The distance of the point P(x, y, z) from the | vedclass

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(A) In a three-dimensional Cartesian coordinate system,the position of a point $P$ is given by its coordinates $(x, y, z)$.The $yz$-plane is defined b

Vedclass · Accepted Answer

$|x|$

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(A) In a three-dimensional Cartesian coordinate system,the position of a point $P$ is given by its coordinates $(x, y, z)$.The $yz$-plane is defined by the equation $x = 0$.The perpendicular distance of any point $(x, y, z)$ from the plane $ax + by + cz + d = 0$ is given by the formula $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$.For the $yz$-plane,the equation is $1x + 0y + 0z + 0 = 0$.Substituting the coordinates of point $P(x, y, z)$ into the distance formula:$d = \frac{|1(x) + 0(y) + 0(z) + 0|}{\sqrt{1^2 + 0^2 + 0^2}}$$d = \frac{|x|}{1} = |x|$.Thus,the distance of the point $P(x, y, z)$ from the $yz$-plane is $|x|$.

The distance of the point $P(x, y, z)$ from the $yz$-plane is:

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