constraint, which I think can be solved with Lagrange multiplier.



The tangent plane is of the form $f_x(p_0)(x-x_0)+f_y(p_0)(y-y_0)+f_z(p_0)(z-z_0)$ where $f_x=2x, f_y=-2y, f_z=4z$. So in that point, we get $4(x-2)+2(y+1)+4(z-1)$. This will be our constraint $g(x,y,z)$



The function we want to optimize is the Euclidian distance squared $(x-0)^2+(y-0)^2+(z-0)^2:=f(x,y,z)$



So with lagrange :$(2x,2y,2z)=lambda(4,2,4)$ so $(x,y,z)=lambda(2,1,2)$.



So $lambda=frac{x}{2}=y=frac{z}{2}$

So $x=z=2y$





My questions are :

1) Is what I did until now correct ?

2) If yes, how am I supposed to find the given point(s) now ? I have everything expressed in term of one variable but how do I find the value of this variable ? I thought about plugging $x$ and $z$ in terms of $y$ into the original surface equation $x^2-y^2+2z^2=5$ but we're asked to find the distance of the tangent plane of that equation at a point, so the tangent plane of that point isn't necessarily member of that surface, or am I wrong ?

Thanks for your help !






real-analysis calculus analysis multivariable-calculus optimization







share|cite|improve this question









$endgroup$




I'm asked to find the distance to the origin of the tangent plane of the surface $x^2-y^2+2z^2=5$ in the point $(2,-1,1)$. This seems to be an optimization problem with constraint, which I think can be solved with Lagrange multiplier.



The tangent plane is of the form $f_x(p_0)(x-x_0)+f_y(p_0)(y-y_0)+f_z(p_0)(z-z_0)$ where $f_x=2x, f_y=-2y, f_z=4z$. So in that point, we get $4(x-2)+2(y+1)+4(z-1)$. This will be our constraint $g(x,y,z)$



The function we want to optimize is the Euclidian distance squared $(x-0)^2+(y-0)^2+(z-0)^2:=f(x,y,z)$



So with lagrange :$(2x,2y,2z)=lambda(4,2,4)$ so $(x,y,z)=lambda(2,1,2)$.



So $lambda=frac{x}{2}=y=frac{z}{2}$

So $x=z=2y$





My questions are :

1) Is what I did until now correct ?

2) If yes, how am I supposed to find the given point(s) now ? I have everything expressed in term of one variable but how do I find the value of this variable ? I thought about plugging $x$ and $z$ in terms of $y$ into the original surface equation $x^2-y^2+2z^2=5$ but we're asked to find the distance of the tangent plane of that equation at a point, so the tangent plane of that point isn't necessarily member of that surface, or am I wrong ?

Thanks for your help !






real-analysis calculus analysis multivariable-calculus optimization




real-analysis calculus analysis multivariable-calculus optimization






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 18:08









PoujhPoujh

6111516




6111516












  • $begingroup$
    You haven’t used your constraint, which is the equation of the tangent plane.
    $endgroup$
    – amd
    Jan 16 at 18:11






  • 1




    $begingroup$
    Note that the distance of a plane from the origin can be read directly from its point-normal equation, so casting this as a Lagrange multiplier problem is a bit overkill.
    $endgroup$
    – amd
    Jan 16 at 18:12










  • $begingroup$
    You’ve found that a critical point of the Lagrangian must have $x=z=2y$. Find a point on the tangent plane that satisfies this condition.
    $endgroup$
    – amd
    Jan 16 at 18:13










  • $begingroup$
    @amd Okay, but still. If we go it that way, what do you mean with " haven’t used your constraint" ? I used it as my constraint.
    $endgroup$
    – Poujh
    Jan 16 at 18:13








  • 1




    $begingroup$
    Sounds like the right approach to me.
    $endgroup$
    – amd
    Jan 16 at 18:25


















  • $begingroup$
    You haven’t used your constraint, which is the equation of the tangent plane.
    $endgroup$
    – amd
    Jan 16 at 18:11






  • 1




    $begingroup$
    Note that the distance of a plane from the origin can be read directly from its point-normal equation, so casting this as a Lagrange multiplier problem is a bit overkill.
    $endgroup$
    – amd
    Jan 16 at 18:12










  • $begingroup$
    You’ve found that a critical point of the Lagrangian must have $x=z=2y$. Find a point on the tangent plane that satisfies this condition.
    $endgroup$
    – amd
    Jan 16 at 18:13










  • $begingroup$
    @amd Okay, but still. If we go it that way, what do you mean with " haven’t used your constraint" ? I used it as my constraint.
    $endgroup$
    – Poujh
    Jan 16 at 18:13








  • 1




    $begingroup$
    Sounds like the right approach to me.
    $endgroup$
    – amd
    Jan 16 at 18:25
















$begingroup$
You haven’t used your constraint, which is the equation of the tangent plane.
$endgroup$
– amd
Jan 16 at 18:11




$begingroup$
You haven’t used your constraint, which is the equation of the tangent plane.
$endgroup$
– amd
Jan 16 at 18:11




1




1




$begingroup$
Note that the distance of a plane from the origin can be read directly from its point-normal equation, so casting this as a Lagrange multiplier problem is a bit overkill.
$endgroup$
– amd
Jan 16 at 18:12




$begingroup$
Note that the distance of a plane from the origin can be read directly from its point-normal equation, so casting this as a Lagrange multiplier problem is a bit overkill.
$endgroup$
– amd
Jan 16 at 18:12












$begingroup$
You’ve found that a critical point of the Lagrangian must have $x=z=2y$. Find a point on the tangent plane that satisfies this condition.
$endgroup$
– amd
Jan 16 at 18:13




$begingroup$
You’ve found that a critical point of the Lagrangian must have $x=z=2y$. Find a point on the tangent plane that satisfies this condition.
$endgroup$
– amd
Jan 16 at 18:13












$begingroup$
@amd Okay, but still. If we go it that way, what do you mean with " haven’t used your constraint" ? I used it as my constraint.
$endgroup$
– Poujh
Jan 16 at 18:13






$begingroup$
@amd Okay, but still. If we go it that way, what do you mean with " haven’t used your constraint" ? I used it as my constraint.
$endgroup$
– Poujh
Jan 16 at 18:13






1




1




$begingroup$
Sounds like the right approach to me.
$endgroup$
– amd
Jan 16 at 18:25




$begingroup$
Sounds like the right approach to me.
$endgroup$
– amd
Jan 16 at 18:25










1 Answer
1






active

oldest

votes


















1












$begingroup$

Hint



The tangent plane is perpendicular to $nabla f$ where $f(x,y,z)=x^2-y^2+2z^2.$ (Why?) So its equation is given by $$2x-2y+4z=D.$$



Moreover, the plane contains the point $(2,-1,1).$ (Why?)



So, you have to get the distance from the origin to a plane.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay, so with your method, which is obviously much simpler than the one I chose, I just need to use the distance from point to plane formula, is that correct ?
    $endgroup$
    – Poujh
    Jan 16 at 18:21










  • $begingroup$
    Yes. When you have got the equation of the plane you only need the distance from the origin to the plane.
    $endgroup$
    – mfl
    Jan 16 at 18:22












  • $begingroup$
    Just to be sure, the point $(2,-1,1)$ lies on the plane because it's stated so in the problem, otherwise they would not give us that point (And they would ask us to find the tangent plane in another point). And if I use your approach, to find D, I just plug that point into the plane equation $2x-2y+4z=D.$ and solve for D Then I use $frac{0+0+0+D}{sqrt{2^2+(-2)^2+4^2}}$, is that correct ?
    $endgroup$
    – Poujh
    Jan 16 at 18:29












  • $begingroup$
    Yes. You need the point $(2,-1,1)$ to get $D$ in $2x-2y+4z=D.$
    $endgroup$
    – mfl
    Jan 16 at 18:41










  • $begingroup$
    @amd Well, I get two different results. With the point plane vesion, I get $D=10$ so distance=$frac{10}{sqrt{24}}$ but with Lagrange, I get $y=frac{5}{9}$, so $sqrt{(2y)^2+y^2+(2y)^2}=sqrt{9y^2}=frac{5}{3}$. Am I missing something ??
    $endgroup$
    – Poujh
    Jan 16 at 18:50














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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Hint



The tangent plane is perpendicular to $nabla f$ where $f(x,y,z)=x^2-y^2+2z^2.$ (Why?) So its equation is given by $$2x-2y+4z=D.$$



Moreover, the plane contains the point $(2,-1,1).$ (Why?)



So, you have to get the distance from the origin to a plane.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay, so with your method, which is obviously much simpler than the one I chose, I just need to use the distance from point to plane formula, is that correct ?
    $endgroup$
    – Poujh
    Jan 16 at 18:21










  • $begingroup$
    Yes. When you have got the equation of the plane you only need the distance from the origin to the plane.
    $endgroup$
    – mfl
    Jan 16 at 18:22












  • $begingroup$
    Just to be sure, the point $(2,-1,1)$ lies on the plane because it's stated so in the problem, otherwise they would not give us that point (And they would ask us to find the tangent plane in another point). And if I use your approach, to find D, I just plug that point into the plane equation $2x-2y+4z=D.$ and solve for D Then I use $frac{0+0+0+D}{sqrt{2^2+(-2)^2+4^2}}$, is that correct ?
    $endgroup$
    – Poujh
    Jan 16 at 18:29












  • $begingroup$
    Yes. You need the point $(2,-1,1)$ to get $D$ in $2x-2y+4z=D.$
    $endgroup$
    – mfl
    Jan 16 at 18:41










  • $begingroup$
    @amd Well, I get two different results. With the point plane vesion, I get $D=10$ so distance=$frac{10}{sqrt{24}}$ but with Lagrange, I get $y=frac{5}{9}$, so $sqrt{(2y)^2+y^2+(2y)^2}=sqrt{9y^2}=frac{5}{3}$. Am I missing something ??
    $endgroup$
    – Poujh
    Jan 16 at 18:50


















1












$begingroup$

Hint



The tangent plane is perpendicular to $nabla f$ where $f(x,y,z)=x^2-y^2+2z^2.$ (Why?) So its equation is given by $$2x-2y+4z=D.$$



Moreover, the plane contains the point $(2,-1,1).$ (Why?)



So, you have to get the distance from the origin to a plane.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay, so with your method, which is obviously much simpler than the one I chose, I just need to use the distance from point to plane formula, is that correct ?
    $endgroup$
    – Poujh
    Jan 16 at 18:21










  • $begingroup$
    Yes. When you have got the equation of the plane you only need the distance from the origin to the plane.
    $endgroup$
    – mfl
    Jan 16 at 18:22












  • $begingroup$
    Just to be sure, the point $(2,-1,1)$ lies on the plane because it's stated so in the problem, otherwise they would not give us that point (And they would ask us to find the tangent plane in another point). And if I use your approach, to find D, I just plug that point into the plane equation $2x-2y+4z=D.$ and solve for D Then I use $frac{0+0+0+D}{sqrt{2^2+(-2)^2+4^2}}$, is that correct ?
    $endgroup$
    – Poujh
    Jan 16 at 18:29












  • $begingroup$
    Yes. You need the point $(2,-1,1)$ to get $D$ in $2x-2y+4z=D.$
    $endgroup$
    – mfl
    Jan 16 at 18:41










  • $begingroup$
    @amd Well, I get two different results. With the point plane vesion, I get $D=10$ so distance=$frac{10}{sqrt{24}}$ but with Lagrange, I get $y=frac{5}{9}$, so $sqrt{(2y)^2+y^2+(2y)^2}=sqrt{9y^2}=frac{5}{3}$. Am I missing something ??
    $endgroup$
    – Poujh
    Jan 16 at 18:50
















1












1








1





$begingroup$

Hint



The tangent plane is perpendicular to $nabla f$ where $f(x,y,z)=x^2-y^2+2z^2.$ (Why?) So its equation is given by $$2x-2y+4z=D.$$



Moreover, the plane contains the point $(2,-1,1).$ (Why?)



So, you have to get the distance from the origin to a plane.






share|cite|improve this answer









$endgroup$



Hint



The tangent plane is perpendicular to $nabla f$ where $f(x,y,z)=x^2-y^2+2z^2.$ (Why?) So its equation is given by $$2x-2y+4z=D.$$



Moreover, the plane contains the point $(2,-1,1).$ (Why?)



So, you have to get the distance from the origin to a plane.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 16 at 18:15









mflmfl

26.9k12142




26.9k12142












  • $begingroup$
    Okay, so with your method, which is obviously much simpler than the one I chose, I just need to use the distance from point to plane formula, is that correct ?
    $endgroup$
    – Poujh
    Jan 16 at 18:21










  • $begingroup$
    Yes. When you have got the equation of the plane you only need the distance from the origin to the plane.
    $endgroup$
    – mfl
    Jan 16 at 18:22












  • $begingroup$
    Just to be sure, the point $(2,-1,1)$ lies on the plane because it's stated so in the problem, otherwise they would not give us that point (And they would ask us to find the tangent plane in another point). And if I use your approach, to find D, I just plug that point into the plane equation $2x-2y+4z=D.$ and solve for D Then I use $frac{0+0+0+D}{sqrt{2^2+(-2)^2+4^2}}$, is that correct ?
    $endgroup$
    – Poujh
    Jan 16 at 18:29












  • $begingroup$
    Yes. You need the point $(2,-1,1)$ to get $D$ in $2x-2y+4z=D.$
    $endgroup$
    – mfl
    Jan 16 at 18:41










  • $begingroup$
    @amd Well, I get two different results. With the point plane vesion, I get $D=10$ so distance=$frac{10}{sqrt{24}}$ but with Lagrange, I get $y=frac{5}{9}$, so $sqrt{(2y)^2+y^2+(2y)^2}=sqrt{9y^2}=frac{5}{3}$. Am I missing something ??
    $endgroup$
    – Poujh
    Jan 16 at 18:50




















  • $begingroup$
    Okay, so with your method, which is obviously much simpler than the one I chose, I just need to use the distance from point to plane formula, is that correct ?
    $endgroup$
    – Poujh
    Jan 16 at 18:21










  • $begingroup$
    Yes. When you have got the equation of the plane you only need the distance from the origin to the plane.
    $endgroup$
    – mfl
    Jan 16 at 18:22












  • $begingroup$
    Just to be sure, the point $(2,-1,1)$ lies on the plane because it's stated so in the problem, otherwise they would not give us that point (And they would ask us to find the tangent plane in another point). And if I use your approach, to find D, I just plug that point into the plane equation $2x-2y+4z=D.$ and solve for D Then I use $frac{0+0+0+D}{sqrt{2^2+(-2)^2+4^2}}$, is that correct ?
    $endgroup$
    – Poujh
    Jan 16 at 18:29












  • $begingroup$
    Yes. You need the point $(2,-1,1)$ to get $D$ in $2x-2y+4z=D.$
    $endgroup$
    – mfl
    Jan 16 at 18:41










  • $begingroup$
    @amd Well, I get two different results. With the point plane vesion, I get $D=10$ so distance=$frac{10}{sqrt{24}}$ but with Lagrange, I get $y=frac{5}{9}$, so $sqrt{(2y)^2+y^2+(2y)^2}=sqrt{9y^2}=frac{5}{3}$. Am I missing something ??
    $endgroup$
    – Poujh
    Jan 16 at 18:50


















$begingroup$
Okay, so with your method, which is obviously much simpler than the one I chose, I just need to use the distance from point to plane formula, is that correct ?
$endgroup$
– Poujh
Jan 16 at 18:21




$begingroup$
Okay, so with your method, which is obviously much simpler than the one I chose, I just need to use the distance from point to plane formula, is that correct ?
$endgroup$
– Poujh
Jan 16 at 18:21












$begingroup$
Yes. When you have got the equation of the plane you only need the distance from the origin to the plane.
$endgroup$
– mfl
Jan 16 at 18:22






$begingroup$
Yes. When you have got the equation of the plane you only need the distance from the origin to the plane.
$endgroup$
– mfl
Jan 16 at 18:22














$begingroup$
Just to be sure, the point $(2,-1,1)$ lies on the plane because it's stated so in the problem, otherwise they would not give us that point (And they would ask us to find the tangent plane in another point). And if I use your approach, to find D, I just plug that point into the plane equation $2x-2y+4z=D.$ and solve for D Then I use $frac{0+0+0+D}{sqrt{2^2+(-2)^2+4^2}}$, is that correct ?
$endgroup$
– Poujh
Jan 16 at 18:29






$begingroup$
Just to be sure, the point $(2,-1,1)$ lies on the plane because it's stated so in the problem, otherwise they would not give us that point (And they would ask us to find the tangent plane in another point). And if I use your approach, to find D, I just plug that point into the plane equation $2x-2y+4z=D.$ and solve for D Then I use $frac{0+0+0+D}{sqrt{2^2+(-2)^2+4^2}}$, is that correct ?
$endgroup$
– Poujh
Jan 16 at 18:29














$begingroup$
Yes. You need the point $(2,-1,1)$ to get $D$ in $2x-2y+4z=D.$
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– mfl
Jan 16 at 18:41




$begingroup$
Yes. You need the point $(2,-1,1)$ to get $D$ in $2x-2y+4z=D.$
$endgroup$
– mfl
Jan 16 at 18:41












$begingroup$
@amd Well, I get two different results. With the point plane vesion, I get $D=10$ so distance=$frac{10}{sqrt{24}}$ but with Lagrange, I get $y=frac{5}{9}$, so $sqrt{(2y)^2+y^2+(2y)^2}=sqrt{9y^2}=frac{5}{3}$. Am I missing something ??
$endgroup$
– Poujh
Jan 16 at 18:50






$begingroup$
@amd Well, I get two different results. With the point plane vesion, I get $D=10$ so distance=$frac{10}{sqrt{24}}$ but with Lagrange, I get $y=frac{5}{9}$, so $sqrt{(2y)^2+y^2+(2y)^2}=sqrt{9y^2}=frac{5}{3}$. Am I missing something ??
$endgroup$
– Poujh
Jan 16 at 18:50




















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