If the sum of the squares of the perpendicular distances of $P (x, y, z)$ from the coordinate axes is $12$ , then the locus of $P$ is:

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x + y + z = 6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x + y + z = 12

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x^{2} + y^{2} + z^{2} = 6$
Let $P (x, y, z)$ be a given point.
Distance of $P$ from $x$ axis $=$ $\sqrt{(y^{2} + z^{2})}$
Distance of $P$ from $y$ axis $=$ $\sqrt{(x^{2} + z^{2})}$
Distance of $P$ from $z$ axis $=$ $\sqrt{(y^{2} + x^{2})}$
Sum of the squares of these distances is given to be $12$ .
i.e. $2 (x^{2} + y^{2} + z^{2}) = 12$
$\Rightarrow x^{2} + y^{2} + z^{2} = 6$

Suggest Corrections

Similar questions

x^{2} + y^{2} + z^{2} = 5, x, y, z \in R,

(x - y)^{2} + (y - z)^{2} + (z - x)^{2}

x + y + z = 0

\frac{1}{x^{2} + y^{2} - z^{2}} + \frac{1}{y^{2} + z^{2} - x^{2}} + \frac{1}{z^{2} + x^{2} - y^{2}}

{(x + y)}^{3} - z^{2} = 4

{(y + z)}^{2} - x^{2} = 9

{(z + x)}^{2} - y^{2} = 36

x + y + z

x, y, z

x + log (x + \sqrt{x^{2} + 1}) = y

y + log (y + \sqrt{y^{2} + 1}) = z

z + log (z + \sqrt{z^{2} + 1}) = x

{(1 - x)}^{2} + {(1 - y)}^{2} + {(1 - z)}^{2}

x^{2} + y^{2} + z^{2} + x + y + z - 1 = 0

x^{2} + y^{2} + z^{2} + x + y + z - 5 = 0

Join BYJU'S Learning Program