
Find the centre and radius of the sphere \[{\vec r^2} - \,\vec r.(4\vec i + 2\vec j - 6\vec k) - 11 = 0\]
Answer
586.2k+ views
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”.
Recently Updated Pages
Two men on either side of the cliff 90m height observe class 10 maths CBSE

Cutting of the Chinese melon means A The business and class 10 social science CBSE

Show an aquatic food chain using the following organisms class 10 biology CBSE

How is gypsum formed class 10 chemistry CBSE

If the line 3x + 4y 24 0 intersects the xaxis at t-class-10-maths-CBSE

Sugar present in DNA is A Heptose B Hexone C Tetrose class 10 biology CBSE

Trending doubts
The average rainfall in India is A 105cm B 90cm C 120cm class 10 biology CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

Indias first jute mill was established in 1854 in A class 10 social science CBSE

Indias first jute mill was established in 1854 in A class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

