Question

find a point on the plane 3x+2y+z-12=0 which is nearest to the origin also find the distance between them

Answer 1

Answer:

$x=\frac{-2 y-z+12}{3}$

Step-by-step explanation:

Step 1: The distance between a given point and its orthogonal projection on a plane in Euclidean space, or the perpendicular distance to the closest point on the plane.

Step 2:  Given two distinct points, we can draw many planes passing through them. Therefore, infinite number of planes can be drawn passing through two distinct points or two points can be common to infinite number of planes.

Step 3: 3 x+2 y+z-12=0

Subtract 2 y+z from both sides

3 x+2 y+z-12-(2 y+z)=0-(2 y+z)

Simplify

3 x-12=-(2 y+z)

Move 12 to the right side

3 x=-2 y-z+12

Divide both sides by 3

$\frac{3 x}{3}=-\frac{2 y}{3}-\frac{z}{3}+\frac{12}{3}$

Simplify

$x=\frac{-2 y-z+12}{3}$

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Answer 2

The distance between a given point and its orthogonal projection on a plane in Euclidean space, or the perpendicular distance to the closest point on the plane.

 Given two distinct points, we can draw many planes passing through them. Therefore, infinite number of planes can be drawn passing through two distinct points or two points can be common to infinite number of planes.

Step 3: 3 x+2 y+z-12=0

Subtract 2 y+z from both sides

3 x+2 y+z-12-(2 y+z)=0-(2 y+z)

Simplify

3 x-12=-(2 y+z)

Move 12 to the right side

3 x=-2 y-z+12

Divide both sides by 3

Simplify

$\frac{2y -z + 12}{3}$ =x

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