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

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

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Solution

We know that the distance between a point $P (x_{1}, y_{1}, z_{1})$ and a plane Ax+By+Cz=D is

$d = \frac{A x_{1} + B y_{1} + C z_{1} - D}{\sqrt{A^{2} + B^{2} + C^{2}}}$ ...(i)

The given point is (-6,0,0) and the plane is 2x -3y+6z-2=0

$∴$ From Eq.(i)

$d = \frac{| 2 \times (- 6) - 3 \times 0 + 6 \times 0 - 2 |}{\sqrt{2^{2} + (- 3)^{2} + 6^{2}}} = \frac{| - 14 |}{\sqrt{4 + 9 + 36}} = \frac{14}{7} = 2 u n i t s$

Note The distance is always positive.

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Similar questions

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

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

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

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

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

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

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

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

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

(- 6, 0, 0)

2 x - 3 y + 6 z = 2

In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(a) (0, 0, 0)

(b) (3, −2, 1)

(d) (−6, 0, 0)

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