Q.Find the perpendicular distance of the origin from the plane $6x+2y+3z-7=0$.

Use $d = \dfrac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}$.

Plane $x + 2y - 2z = 9$, point $(2,3,-5)$:

$d = \frac{|1(2) + 2(3) - 2(-5) - 9|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{|2 + 6 + 10 - 9|}{\sqrt{9}} = \frac{9}{3} = 3.$

Distance $= \dfrac{13}{\sqrt{21}} = \dfrac{13\sqrt{21}}{21}$.

The distance of a point $(x_0, y_0, z_0)$ from the plane $ax + by + cz = d$ is

$D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}.$

Here the plane is $x - 2y + 4z = 9$ and the point is $(2, 1, -1)$:

$ax_0 + by_0 + cz_0 - d = (1)(2) + (-2)(1) + (4)(-1) - 9 = 2 - 2 - 4 - 9 = -13.$

$\sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}.$

$D = \frac{|-13|}{\sqrt{21}} = \frac{13}{\sqrt{21}} = \frac{13\sqrt{21}}{21}.$

Tests point-to-plane distance formula; the distance is $3$.

Distance $=\dfrac{11}{\sqrt{61}}$.

For a plane $ax+by+cz=d$, the distance from the origin is $\dfrac{|d|}{\sqrt{a^{2}+b^{2}+c^{2}}}$. Here $a=3,b=-4,c=6,d=11$:

$\dfrac{|11|}{\sqrt{9+16+36}}=\dfrac{11}{\sqrt{61}}.$

Distance from point $(x_1,y_1,z_1)$ to plane $ax+by+cz+d=0$ is $\dfrac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$.

Plane: $2x+y-2z+1=0$, point $(0,-1,3)$.

Distance $= \dfrac{|2(0)+1(-1)-2(3)+1|}{\sqrt{2^2+1^2+(-2)^2}} = \dfrac{|0-1-6+1|}{\sqrt{4+1+4}} = \dfrac{|-6|}{\sqrt9} = \dfrac{6}{3} = 2$.

Rewrite the second plane with the same normal as the first, then use the parallel-plane distance formula.

Given planes $2x+3y+4z=4$ and $4x+6y+8z=12$. Dividing the second equation by 2:

$2x+3y+4z = 6$

Both planes now share the normal vector $(2,3,4)$, so they are parallel with equations $2x+3y+4z=4$ and $2x+3y+4z=6$.

Distance between two parallel planes $ax+by+cz=d_1$ and $ax+by+cz=d_2$ is $\dfrac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}}$:

$d = \dfrac{|6-4|}{\sqrt{2^2+3^2+4^2}} = \dfrac{2}{\sqrt{4+9+16}} = \dfrac{2}{\sqrt{29}}$

The $XY$-plane is $z=0$; the perpendicular distance from any point to this plane is simply the magnitude of its $z$-coordinate.

For $P(x,y,z)$, the distance to the plane $z=0$ (the $XY$-plane) is $|z|$, written simply as $z$ among the given choices.

Length of the perpendicular from the origin to $\vec r\cdot\vec N=d$ is $\dfrac{|d|}{|\vec N|}$.

The vector equation of a plane in the form $\vec r\cdot\vec N=d$ (where $\vec N$ is a normal vector to the plane, not necessarily a unit vector) can be converted to normal form by dividing throughout by $|\vec N|$:

$\vec r\cdot\frac{\vec N}{|\vec N|}=\frac{d}{|\vec N|}.$

This is the standard normal form $\vec r\cdot\hat n=p$, where $\hat n=\vec N/|\vec N|$ is the unit normal and $p$ is precisely the perpendicular distance of the plane from the origin.

So the perpendicular distance from the origin is

$p=\frac{|d|}{|\vec N|}$

(absolute value, since distance is non-negative).

Use the point-to-plane distance formula $\left|\dfrac{Ax_0+By_0+Cz_0+D}{\sqrt{A^2+B^2+C^2}}\right|$.

Treat the given plane $3x+4y-20=0$ as a plane in 3D space: $3x+4y+0\cdot z-20=0$, i.e. $A=3,B=4,C=0,D=-20$.

For the point $(0,0,-7)$:

$\text{Distance} = \frac{|3(0)+4(0)+0(-7)-20|}{\sqrt{3^2+4^2+0^2}} = \frac{|-20|}{\sqrt{25}} = \frac{20}{5} = 4$

standard formula for distance of a point from a plane in vector form

The distance of a point with position vector $\vec a$ from the plane $\vec r\cdot\vec n=q$ is $d=\dfrac{|\vec a\cdot\vec n-q|}{|\vec n|}$ — dividing by $|\vec n|$ is essential because $\vec n$ need not be a unit vector.

distance-of-a-point-from-a-plane formula

The perpendicular distance of a point with position vector $\vec a$ from the plane $\vec r\cdot\vec n=q$ is obtained by substituting $\vec a$ in place of $\vec r$ in the plane's equation and dividing by $|\vec n|$:

$d=\dfrac{|\vec a\cdot\vec n-q|}{|\vec n|}$