Question

The equation \[{{\left| r \right|}^{2}}-r.\left( 2i+4j-2k \right)-10=0\]represents a
A Circle
B Plane
C Sphere of radius 4
D Sphere of radius 3
E None of these

Answer 1

Hint: The equation of the circle is such in which only two coordinates are present or can say the circle is what we can draw at the plane but when we add a third coordinate in the equation of circle it will represent a sphere. The general equation of the circle is \[{{(x-\alpha )}^{2}}+{{(y-\beta )}^{2}}=\text{ }{{r}^{2}}\], where x and y are the two coordinates, \[\alpha \] and \[\beta \]represent the central point and r is the radius of the circle with the center is 0. The general equation of the sphere is given as \[{{(x-\alpha )}^{2}}+{{(y-\beta )}^{2}}+{{(z-\gamma )}^{2}}={{r}^{2}}\], where r is the radius of the sphere, (x, y, and z) represents the three coordinates or space, and alpha beta gamma represents the position of center.

Complete step by step solution: In this question, the given equation is \[{{\left| r \right|}^{2}}-r.\left( 2i+4j-2k \right)-10=0\]and we need to determine what this equation represents.
By putting the value of r in the given equation and modifying the equation such as
\[{{\left| r \right|}^{2}}-r.\left( 2i+4j-2k \right)-10=0\]
Let \[r\text{ }=\text{ }xi\text{ }+\text{ }yj\text{ }+\text{ }zk~\](position vector of any point)
\[\left| r \right|\text{ }=\text{ }\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\]
And
\[{{\left| \text{ }r \right|}^{2}}\text{ }=\text{ }{{x}^{2}}\text{ }+\text{ }{{y}^{2}}\text{ }+\text{ }{{z}^{2}}\]
Putting the value of r and \[{{\left| r \right|}^{2}}\]we will get
\[{{\left| r \right|}^{2}}-r.\left( 2i+4j-2k \right)-10=0\]
\[\left( {{x}^{2}}\text{ }+\text{ }{{y}^{2}}\text{ }+\text{ }{{z}^{2}} \right)\text{ }\text{ }\left( xi\text{ }+\text{ }yj\text{ }+\text{ }zk \right)\left( 2i+4j-2k \right)\text{ }\text{ }10\text{ }=\text{ }0\]
Or
\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-(2x+4y-2z)-10=0\]
Or
\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2x-4y+2z=\text{ }10\]
Adding 6 on both sides we will get
\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2x-4y+2z\text{ }+\text{ }6=\text{ }16\]
Rearranging such as
\[{{\left( x\text{ }\text{ }1 \right)}^{2}}\text{ }+\text{ }{{\left( y\text{ }\text{ }2 \right)}^{2}}\text{ }+\text{ }{{\left( z\text{ }+\text{ }1 \right)}^{2}}\text{ }=\text{ }{{4}^{2}}\]
The above equation just matches with the general equation of the sphere which is given as
\[{{(x-\alpha )}^{2}}+{{(y-\beta )}^{2}}+{{(z-\gamma )}^{2}}={{r}^{2}}\]
Comparing the above two equations with general equation we will get \[\alpha \text{ }=\text{ }1\], \[\beta \text{ }=\text{ }2\], \[\gamma \text{ }=\text{ }-1\]and r is 4
This, the given equation is the equation of sphere of radius 4

Option ‘C’ is correct

Note: It is important to note that when the \[\alpha \], \[\beta \], and \[\gamma \]in the equation of sphere will equal to zero and the equation of sphere is given as \[{{x}^{2}}\text{ }+\text{ }{{y}^{2}}\text{ }+\text{ }{{z}^{2}}\text{ }=\text{ }r2\], then the center of the sphere is at origin whose coordinates are (0 0 0). Also when the alpha and beta will be zero then the center of the circle will be at the origin such as (0 0).