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Find the centre and radius of the sphere \[{\vec r^2} - \,\vec r.(4\vec i + 2\vec j - 6\vec k) - 11 = 0\]
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Verified
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Hint: A sphere with center \[\left( {a,b,c} \right)\] and radius r has the equation \[\left( {x - a} \right){\text{ }} + {\text{ }}\left( {y - b} \right){\text{ }} + {\text{ }}{\left( {z - c} \right)^2} = {r^2}\]
Or x2 + y2 +z2 + 2ux + 2vy + 2wz+d=0
Where center of sphere is y\[ = {\text{ }}\left( { - u, - v, - w} \right)\]
And radius of sphere is
= \[\sqrt {{u^2} + {v^2} + {w^2} - d} \]
Complete step by step answer:
Let
\[\vec r = x\vec i + j\vec y + z\vec k\,\], and \[r = \sqrt {{x^2} + {y^2} + {z^2}} \]
\[ = {r^2} = {x^2} + {y^2} + {z^2}\] [squaring both sides]
Now dot product of \[\vec r.(4\vec i + 2\vec j - 6\vec k)\]is:
\[ = 4\vec r.\vec i + 2\vec r.\vec j - 6\vec r.\vec k\]
\[ = 4x + 2y - 6z\]----(2) \[[\vec r.\vec i = x,\,\,\vec r.\vec j = y,\,\vec r.\vec k = z]\]
Our given equation is,
\[{\vec r^2} + \vec r.(4\vec i + 2\vec j - 6\vec k) - 11 = 0\]-----(3)
Using value of 1 and 2 in equation 3
\[{x^2} + {y^2} + {z^2} - (4x + 2y - 6z) - 11 = 0\]
\[{x^2} + {y^2} + {z^2} - 4x + 2y - 6z - 11 = 0\]
General equation of the sphere is
\[{x^2} + {y^2} + {z^2} + 2ux + 2vy{\text{ }} + 2wz + d{\text{ }} = 0\;\;\;\;\;\;\;\;\;\]----(5)
On comparing 4 with 5, we get;
\[2u = - 4,{\text{ }}2v = - 2,2w = 6\]and \[d = - 11\]
= \[u{\text{ }} - \] \[\dfrac{4}{2} = - 2,v = - \dfrac{2}{2} = 1,w = \dfrac{6}{2} = 3and\] \[d = - 1\]
\[u = - 2v = - 1,w = 3and\] \[d = - 11\]
Center \[ = \left( {{\text{ }} - u{\text{ }},v, - w} \right)\;\]\[ = \]\[\left( { - ( - 2)} \right), - \left( { - 1} \right), - 3\])
\[ = \left( {2,{\text{ }}1, - 3} \right)\]
And radius \[ = \] \[\sqrt {{u^2} + {v^2} + {w^2} - d} \]
Put \[u = {\text{ }} - 2,v = {\text{ }} - 1,{\text{ }}w = 3\]and \[d = {\text{ }} - 11\]
\[ = \sqrt {{{( - 2)}^2} + {{( - 1)}^2} + {{(3)}^2} + 11} \]
\[ = \sqrt {4 + 1 + 9 + 11} \]
\[ = \sqrt {14 + 11} \]
\[ = \sqrt {25} \]
\[ = 5\]
Hence, are the center and radius of sphere.
Note: a sphere is a three-dimension shape and it is mathematically defined as a set of points from the given point called “center” with an equal distance called radius “r” in the three-dimensional space of Euclidean space. The diameter “d’ is twice the radius. The pair of points that connect the opposite sides of a sphere is called “antipodes”. The sphere is sometimes interchangeably called “ball”.
The important properties of the sphere are:
A sphere is perfectly symmetrical.
It is not a polyhedron.
All the points on the surface are equidistant from the center.
It does not have a surface of centers.
It has constant mean curvature.
It has a constant width and circumference.
“while comparing the equation with the general equation we must take care of the signs”.
Or x2 + y2 +z2 + 2ux + 2vy + 2wz+d=0
Where center of sphere is y\[ = {\text{ }}\left( { - u, - v, - w} \right)\]
And radius of sphere is
= \[\sqrt {{u^2} + {v^2} + {w^2} - d} \]
Complete step by step answer:
Let
\[\vec r = x\vec i + j\vec y + z\vec k\,\], and \[r = \sqrt {{x^2} + {y^2} + {z^2}} \]
\[ = {r^2} = {x^2} + {y^2} + {z^2}\] [squaring both sides]
Now dot product of \[\vec r.(4\vec i + 2\vec j - 6\vec k)\]is:
\[ = 4\vec r.\vec i + 2\vec r.\vec j - 6\vec r.\vec k\]
\[ = 4x + 2y - 6z\]----(2) \[[\vec r.\vec i = x,\,\,\vec r.\vec j = y,\,\vec r.\vec k = z]\]
Our given equation is,
\[{\vec r^2} + \vec r.(4\vec i + 2\vec j - 6\vec k) - 11 = 0\]-----(3)
Using value of 1 and 2 in equation 3
\[{x^2} + {y^2} + {z^2} - (4x + 2y - 6z) - 11 = 0\]
\[{x^2} + {y^2} + {z^2} - 4x + 2y - 6z - 11 = 0\]
General equation of the sphere is
\[{x^2} + {y^2} + {z^2} + 2ux + 2vy{\text{ }} + 2wz + d{\text{ }} = 0\;\;\;\;\;\;\;\;\;\]----(5)
On comparing 4 with 5, we get;
\[2u = - 4,{\text{ }}2v = - 2,2w = 6\]and \[d = - 11\]
= \[u{\text{ }} - \] \[\dfrac{4}{2} = - 2,v = - \dfrac{2}{2} = 1,w = \dfrac{6}{2} = 3and\] \[d = - 1\]
\[u = - 2v = - 1,w = 3and\] \[d = - 11\]
Center \[ = \left( {{\text{ }} - u{\text{ }},v, - w} \right)\;\]\[ = \]\[\left( { - ( - 2)} \right), - \left( { - 1} \right), - 3\])
\[ = \left( {2,{\text{ }}1, - 3} \right)\]
And radius \[ = \] \[\sqrt {{u^2} + {v^2} + {w^2} - d} \]
Put \[u = {\text{ }} - 2,v = {\text{ }} - 1,{\text{ }}w = 3\]and \[d = {\text{ }} - 11\]
\[ = \sqrt {{{( - 2)}^2} + {{( - 1)}^2} + {{(3)}^2} + 11} \]
\[ = \sqrt {4 + 1 + 9 + 11} \]
\[ = \sqrt {14 + 11} \]
\[ = \sqrt {25} \]
\[ = 5\]
Hence, are the center and radius of sphere.
Note: a sphere is a three-dimension shape and it is mathematically defined as a set of points from the given point called “center” with an equal distance called radius “r” in the three-dimensional space of Euclidean space. The diameter “d’ is twice the radius. The pair of points that connect the opposite sides of a sphere is called “antipodes”. The sphere is sometimes interchangeably called “ball”.
The important properties of the sphere are:
A sphere is perfectly symmetrical.
It is not a polyhedron.
All the points on the surface are equidistant from the center.
It does not have a surface of centers.
It has constant mean curvature.
It has a constant width and circumference.
“while comparing the equation with the general equation we must take care of the signs”.
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