Q.Find the distance of the point $(4,-5,6)$ from the plane $\vec{r}\cdot(4\vec{i}-4\vec{j}+7\vec{k})=-6$.

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

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

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 $(2, 1, -1)$:

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

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 $a=3,b=-4,c=6,d=11$:

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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 $(0,-1,3)$.

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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 $\dfrac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}}$: …

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

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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 $|\vec N|$:

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

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 $(0,0,-7)$:

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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 becaus …

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 $|\vec n|$:

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