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Question: If the distance of P from \(\left( {1,1,1} \right)\) which is equal to double the distance of P from...

If the distance of P from (1,1,1)\left( {1,1,1} \right) which is equal to double the distance of P from y-axis, then the locus P is
A. 3x2−y2+3z2−2x−2y−2z−3=03{x^2} - {y^2} + 3{z^2} - 2x - 2y - 2z - 3 = 0
B. 3x2−y2+3mz2−2x−2y−2z−3=03{x^2} - {y^2} + 3m{z^2} - 2x - 2y - 2z - 3 = 0
C. 3x2+3y2+3z2−2x−2y−2z−3=03{x^2} + 3{y^2} + 3{z^2} - 2x - 2y - 2z - 3 = 0
D. 3x2−y2+3z2+2x+2y+2z−3=03{x^2} - {y^2} + 3{z^2} + 2x + 2y + 2z - 3 = 0

Explanation

Solution

Locus is basically a set of all the points where the location of those points satisfies under some specific condition. For example, we are having some points and we have to find the location of all the points that mean the set of all the points under some specific condition, then that set of all the points is known as locus.

Complete step-by-step answer:
We have to find the distance of P from (1,1,1)\left( {1,1,1} \right) which is equal to double the distance of P from y-axis
Therefore, let us assume the coordinates of P such that P(x,y,z)\left( {x,y,z} \right)
We are given that distance of P from (1,1,1)\left( {1,1,1} \right)
Therefore, distance of P(x,y,z)\left( {x,y,z} \right) from (1,1,1)\left( {1,1,1} \right) is
\Rightarrow $$$\sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 1} \right)}^2} + {{\left( {z - 1} \right)}^2}} $ \Rightarrow \sqrt {{x^2} + 1 - 2x + {y^2} + 1 - 2y + {z^2} + 1 - 2z} $ $$ \Rightarrow \sqrt {{x^2} + {y^2} + {z^2} - 2x - 2y - 2z + 3} Using formula:$${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$$ Also, distance of P from y-axis is given by\sqrt {{x^2} + {y^2}} NowaccordingtothegivenconditionthedistanceofPfrom Now according to the given condition the distance of P from\left( {1,1,1} \right)is equal to double the distance of P from y-axis. Therefore $$ \Rightarrow $$$\sqrt {{x^2} + {y^2} + {z^2} - 2x - 2y - 2z + 3} = 2\sqrt {{x^2} + {y^2}}
Doing square on both sides,
\Rightarrow $$$${x^2} + {y^2} + {z^2} - 2x - 2y - 2z + 3 = 4\left( {{x^2} + {z^2}} \right)
Square root and square cancel each other
\Rightarrow $$$${x^2} + {y^2} + {z^2} - 2x - 2y - 2z + 3 = 4{x^2} + 4{y^2}
\Rightarrow $$$${x^2} + {y^2} + {z^2} - 2x - 2y - 2z + 3 - 4{x^2} - 4{y^2} = 0
Taking all the terms to left hand side & arranging
⇒\Rightarrow −3x2+y2−3z2−2x−2y−2z+3=0- 3{x^2} + {y^2} - 3{z^2} - 2x - 2y - 2z + 3 = 0
Taking minus sign common on both sides & cancels
\Rightarrow $$$$3{x^2} - {y^2} + 3{z^2} + 2x + 2y + 2z - 3 = 0

So, option (D) is correct.

Note: Distance formula between 22 points (p,q)\left( {p,q} \right) and (u,v)\left( {u,v} \right) is (p−u)2+(q−v)2\sqrt {{{\left( {p - u} \right)}^2} + {{\left( {q - v} \right)}^2}} . If points are in 2d2d, then distance formula for 22 points (x1,y1) & (x2,y2)\left( {{x_1},{y_1}} \right){\text{ }}\& {\text{ }}\left( {{x_2},{y_2}} \right) is (x1−x2)2+(y1−y2)2\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} . If points are in 3d, then distance formula for 33 points is (x1−x2)2+(y1−y2)2+(z1−z2)2\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2} + {{\left( {{z_1} - {z_2}} \right)}^2}} for 2 points.(x1,y1,z1) & (x2,y2,z2)\left( {{x_1},{y_1},{z_1}} \right){\text{ }}\& {\text{ }}\left( {{x_2},{y_2},{z_2}} \right).$$