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 point $(- 6,, 0, 0)$
from the plane $2 x - 3 y + 6 z = 2$
$\begin{matrix} ∣ ∣ ∣ ∣ ∣ \frac{2 (- 6) - 3 (0) + 6 (0) - 2}{{\sqrt{{(2)}^{2} + {(- 3)}^{2} + {(6)}^{2}}}} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \frac{- 12 - 2}{\sqrt{4 + 9 + 36}} ∣ ∣ ∣ = \frac{14}{\begin{matrix} \sqrt{49} = \frac{14}{7} = 2 \end{matrix}} \end{matrix}$

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Similar questions

(- 6, 0, 0)

2 x - 3 y + 6 z = 2

(- 6, 0, 0)

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:

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

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

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:

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

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

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

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

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