The sum of the squares of the perpendicular d | vedclass

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(A) Let the point be $P(x, y, z)$.$(a)$ The perpendicular distance $d_1$ of point $P$ from the $x$-axis is $\sqrt{y^2 + z^2}$. So,$d_1^2 = y^2 + z^2$.

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$2$

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(A) Let the point be $P(x, y, z)$.$(a)$ The perpendicular distance $d_1$ of point $P$ from the $x$-axis is $\sqrt{y^2 + z^2}$. So,$d_1^2 = y^2 + z^2$.$(b)$ The perpendicular distance $d_2$ of point $P$ from the $y$-axis is $\sqrt{x^2 + z^2}$. So,$d_2^2 = x^2 + z^2$.$(c)$ The perpendicular distance $d_3$ of point $P$ from the $z$-axis is $\sqrt{x^2 + y^2}$. So,$d_3^2 = x^2 + y^2$.The distance $d$ of point $P$ from the origin $(0, 0, 0)$ is $\sqrt{x^2 + y^2 + z^2}$. So,$d^2 = x^2 + y^2 + z^2$.The sum of the squares of the perpendicular distances is $d_1^2 + d_2^2 + d_3^2 = (y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 2(x^2 + y^2 + z^2)$.Given that this sum is $k$ times the square of the distance from the origin,we have $2(x^2 + y^2 + z^2) = k(x^2 + y^2 + z^2)$.Therefore,$k = 2$.

The sum of the squares of the perpendicular distances of a point $(x, y, z)$ from the coordinate axes is $k$ times the square of the distance of the point from the origin. Then $k=$

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