Question

The locus of a point whose distance from $(1,2,3)$ is equal to its distance from the $xy$-plane is

• A. $x^2+y^2+z^2-2x-4y-6z+14=0$
• B. $x^2+y^2-2x+14=0$
• C. $x^2+y^2-2x-4y-6z+14=0$
• D. $y^2-4y-6x+14=0$

Answer 1

Answer: (C)

$x^2+y^2-2x-4y-6z+14=0$

Let the point be $P(x,y,z)$

So its distance from $xy-$ plane is $d_1=|z|$ (modulus of $z$ -coordinate )

And its distance of this point from the point $(1,2,3)$ is $d_2=\sqrt{(x-1{)}^2+(y-2{)}^2+(z-3{)}^2}$

Now, it is given that $d_1=d_2$.

$\Rightarrow d_1^2=d_2^2$

$\Rightarrow (x-1{)}^2+(y-2{)}^2+(z-3{)}^2=z^2$

$\Rightarrow x^2+y^2-2x-4y-6z+14=0$ (expand above equation)