The locus of a point, such that the sum of the squares of its distances from the planes x+y+z=0, x-z=0 and x -2 y + z =0 is 9, is

Question

The locus of a point, such that the sum of the squares of its distances from the planes x+y+z=0, x-z=0 and x -2 y + z =0 is 9, is (A) x2+y2+z2=3 (B) x2+y2+z2=6 (C) x2+y2+z2=9 (D) x2+y2+z2=12

Tardigrade Admin · Accepted Answer

Q. The locus of a point, such that the sum of the squares of its distances from the planes $x + y + z = 0, x - z = 0$ and $x - 2 y + z = 0$ is $9$ , is

$x^{2} + y^{2} + z^{2} = 3$

$x^{2} + y^{2} + z^{2} = 6$

$x^{2} + y^{2} + z^{2} = 9$

$x^{2} + y^{2} + z^{2} = 12$

Let the variable point be $(α, β, γ)$ then according to question
$(\frac{∣ α + β + γ ∣}{3})^{2} + (\frac{∣ α - γ ∣}{2})^{2} + (\frac{∣ α - 2 β + γ ∣}{6})^{2} = 9$
$\Rightarrow α^{2} + β^{2} + γ^{2} = 9$ .
So, the locus of the point is $x^{2} + y^{2} + z^{2} = 9$

Q. The locus of a point, such that the sum of the squares of its distances from the planes x+y+z=0,x−z=0 and x−2y+z=0 is 9, is

x2+y2+z2=3

x2+y2+z2=6

x2+y2+z2=9

x2+y2+z2=12

Let the variable point be (α,β,γ) then according to question (3​∣α+β+γ∣​)2+(2​∣α−γ∣​)2+(6​∣α−2β+γ∣​)2=9 ⇒α2+β2+γ2=9. So, the locus of the point is x2+y2+z2=9

Q. The locus of a point, such that the sum of the squares of its distances from the planes $x + y + z = 0, x - z = 0$ and $x - 2 y + z = 0$ is $9$ , is

$x^{2} + y^{2} + z^{2} = 3$

$x^{2} + y^{2} + z^{2} = 6$

$x^{2} + y^{2} + z^{2} = 9$

$x^{2} + y^{2} + z^{2} = 12$

Let the variable point be $(α, β, γ)$ then according to question
$(\frac{∣ α + β + γ ∣}{3})^{2} + (\frac{∣ α - γ ∣}{2})^{2} + (\frac{∣ α - 2 β + γ ∣}{6})^{2} = 9$
$\Rightarrow α^{2} + β^{2} + γ^{2} = 9$ .
So, the locus of the point is $x^{2} + y^{2} + z^{2} = 9$