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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$|x|$

Vedclass · Answer

(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:

Similar Questions

Find the coordinates of a point on the $y$-axis which is at a distance of $5\sqrt{2}$ from the point $P(3, -2, 5)$.

Which of the following is at a distance of $\sqrt{10}$ from the point $(1, 2, 3)$?

Find the equation of the set of points $P$ such that $PA^{2} + PB^{2} = 2k^{2}$,where $A$ and $B$ are the points $(3, 4, 5)$ and $(-1, 3, -7)$,respectively.

If the sum of the squares of the distances of a point $P(x, y, z)$ from the three co-ordinate axes is $324$,then the distance of point $P$ from the origin is ....

The shortest distance of the point $(a, b, c)$ from the $X-$axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module