Find an equation for the plane consisting of all points that are equidistant from the points $(4, 0, -2)$ and $(6, 14, 0)$.

Neetesh Kumar | December 16, 2024

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This is the solution to Math 1C
Assignment: 12.5 Question Number 25
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Step-by-step solution:

We are tasked with finding the equation of the plane that consists of all points equidistant from the points $A(4, 0, -2)$ and $B(6, 14, 0)$. To do this, we follow the steps below.

Step 1: Midpoint of $A$ and $B$

The plane equidistant from two points $A$ and $B$ will pass through the midpoint of these points.

The formula for the midpoint of two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is:

$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)$

Substitute $A(4, 0, -2)$ and $B(6, 14, 0)$:

$M = \left( \frac{4 + 6}{2}, \frac{0 + 14}{2}, \frac{-2 + 0}{2} \right)$

Simplify:

$M = (5, 7, -1)$

The midpoint is $M(5, 7, -1)$.

Step 2: Direction Vector between $A$ and $B$

The direction vector $\vec{AB}$ between the two points is given by:

$\vec{AB} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)$

Substitute $A(4, 0, -2)$ and $B(6, 14, 0)$:

$\vec{AB} = (6 - 4, 14 - 0, 0 - (-2))$

Simplify:

$\vec{AB} = (2, 14, 2)$

Thus, the direction vector $\vec{AB}$ is $(2, 14, 2)$.

Step 3: Normal Vector of the Plane

The plane that is equidistant from $A$ and $B$ is perpendicular to the direction vector $\vec{AB}$. Therefore, the normal vector $\vec{n}$ of the plane is:

$\vec{n} = (2, 14, 2)$

Step 4: Equation of the Plane

The general form of a plane equation is:

$n_1 (x - x_0) + n_2 (y - y_0) + n_3 (z - z_0) = 0$

Where:

• $(n_1, n_2, n_3)$ is the normal vector,
• $(x_0, y_0, z_0)$ is a point on the plane.

Here:

• Normal vector $\vec{n} = (2, 14, 2)$,
• Point on the plane $M(5, 7, -1)$.

Substitute these into the equation:

$2(x - 5) + 14(y - 7) + 2(z + 1) = 0$

Step 5: Simplify the Equation

Expand the terms:

$2x - 10 + 14y - 98 + 2z + 2 = 0$

Combine like terms:

$2x + 14y + 2z - 106 = 0$

Divide through by $2$ to simplify:

$x + 7y + z - 53 = 0$

Final Answer:

The equation of the plane is:
$\boxed{x + 7y + z - 53 = 0}$

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