Problem 70 - Thomas Calculus

Question

Find a plane through the origin that meets the plane \(M : 2 x+\) \(3 y+z=12\) in a right angle. How do you know that your plane is perpendicular to \(M ?\)

Step-by-Step Solution

Answer

The plane is \(3x - 2y = 0\), perpendicular by dot product zero.

Step 1: Understand the Problem

We need to find the equation of a plane through the origin that is perpendicular to another given plane. The given plane equation is \(M: 2x + 3y + z = 12\).

Step 2: Identify the Normal Vector of Plane M

Every plane equation of the form \(ax + by + cz = d\) has a normal vector \(\langle a, b, c \rangle\). Therefore, the normal vector for plane \(M\) is \(\langle 2, 3, 1 \rangle\).

Step 3: Use Orthogonality Condition

For two planes to be perpendicular, the normal vectors of these planes must be orthogonal. This means their dot product should be zero. Let \(\langle a, b, c \rangle\) be the normal vector of the plane we need to find. We require that \(\langle a, b, c \rangle \cdot \langle 2, 3, 1 \rangle = 0\).

Step 4: Write the Dot Product Equation

The dot product of the normal vectors \(\langle a, b, c \rangle\) and \(\langle 2, 3, 1 \rangle\) is \(2a + 3b + c = 0\).

Step 5: Find a Specific Normal Vector

We need any non-zero vector that satisfies the equation \(2a + 3b + c = 0\). One choice is to let \(a = 3\), \(b = -2\), and \(c = 0\), which satisfies \(2(3) + 3(-2) = 0\).

Step 6: Write the Plane Equation through the Origin

Using the normal vector \(\langle 3, -2, 0 \rangle\), the equation of the plane through the origin is \(3x - 2y + 0z = 0\) or simply \(3x - 2y = 0\).

Step 7: Verification

Verify that the dot product of the normal vectors \(\langle 3, -2, 0 \rangle\) and \(\langle 2, 3, 1 \rangle\) is indeed zero: \(3(2) + (-2)(3) + 0(1) = 6 - 6 = 0\), confirming the planes are perpendicular.

Key Concepts

Normal Vector

A normal vector is a fundamental element in the field of geometry, especially when dealing with planes in space. Imagine a plane just floating in three-dimensional space. A normal vector is a vector that points out perpendicularly from this plane. It acts like an arrow sticking straight out.

The equation of a plane is typically expressed in the form:

• \(ax + by + cz = d\)

Here, the normal vector is represented by the coefficients \(\langle a, b, c \rangle\). This vector defines the orientation of the plane in space. From the exercise, the plane \(M\) is given by \(2x + 3y + z = 12\), so its normal vector is \(\langle 2, 3, 1 \rangle\). Understanding this provides a critical foundation for solving perpendicular plane problems.

Normal vectors are at the heart of the concept of orthogonality, as they ensure that different planes intersect at right angles, essentially dictating the orientation of these planes.

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers—usually vectors—and returns a single number. It is a crucial concept when evaluating the angle between two vectors.

For vectors \(\mathbf{A} = \langle a_1, b_1, c_1 \rangle\) and \(\mathbf{B} = \langle a_2, b_2, c_2 \rangle\), the dot product is calculated as:

• \(a_1a_2 + b_1b_2 + c_1c_2\)

This result gives insight into the relationship between the two vectors. Specifically, if the dot product is zero, the vectors are orthogonal, meaning they are perpendicular to each other.

In our context, to confirm two planes are perpendicular, you calculate the dot product of their normal vectors. So, for \(\langle a, b, c \rangle\) and \(\langle 2, 3, 1 \rangle\), the condition \(2a + 3b + c = 0\) ensures they are orthogonally aligned. This principle simplifies the problem of finding perpendicular planes.

Plane Equation

A plane equation is the mathematical representation of a flat, two-dimensional surface in three-dimensional space. It is defined in the general form:

• \(ax + by + cz = d\)

The coefficients \(a, b,\) and \(c\) determine the orientation of the plane. The variable \(d\) sets its position relative to the origin.

When the plane goes through the origin, as in our exercise, \(d\) becomes zero. Therefore, the equation simplifies to \(ax + by + cz = 0\). This situation arises because no shift in position from the origin is needed.

For the exercise problem, we derived the plane through the origin using a normal vector \(\langle 3, -2, 0 \rangle\), arriving at the equation \(3x - 2y = 0\). This approach highlights how a plane's equation is tied to its normal vector and showcases its direct influence on the plane's spatial orientation.

Orthogonal Vectors

Orthogonal vectors are vectors that intersect at a right angle, which is the simplest form of perpendicularity. When two vectors are orthogonal, their dot product is zero, making them valuable in various mathematical applications, including geometry and vector calculus.

In our original exercise, ensuring that two planes meet at a right angle involves finding orthogonal normal vectors for these planes. By using the condition \(\langle a, b, c \rangle \cdot \langle 2, 3, 1 \rangle = 0\), we mathematically express the idea of orthogonality.

This means any vector satisfying the equation \(2a + 3b + c = 0\) is orthogonal to \(\langle 2, 3, 1 \rangle\). For instance, the vector \(\langle 3, -2, 0 \rangle\) fulfills this requirement, thus allowing us to find the perpendicular plane through the origin. By understanding orthogonal vectors, you gain insight into how different geometric entities relate to each other in space.