Find the distance of the point $(- 6, 0, 0)$ from the plane $2 x - 3 y + 6 z = 2$ .

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Solution

Distance of a point $(a_{1}, b_{1}, c_{1})$ from a plane $a x + b y + c z + d = 0$ is given by $d = ∣ ∣ ∣ \frac{a a_{1} + b b_{1} + c c_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} ∣ ∣ ∣$
$\Rightarrow$ Here, $d = ∣ ∣ ∣ \frac{- 12 - 2}{\sqrt{4 + 9 + 36}} ∣ ∣ ∣ = \frac{14}{\sqrt{49}} = 2$ .

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Similar questions

Q. Find the distance of the point

(- 6, 0, 0)

from the plane

2 x - 3 y + 6 y = 2

In the following example find the distance of each of the given points from the corresponding given plane:

$\begin{matrix} p o i n t & P l a n e (- 6, 0, 0) & 2 x - 3 y + 6 z - 2 = 0 \end{matrix}$

Q. The distance between the planes

2 x - 3 y + 6 z + 12 = 0

and

2 x - 3 y + 6 z - 2 = 0

In the following exercises find the distance of each of the given points from the corresponding given plane:

$\begin{matrix} p o i n t & P l a n e (a) (0, 0, 0) & 3 x - 4 y + 12 z = 3 \end{matrix}$

$\begin{matrix} p o i n t & P l a n e (b) (3, - 2, 1) & 2 x - y + 2 z + 3 = 0 \end{matrix}$

$\begin{matrix} p o i n t & P l a n e (c) (2, 3, - 5) & x + 2 y - 2 z = 9 \end{matrix}$

$\begin{matrix} p o i n t & P l a n e (d) (- 6, 0, 0) & 2 x - 3 y + 6 z - 2 = 0 \end{matrix}$

A line having direction ratios $3, 4, 5$ cuts $2$ planes $2 x - 3 y + 6 z - 12 = 0$ and $2 x - 3 y + 6 z + 2 = 0$ at point P & Q, then Find length of PQ

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Find the distance of the point (−6,0,0) from the plane 2x−3y+6z=2.

Distance of a point (a1,b1,c1) from a plane ax+by+cz+d=0 is given by d=∣∣∣aa1+bb1+cc1+d√a2+b2+c2∣∣∣⇒ Here, d=∣∣∣−12−2√4+9+36∣∣∣=14√49=2.

Find the distance of the point $(- 6, 0, 0)$ from the plane $2 x - 3 y + 6 z = 2$ .