Problem 71 - University Calculus: Early Transcendentals

Question

Is the line \(x=1-2 t, y=2+5 t, z=-3 t\) parallel to the plane \(2 x+y-z=8 ?\) Give reasons for your answer.

Step-by-Step Solution

Answer

The line is not parallel to the plane.

Step 1: Identify the direction vector of the line

The line is given in parametric form: \( x = 1 - 2t, y = 2 + 5t, z = -3t \). The direction vector can be extracted from the coefficients of \( t \), which are \( (-2, 5, -3) \).

Step 2: Find the normal vector of the plane

The plane equation is \( 2x + y - z = 8 \). The normal vector of the plane can be derived from the coefficients of \( x, y, \) and \( z \), which are \( (2, 1, -1) \).

Step 3: Check if the direction vector is orthogonal to the normal vector

To determine if the line is parallel to the plane, check if the dot product of the direction vector \((-2, 5, -3)\) and the normal vector \((2, 1, -1)\) is zero. Calculate: \[ (-2) \cdot 2 + 5 \cdot 1 + (-3) \cdot (-1) = -4 + 5 + 3 = 4 \]

Step 4: Analyze the result of the dot product

Since the dot product is not zero (equals 4), the direction vector is not orthogonal to the normal vector. Thus, the line is not parallel to the plane.

Key Concepts

Direction Vector

The direction vector of a line is an essential concept in vector calculus, especially when working with parametric equations. When you have a line given in parametric form, such as \( x = 1 - 2t, y = 2 + 5t, z = -3t \), the direction vector can be directly obtained from the coefficients of \( t \). Here, it is \((-2, 5, -3)\). This vector indicates the direction in which the line extends in space.

Understanding the direction vector is crucial because it allows us to analyze the orientation of the line, determining how it moves as \( t \) varies. It does not matter where the line starts or ends; the direction vector purely describes its orientation.

An easy way to remember this is to think of the line as an arrow, where the initial point does not affect the direction of the arrow, only its head does. So, when analyzing spatial problems, always first extract and interpret the direction vector.

Plane Equation

When dealing with planes in vector calculus, a plane equation is often given in the form of \( ax + by + cz = d \). For our problem, the plane equation is \( 2x + y - z = 8 \). Here, the coefficients \( a = 2 \), \( b = 1 \), and \( c = -1 \) are vital components.

The plane equation provides a concise way to represent a two-dimensional surface in three-dimensional space. It allows us to see how changes in \( x, y, \) and \( z \) relate to each other within that surface.

It's important to understand that every point \((x, y, z)\) that satisfies this equation lies on the plane. The beauty of a plane equation in vector calculus is its ability to encapsulate infinite points in space through a simple algebraic expression. This makes it a powerful tool for solving geometric problems.

Normal Vector

A normal vector is a key concept when discussing planes. It is the vector that is perpendicular to the plane. For our given plane equation \( 2x + y - z = 8 \), the normal vector can be extracted directly from the coefficients \((2, 1, -1)\).

Grasping the role of the normal vector helps in understanding how the plane is oriented in 3D space. It points directly away from the surface of the plane and is essential in calculations involving angles and distances to other geometric entities.

The normal vector is indispensable for operations like checking orthogonality, which is crucial in determining parallelism between lines and planes. Recognizing this vector's direction helps one visualize and solve spatial problems effectively.

Dot Product

The dot product is a fundamental operation in vector calculus which applies to pairs of vectors. It provides a measure of how much two vectors "point" in the same direction. Mathematically, the dot product of two vectors \((a, b, c)\) and \((d, e, f)\) is given by the formula \( a \cdot d + b \cdot e + c \cdot f \).

For the direction vector \((-2, 5, -3)\) and the normal vector \((2, 1, -1)\), calculating the dot product reveals whether these vectors are orthogonal. Plugging in the values, we get \((-2)\cdot 2 + 5\cdot 1 + (-3)\cdot (-1) = 4\).

The result, 4, indicates the vectors are not orthogonal. Orthogonal vectors have a dot product equal to zero. Thus, since our result is not zero, the line is not parallel to the plane. Understanding the dot product aids significantly in analyzing vector alignments in space.