Question

If $A{P}^2-B{P}^2=18$, where A ≡ (1, 2, –3) and B ≡ (3, –2, 1). Then the locus of point P is

• A. $2x+3y+4=9$
• B. $2x+4y+4z=9$
• C. $2x–3y+4z=9$
• D. $2x–4y+4z=9$

Answer 1

The correct option is D $2x–4y+4z=9$

We want to find the locus of P. That is, the set of Ps satisfying the given condition. For that, let the point P be (x, y, z). Using distance formula, we get

$A{P}^2=(x-1{)}^2+(y-2{)}^2+(z+3{)}^2$
$B{P}^2=(x-3{)}^2+(y+2{)}^2+(z-1{)}^2$

Given that $A{P}^2-B{P}^2=18$

$\Rightarrow (x-1{)}^2+(y-2{)}^2+(z+3{)}^2-\{(x-3{)}^2+(y+2{)}^2+(z--1{)}^2\}=18$

$\Rightarrow x^2+1-2x+y^2+4-4y+z^2+9+6z-\{x^2+9-6x+y^2+4+4y+z^2+1-2z\}=18$
$\Rightarrow x^2+y^2+z^2-2x-4y+6z+14-x^2-y^2-z^2+6x-4y+2z-14=28$

$\Rightarrow 4x–8y+8z=18$
$\Rightarrow 2x–4y+4z=9$

Locus of the point P is 2x – 4y + 4z = 9