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The sum of the squares of the perpendicular distances of a point $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ from the coordinate axes is $\mathrm{k}$ times the square of the distance of the point from the origin.
Then $\mathrm{k}=$
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Then $\mathrm{k}=$
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Verified Answer
The correct answer is:
$2$
Distance of $(x, y, z)$
(a) From $x$ axis i.e. $(x, 0,0)$
$d=\sqrt{y^2+z^2}$
(b) From $y$ axis i.e. $(0, y, 0)$
$d_2=\sqrt{x^2+z^2}$
(c) From $z$ axis i.e. $(0,0, z)$
$d_3=\sqrt{x^2+y^2}$
Distance of $(x, y, z)$ from origin
$d=\sqrt{x^2+y^2+z^2}$
$\begin{array}{r}\therefore \quad d_1^2+d_2^2+d_3^2=y^2+z^2+x^2+z^2+x^2+y^2 \\ =2\left(x^2+y^2+z^2\right)=2 d^2\end{array}$
$\therefore \quad K=2$.
(a) From $x$ axis i.e. $(x, 0,0)$
$d=\sqrt{y^2+z^2}$
(b) From $y$ axis i.e. $(0, y, 0)$
$d_2=\sqrt{x^2+z^2}$
(c) From $z$ axis i.e. $(0,0, z)$
$d_3=\sqrt{x^2+y^2}$
Distance of $(x, y, z)$ from origin
$d=\sqrt{x^2+y^2+z^2}$
$\begin{array}{r}\therefore \quad d_1^2+d_2^2+d_3^2=y^2+z^2+x^2+z^2+x^2+y^2 \\ =2\left(x^2+y^2+z^2\right)=2 d^2\end{array}$
$\therefore \quad K=2$.
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