Find the point that is at distance $1$ from $(0,0,0)$ and at distance $3$ from $(1,2,3)$ that is closest to...












1












$begingroup$


I have this question :




Find the point that is at distance $1$ from $(0,0,0)$ and at distance $3$ from $(1,2,3)$ that is closest to $(5,-2,4)$.




Here is my failed attempt.



enter image description here



I used Lagrange multipliers and the fact that the determinant of matrix is 0 when we have non trivial solution. the quad equation went bad






geometry optimization analytic-geometry nonlinear-optimization lagrange-multiplier







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I have this question :




    Find the point that is at distance $1$ from $(0,0,0)$ and at distance $3$ from $(1,2,3)$ that is closest to $(5,-2,4)$.




    Here is my failed attempt.



    enter image description here



    I used Lagrange multipliers and the fact that the determinant of matrix is 0 when we have non trivial solution. the quad equation went bad






    geometry optimization analytic-geometry nonlinear-optimization lagrange-multiplier







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have this question :




      Find the point that is at distance $1$ from $(0,0,0)$ and at distance $3$ from $(1,2,3)$ that is closest to $(5,-2,4)$.




      Here is my failed attempt.



      enter image description here



      I used Lagrange multipliers and the fact that the determinant of matrix is 0 when we have non trivial solution. the quad equation went bad






      geometry optimization analytic-geometry nonlinear-optimization lagrange-multiplier







      share|cite|improve this question











      $endgroup$




      I have this question :




      Find the point that is at distance $1$ from $(0,0,0)$ and at distance $3$ from $(1,2,3)$ that is closest to $(5,-2,4)$.




      Here is my failed attempt.



      enter image description here



      I used Lagrange multipliers and the fact that the determinant of matrix is 0 when we have non trivial solution. the quad equation went bad






      geometry optimization analytic-geometry nonlinear-optimization lagrange-multiplier




      geometry optimization analytic-geometry nonlinear-optimization lagrange-multiplier






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 '18 at 21:52









      Batominovski

      33k33293




      33k33293










      asked Dec 6 '18 at 21:47









      Razi AwadRazi Awad

      124




      124






















          1 Answer
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          $begingroup$

          You have done great so far. The only thing left is to substitute $x=dfrac{57z-33}{17}$ and $y=dfrac{42-54z}{17}$ into $x^2+y^2+z^2=1$. In other words, you have
          $$frac{2(3227z^2-4149z+1282)}{289}=0,.$$
          From this horrendous quadratic equation, you solve it using the famous quadratic formula, but not without pain, and obtain
          $$z=frac{4149pm17sqrt{2305}}{6454},.$$
          That is,
          $$x=frac{3(461pm19sqrt{2305})}{6454}$$
          and
          $$y=frac{3(461mp9sqrt{2305})}{3227},.$$
          Therefore, the distances to $(5,-2,4)$ from these optimizing points are $$sqrt{frac{283mpsqrt{2305}}{7}},.$$
          Thus, the minimizing point is
          $$begin{align}(x,y,z)&={smallleft(frac{3(461+19sqrt{2305})}{6454},frac{3(461-9sqrt{2305})}{3227},frac{4149+17sqrt{2305}}{6454}right)}\&approx(0.638,0.027,0.769),,end{align}$$
          yielding the minimum distance $sqrt{dfrac{283-sqrt{2305}}{7}}approx5.794$. On the other hand, the maximizing point is
          $$begin{align}(x,y,z)&={smallleft(frac{3(461-19sqrt{2305})}{6454},frac{3(461+9sqrt{2305})}{3227},frac{4149-17sqrt{2305}}{6454}right)}\&approx(-0.210,0.830,0.516),,end{align}$$
          yielding the maximum distance $sqrt{dfrac{283+sqrt{2305}}{7}}approx6.877$.






          share|cite|improve this answer











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            $begingroup$

            You have done great so far. The only thing left is to substitute $x=dfrac{57z-33}{17}$ and $y=dfrac{42-54z}{17}$ into $x^2+y^2+z^2=1$. In other words, you have
            $$frac{2(3227z^2-4149z+1282)}{289}=0,.$$
            From this horrendous quadratic equation, you solve it using the famous quadratic formula, but not without pain, and obtain
            $$z=frac{4149pm17sqrt{2305}}{6454},.$$
            That is,
            $$x=frac{3(461pm19sqrt{2305})}{6454}$$
            and
            $$y=frac{3(461mp9sqrt{2305})}{3227},.$$
            Therefore, the distances to $(5,-2,4)$ from these optimizing points are $$sqrt{frac{283mpsqrt{2305}}{7}},.$$
            Thus, the minimizing point is
            $$begin{align}(x,y,z)&={smallleft(frac{3(461+19sqrt{2305})}{6454},frac{3(461-9sqrt{2305})}{3227},frac{4149+17sqrt{2305}}{6454}right)}\&approx(0.638,0.027,0.769),,end{align}$$
            yielding the minimum distance $sqrt{dfrac{283-sqrt{2305}}{7}}approx5.794$. On the other hand, the maximizing point is
            $$begin{align}(x,y,z)&={smallleft(frac{3(461-19sqrt{2305})}{6454},frac{3(461+9sqrt{2305})}{3227},frac{4149-17sqrt{2305}}{6454}right)}\&approx(-0.210,0.830,0.516),,end{align}$$
            yielding the maximum distance $sqrt{dfrac{283+sqrt{2305}}{7}}approx6.877$.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              You have done great so far. The only thing left is to substitute $x=dfrac{57z-33}{17}$ and $y=dfrac{42-54z}{17}$ into $x^2+y^2+z^2=1$. In other words, you have
              $$frac{2(3227z^2-4149z+1282)}{289}=0,.$$
              From this horrendous quadratic equation, you solve it using the famous quadratic formula, but not without pain, and obtain
              $$z=frac{4149pm17sqrt{2305}}{6454},.$$
              That is,
              $$x=frac{3(461pm19sqrt{2305})}{6454}$$
              and
              $$y=frac{3(461mp9sqrt{2305})}{3227},.$$
              Therefore, the distances to $(5,-2,4)$ from these optimizing points are $$sqrt{frac{283mpsqrt{2305}}{7}},.$$
              Thus, the minimizing point is
              $$begin{align}(x,y,z)&={smallleft(frac{3(461+19sqrt{2305})}{6454},frac{3(461-9sqrt{2305})}{3227},frac{4149+17sqrt{2305}}{6454}right)}\&approx(0.638,0.027,0.769),,end{align}$$
              yielding the minimum distance $sqrt{dfrac{283-sqrt{2305}}{7}}approx5.794$. On the other hand, the maximizing point is
              $$begin{align}(x,y,z)&={smallleft(frac{3(461-19sqrt{2305})}{6454},frac{3(461+9sqrt{2305})}{3227},frac{4149-17sqrt{2305}}{6454}right)}\&approx(-0.210,0.830,0.516),,end{align}$$
              yielding the maximum distance $sqrt{dfrac{283+sqrt{2305}}{7}}approx6.877$.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                You have done great so far. The only thing left is to substitute $x=dfrac{57z-33}{17}$ and $y=dfrac{42-54z}{17}$ into $x^2+y^2+z^2=1$. In other words, you have
                $$frac{2(3227z^2-4149z+1282)}{289}=0,.$$
                From this horrendous quadratic equation, you solve it using the famous quadratic formula, but not without pain, and obtain
                $$z=frac{4149pm17sqrt{2305}}{6454},.$$
                That is,
                $$x=frac{3(461pm19sqrt{2305})}{6454}$$
                and
                $$y=frac{3(461mp9sqrt{2305})}{3227},.$$
                Therefore, the distances to $(5,-2,4)$ from these optimizing points are $$sqrt{frac{283mpsqrt{2305}}{7}},.$$
                Thus, the minimizing point is
                $$begin{align}(x,y,z)&={smallleft(frac{3(461+19sqrt{2305})}{6454},frac{3(461-9sqrt{2305})}{3227},frac{4149+17sqrt{2305}}{6454}right)}\&approx(0.638,0.027,0.769),,end{align}$$
                yielding the minimum distance $sqrt{dfrac{283-sqrt{2305}}{7}}approx5.794$. On the other hand, the maximizing point is
                $$begin{align}(x,y,z)&={smallleft(frac{3(461-19sqrt{2305})}{6454},frac{3(461+9sqrt{2305})}{3227},frac{4149-17sqrt{2305}}{6454}right)}\&approx(-0.210,0.830,0.516),,end{align}$$
                yielding the maximum distance $sqrt{dfrac{283+sqrt{2305}}{7}}approx6.877$.






                share|cite|improve this answer











                $endgroup$



                You have done great so far. The only thing left is to substitute $x=dfrac{57z-33}{17}$ and $y=dfrac{42-54z}{17}$ into $x^2+y^2+z^2=1$. In other words, you have
                $$frac{2(3227z^2-4149z+1282)}{289}=0,.$$
                From this horrendous quadratic equation, you solve it using the famous quadratic formula, but not without pain, and obtain
                $$z=frac{4149pm17sqrt{2305}}{6454},.$$
                That is,
                $$x=frac{3(461pm19sqrt{2305})}{6454}$$
                and
                $$y=frac{3(461mp9sqrt{2305})}{3227},.$$
                Therefore, the distances to $(5,-2,4)$ from these optimizing points are $$sqrt{frac{283mpsqrt{2305}}{7}},.$$
                Thus, the minimizing point is
                $$begin{align}(x,y,z)&={smallleft(frac{3(461+19sqrt{2305})}{6454},frac{3(461-9sqrt{2305})}{3227},frac{4149+17sqrt{2305}}{6454}right)}\&approx(0.638,0.027,0.769),,end{align}$$
                yielding the minimum distance $sqrt{dfrac{283-sqrt{2305}}{7}}approx5.794$. On the other hand, the maximizing point is
                $$begin{align}(x,y,z)&={smallleft(frac{3(461-19sqrt{2305})}{6454},frac{3(461+9sqrt{2305})}{3227},frac{4149-17sqrt{2305}}{6454}right)}\&approx(-0.210,0.830,0.516),,end{align}$$
                yielding the maximum distance $sqrt{dfrac{283+sqrt{2305}}{7}}approx6.877$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 6 '18 at 22:29

























                answered Dec 6 '18 at 22:23









                BatominovskiBatominovski

                33k33293




                33k33293






























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