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Find an equation for the plane consisting of all points that are equidistant from the points (1,0,-2) and (3,4,0).

Find An Equation For The Plane Consisting Of All Points That Are Equidistant From The Points

This problem aims to familiarize us with geometrical calculations. The concept required to solve this problem is the distance formula in 3-dimensional space, and some square and cubic algebraic formulas.

The formula for distance states that the distance between two points in xyz-space is the sum of the squares of the differences between similar xyz coordinates under a square root. Let’s say we have points:

\[ P_1 = (x_1,y_1,z_1)\space and\space P_2 = (x_2,y_2,z_2)\]

The total distance between $P_1$ and $P_2$ is yielded as:

\[ d(P_1,P_2) = \sqrt{(x_2 x_1)^2 + (y_2 y_1)^2 + (z_2 z_1)^2}\]

Expert Answer

Given points are $(1,0,-2)$ and $(3,4,0)$.

We have to generate an equation for the plane consisting of all points that are equidistant from the points $(1,0,-2)$ and $(3,4,0)$.

Let’s assume the point $(x,y,z)$ on the plane that is equidistant from the given points. To calculate the distance of the given points with the $(x,y,z)$, we will use the distance formula.

Distance Formula is given as:

\[ \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 +(z_2 – z_1)^2 } \]

Applying this formula on points $(x,y,z)$ and $(1,0,-2)$ to calculate the distance:

\[ \sqrt{(x – 1)^2 + (y – 0)^2 +(z + 2)^2 } \]

Expanding the expression using the algebraic formulas:

$(a-b)^2=a^2+b^2-2ab$

$(a+b)^2=a^2+b^2+2ab$

\[\sqrt{(x^2 -2x +1) + y^2 +(z^2 +4z+4)}\]

\[\sqrt{(x^2 + y^2 + z^2 -2x +4z +5)}\]

Now calculating the distance of the point $(3,4,0)$ with the $(x,y,z)$.

\[\sqrt{(x – 3)^2 + (y – 4)^2 + z^2 }\]

Expanding the expression using the algebraic formulas:

\[\sqrt{(x^2 -6x +9) + (y^2 -8y+16) + z^2 }\]

\[\sqrt{(x^2 + y^2 + z^2 -6x – 8y + 25)}\]

As both distances are equidistant, equating them and then simplifying:

\[\sqrt{(x^2 + y^2 + z^2 -2x +4z +5)} = \sqrt{(x^2 + y^2 + z^2 -6x – 8y + 25)}\]

The expression is re-written as:

\[x^2 + y^2 + z^2 -2x +4z +5 = x^2 + y^2 + z^2 -6x -8y + 25\]

\[ \cancel{x^2}+\cancel{y^2}+\cancel{z^2}-2x+4z+5 = \cancel{x^2}+\cancel{y^2}+\cancel{z^2}-6x-8y+25 \]

\[-2x+4z+5=-6x-8y+25 \]

\[-2x+6x +8y+4z +5-25 = 0 \]

\[4x +8y+4z -20=0\]

Dividing the equation with $4$:

\[x+2y+z=5\]

Numerical Answer

So the equation of the plane that consists of all the points that are equidistant from the given points is calculated to be:

$(1,0,-2)$ and $(3,4,0)$ is $ x +2y+z = 5$.

Example

What is the equation of the plane consisting of all points that are equidistant from $(-5, 5, -3)$ and $(4,5,3)$?

Calculating the distance between $(x,y,z)$ and $(-5,5,-3)$:

\[ \sqrt{(x + 5)^2 + (y – 5)^2 +(z + 3)^2 } \]

\[ \sqrt{(x^2 + y^2 + z^2 +10x -10y +6z + 59)} \]

Now calculating the distance between $(4,5,3)$ with $(x,y,z)$.

\[ \sqrt{(x – 4)^2 + (y – 5)^2 + (z-3)^2 } \]

\[ \sqrt{(x^2 + y^2 + z^2 -8x – 10y -6z+ 50)} \]

As both distances are equidistant, putting them equal to each other and simplifying:

\[ \sqrt{(x^2 + y^2 + z^2 +10x -10y +6z + 59)} = \sqrt{(x^2 + y^2 + z^2 -8x – 10y -6z+ 50)} \]

Re-writing:

\[ 10x + 8x -10y + 10y +6z +6z +59 -50 = 0 \]

\[ 6x + 4z = -3 \]

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