Problem 20 - Precalculus Mathematics for Calculus

Question

A plane has normal vector \(n\) and passes through the point \(P\). (a) Find an equation for the plane. (b) Find the intercepts and sketch a graph of the plane. $$\mathbf{n}=\mathbf{i}+4 \mathbf{j}, \quad P(1,0,-9)$$

Step-by-Step Solution

Answer

Equation: \(x + 4y = 1\); X-intercept: (1,0,0); Y-intercept: (0,1/4,0).

Step 1: Understand the normal vector

The normal vector to the plane is given by \( \mathbf{n} = \mathbf{i} + 4 \mathbf{j} \). This means the direction of the plane is perpendicular to the vector \((1, 4, 0)\).

Step 2: Use the point and normal vector to form the plane equation

The plane equation can be derived from the point-normal form: \( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \), where \((a, b, c)\) are the components of the normal vector, and \((x_1, y_1, z_1)\) is the given point. Substituting \(a = 1\), \(b = 4\), \(c = 0\), and \((x_1, y_1, z_1) = (1, 0, -9)\), we get \(1(x - 1) + 4(y - 0) + 0(z + 9) = 0\).

Step 3: Simplify the plane equation

Simplify the equation: \(x - 1 + 4y = 0\) becomes \(x + 4y = 1\). This is the equation of the plane.

Step 4: Find the X and Y intercepts of the plane

For the x-intercept, set \(y = 0\) and solve for \(x\): \(x = 1\). For the y-intercept, set \(x = 0\) and solve for \(y\): \(4y = 1\) gives \(y = \frac{1}{4}\). The x-intercept is at \((1,0,0)\), and the y-intercept is at \((0,\frac{1}{4},0)\).

Step 5: Sketch the graph

Sketch the plane in the XYZ-coordinate system. Use the intercepts to plot the plane's position: it crosses the x-axis at \((1,0,0)\) and the y-axis at \((0,\frac{1}{4},0)\). The plane appears parallel to the xy-plane with its orientation guided by the intercepts and given normal vector.

Key Concepts

Normal Vector

In geometry, the concept of a normal vector is fundamental when dealing with planes. A normal vector is a vector that is perpendicular to a plane or surface. Its components are used to describe the tilt or orientation of the plane in three-dimensional space. In our exercise, the normal vector is given as \( \mathbf{n} = \mathbf{i} + 4 \mathbf{j} \), which corresponds to the vector components \((1, 4, 0)\). This vector does not affect the z-axis, implying that the plane is "flat" along that dimension and acts as a guide for the plane's position.

• Understanding the role of the normal vector helps in visualizing the plane's tilt; here, since the z-component is zero, it's parallel to the xy-plane.
• A normal vector is crucial for determining the equation of the plane, typically by using it alongside a given point on the plane to form what is called the "point-normal form."
• It's also important as it defines what is "upwards" or "outwards," providing a clear direction perpendicular to the plane.

Intercepts

Intercepts are the points where a geometric figure intersects the axes in a coordinate system. Finding the intercepts of a plane can help in sketching and understanding its position in 3D space. For this specific plane, given the equation \(x + 4y = 1\), we identify the intercepts by considering where the plane crosses each axis:

• To find the x-intercept, set \(y = 0\) and solve for \(x\), yielding the intercept point \((1, 0, 0)\).
• To find the y-intercept, set \(x = 0\) and solve for \(y\). This gives \(y = \frac{1}{4}\), leading to the intercept point \((0, \frac{1}{4}, 0)\).

These intercepts indicate that the plane crosses the x-axis at 1 and the y-axis at \(\frac{1}{4}\). Knowing these points helps us visualize the plane's orientation and how it is situated within the coordinate system. It's particularly essential when drawing the plane or trying to comprehend its spatial dynamics.

Point-Normal Form

The point-normal form of a plane equation is crucial as it provides a method to define a plane using a point on the plane and its normal vector. The form of this equation is:\[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \]Here, \((a, b, c)\) are the components of the normal vector, and \((x_1, y_1, z_1)\) is a point on the plane. Our exercise specifies the normal vector \((1, 4, 0)\) and passes through the point \((1, 0, -9)\). Substituting these values into the point-normal form, we prepend variables \(x, y, z\) to the given coordinates:- \(1(x-1) + 4(y-0) + 0(z+9) = 0\)This simplifies to \(x + 4y = 1\). The simplicity of the point-normal form derives from how it integrates both a spatial location (via the point) and orientation (through the normal vector) of the plane. It's particularly beneficial when you need to derive a precise, equation-based representation of a plane with given data.