Problem 68 - University Calculus: Early Transcendentals

Question

Use the component form to generate an equation for the plane through \(P_{1}(4,1,5)\) normal to \(\mathbf{n}_{1}=\mathbf{i}-2 \mathbf{j}+\mathbf{k} .\) Then generate another equation for the same plane using the point \(P_{2}(3,-2,0)\) and the normal vector \(\mathbf{n}_{2}=-\sqrt{2} \mathbf{i}+2 \sqrt{2} \mathbf{j}-\sqrt{2} \mathbf{k}\)

Step-by-Step Solution

Answer

The plane's equation is \(x - 2y + z - 7 = 0\).

Step 1: Understanding the Problem

We are given a point \(P_1(4,1,5)\) and a normal vector \(\mathbf{n}_1 = \mathbf{i} - 2 \mathbf{j} + \mathbf{k}\). We need to find the equation of a plane that goes through the point \(P_1\) and is perpendicular to \(\mathbf{n}_1\).

Step 2: Equation of a Plane with Normal Vector

The general equation of a plane with a normal vector \(\mathbf{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) and passing through a point \(P(x_1, y_1, z_1)\) is:\[a(x - x_1) + b(y - y_1) + c(z - z_1) = 0\]

Step 3: Applying the Equation to First Case

For \(\mathbf{n}_1 = \mathbf{i} - 2\mathbf{j} + \mathbf{k}\), we have \(a = 1\), \(b = -2\), \(c = 1\), and for point \(P_1(4,1,5)\), we substitute into the plane equation:\[1(x - 4) - 2(y - 1) + 1(z - 5) = 0\]Simplifying, we get:\[x - 4 - 2y + 2 + z - 5 = 0\]\[x - 2y + z - 7 = 0\]

Step 4: Setting up for Second Case

Now, we use the given point \(P_2(3,-2,0)\) and normal vector \(\mathbf{n}_2 = -\sqrt{2} \mathbf{i} + 2\sqrt{2} \mathbf{j} - \sqrt{2} \mathbf{k}\) to create another equation for the same plane.

Step 5: Applying the Equation to Second Case

Using the normal vector \(\mathbf{n}_2\), we have \(a = -\sqrt{2}\), \(b = 2\sqrt{2}\), and \(c = -\sqrt{2}\). For point \(P_2(3,-2,0)\), the plane equation becomes:\[-\sqrt{2}(x - 3) + 2\sqrt{2}(y + 2) - \sqrt{2}(z - 0) = 0\]Simplifying, we get:\[-\sqrt{2}x + 3\sqrt{2} + 2\sqrt{2}y + 4\sqrt{2} - \sqrt{2}z = 0\]Divide the whole equation by \(-\sqrt{2}\) to simplify:\[x - 2y + z - 7 = 0\]

Step 6: Comparing Both Equations

Interestingly, both equations result in the same plane equation: \(x - 2y + z - 7 = 0\). This confirms that the two different methods using different points and normal vectors yielded equivalent equations for the same plane.

Key Concepts

Normal Vector

A normal vector is a vector that is perpendicular to a certain surface or plane. In the context of three-dimensional geometry, it plays a key role in defining the orientation of a plane. Imagine standing on an infinitely large flat floor. If you were to hold a stick straight up, that stick would resemble a normal vector to the floor.

In coordinate geometry, a plane's normal vector is represented mathematically as \(\mathbf{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\). This is a vector notation, indicating the direction of the vector in terms of its components along the x, y, and z axes.

• \(a, b, c\) are the coefficients that describe how much the plane tilts in the direction of each axis.
• The plane is perpendicular to this vector.

Understanding the normal vector is crucial for writing equations of planes, as it dictates the plane's orientation.

Component Form

The component form is a way to express vectors using their constituent parts along each axis: x, y, and z. It's particularly useful in describing both forces in physics and equations in mathematics.

When we have a vector, like the normal vector \(\mathbf{n} = \mathbf{i} - 2 \mathbf{j} + \mathbf{k}\), its component form is \((1, -2, 1)\). This means:

• The vector has a magnitude of 1 in the direction of the x-axis.
• A magnitude of -2 in the direction of the y-axis.
• And a magnitude of 1 in the direction of the z-axis.

In plane geometry, once we know the component form of a normal vector, we can plug these values into the general formula for a plane. It allows us to understand how inclined or skewed a plane is, relative to each coordinate axis.

Coordinate Geometry

Coordinate geometry, often called analytic geometry, uses algebraic methods to solve geometric problems. Within this discipline, equations and formulas are employed to represent geometric figures like lines, circles, and planes.

Understanding equations of planes involves linking algebra and geometry so that we can visualize planes in three-dimensional space through equations like \(a(x-x_1)+b(y-y_1)+c(z-z_1)=0\). Here:

• \(P(x_1, y_1, z_1)\) represents a specific point on the plane.
• \(a, b, c\) are components of the normal vector, defining the plane's tilt.

By adjusting these variables, you can define different planes. Thus, coordinate geometry makes it possible to represent complex shapes and solve related problems using simpler mathematical equations.

Point in Space

A point in space is defined by its position on the coordinate axes, described using three coordinates: x, y, and z. Each coordinate signifies a point's location relative to other points, similar to a dot in the vast 3D landscape.

The point \(P_1(4,1,5)\) signifies:

• An x-coordinate of 4, indicating how far along the x-axis it lies.
• A y-coordinate of 1, showing its distance along the y-axis.
• A z-coordinate of 5, indicating elevation from the origin along the z-axis.

In defining a plane, having a specific point ensures that the plane will pass through that point, making the plane uniquely determined in conjunction with its normal vector. The combination of these two elements fully describes the orientation and location of the plane in three-dimensional space.