If the sum of the squares of the distance of a point from the three coordinate axes be 36, then its distance from the origin is

Question

If the sum of the squares of the distance of a point from the three coordinate axes be 36, then its distance from the origin is (A) 5 (B) 3 √2 (C) √21 (D) 3 √3

Tardigrade Admin · Accepted Answer

Let P(x, y, z) Now under given condition, we get [√x2+y2]2+[√y2+z2]2+[√z2+x2]2=36 ⇒ x2+y2+z2=18 Then distance from origin to the point (x, y, z) is √x2+y2+z2=√18=3 √2

Q. If the sum of the squares of the distance of a point from the three coordinate axes be $36$ , then its distance from the origin is

$5$

$32$

$21$

$33$

Let $P (x, y, z)$ Now under given condition, we get
$[x^{2} + y^{2}]^{2} + [y^{2} + z^{2}]^{2} + [z^{2} + x^{2}]^{2} = 36$
$\Rightarrow x^{2} + y^{2} + z^{2} = 18$
Then distance from origin to the point $(x, y, z)$ is
$x^{2} + y^{2} + z^{2} = 18 = 32$

Q. If the sum of the squares of the distance of a point from the three coordinate axes be 36, then its distance from the origin is

5

32​

21​

33​

Let P(x,y,z) Now under given condition, we get [x2+y2​]2+[y2+z2​]2+[z2+x2​]2=36 ⇒x2+y2+z2=18 Then distance from origin to the point (x,y,z) is x2+y2+z2​=18​=32​

Q. If the sum of the squares of the distance of a point from the three coordinate axes be $36$ , then its distance from the origin is

$5$

$32$

$21$

$33$

Let $P (x, y, z)$ Now under given condition, we get
$[x^{2} + y^{2}]^{2} + [y^{2} + z^{2}]^{2} + [z^{2} + x^{2}]^{2} = 36$
$\Rightarrow x^{2} + y^{2} + z^{2} = 18$
Then distance from origin to the point $(x, y, z)$ is
$x^{2} + y^{2} + z^{2} = 18 = 32$