The equation of the locus of a point P(x, y, | vedclass

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(A) The distance of a point $P(x, y, z)$ from the $X$-axis is given by $\sqrt{y^2+z^2}$.The distance of the point $P(x, y, z)$ from the plane $x+z-1=0

Vedclass · Accepted Answer

$x^2-2 y^2-z^2+2 x z-2 x-2 z+1=0$

Vedclass · Answer

(A) The distance of a point $P(x, y, z)$ from the $X$-axis is given by $\sqrt{y^2+z^2}$.The distance of the point $P(x, y, z)$ from the plane $x+z-1=0$ is given by $\frac{|x+z-1|}{\sqrt{1^2+0^2+1^2}} = \frac{|x+z-1|}{\sqrt{2}}$.According to the problem,these distances are equal:$\sqrt{y^2+z^2} = \frac{|x+z-1|}{\sqrt{2}}$Squaring both sides,we get:$y^2+z^2 = \frac{(x+z-1)^2}{2}$$2(y^2+z^2) = x^2+z^2+1+2xz-2x-2z$$2y^2+2z^2 = x^2+z^2+1+2xz-2x-2z$Rearranging the terms,we get:$x^2-2y^2-z^2+2xz-2x-2z+1=0$Thus,the correct option is $A$.

The equation of the locus of a point $P(x, y, z)$ such that its distance from the $X$-axis is equal to its distance from the plane $x+z=1$ is

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