Find an equation of the plane with $x$-intercept $a$, $y$-intercept $b$, and $z$-intercept $c$.

Neetesh Kumar | December 16, 2024

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This is the solution to Math 1C
Assignment: 12.5 Question Number 26
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Step-by-step solution:

We need to find the equation of the plane with $x$-intercept $a$, $y$-intercept $b$, and $z$-intercept $c$.

Step 1: General form of the plane equation

The general form of the equation of a plane is: $Ax + By + Cz = D$ where $A$, $B$, and $C$ are constants representing the normal vector of the plane, and $D$ is a constant.

Step 2: Intercepts of the plane

The plane intersects the $x$-axis at $(a, 0, 0)$, the $y$-axis at $(0, b, 0)$, and the $z$-axis at $(0, 0, c)$.

The plane passing through these three points has the following equation:

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

Step 3: Verification

1. At the $x$-intercept $(a, 0, 0)$:

   $\frac{a}{a} + \frac{0}{b} + \frac{0}{c} = 1$ This holds true.

2. At the $y$-intercept $(0, b, 0)$: $\frac{0}{a} + \frac{b}{b} + \frac{0}{c} = 1$ This also holds true.

3. At the $z$-intercept $(0, 0, c)$:

   $\frac{0}{a} + \frac{0}{b} + \frac{c}{c} = 1$ This holds true as well.

Thus, the equation is consistent for all three intercepts.

Final Answer:

The equation of the plane with $x$-intercept $a$, $y$-intercept $b$, and $z$-intercept $c$ is:

$\boxed{\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1}$

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