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show that line \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}\) is parallel to plane \(x-2 y+z=6\)

Short Answer

Expert verified
The line is parallel to the plane.

Step by step solution

01

Determine the Direction Vector of the Line

The given line equation is \( \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{4} \). The direction vector \( \mathbf{d} \) of the line can be directly read from the numerators of the equalities: \( \mathbf{d} = \langle 2, 3, 4 \rangle \).
02

Identify the Normal Vector of the Plane

The plane equation is \( x - 2y + z = 6 \). From the coefficients of \( x \), \( y \), and \( z \), the normal vector \( \mathbf{n} \) to the plane is \( \langle 1, -2, 1 \rangle \).
03

Check for Parallelism via the Dot Product

A line is parallel to a plane if the direction vector of the line is perpendicular to the normal vector of the plane. This is checked by computing the dot product between \( \mathbf{d} \) and \( \mathbf{n} \). The dot product is given by: \( \mathbf{d} \cdot \mathbf{n} = 2 \cdot 1 + 3 \cdot (-2) + 4 \cdot 1 = 2 - 6 + 4 \).
04

Evaluate the Dot Product

Compute \( 2 - 6 + 4 = 0 \). Since the dot product is 0, the direction vector \( \mathbf{d} \) is perpendicular to the normal vector \( \mathbf{n} \).
05

Conclude the Relation Between the Line and the Plane

Since the direction vector is perpendicular to the normal vector, the line is parallel to the plane. Therefore, the line \( \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{4} \) is parallel to the plane \( x - 2y + z = 6 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direction Vector
To understand if a line is parallel to a plane, a crucial point is identifying the direction vector of the line. A direction vector indicates the direction in which a line extends. Just like the name suggests, it tells us the way! In the context of a line given in parametric form, the direction vector can usually be determined directly from the line's equation.
For example, consider the line represented by \( \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{4} \). The direction vector \( \mathbf{d} \) of this line is \( \langle 2, 3, 4 \rangle \).
The components of this vector (2, 3, and 4) come directly from the denominators in the line's parametrization. This direction vector represents the orientation and path of the line in three-dimensional space, which helps in determining whether lines are parallel or intersecting certain planes.
Normal Vector
In geometry, particularly when dealing with planes, the concept of a normal vector is fundamental. A normal vector is perpendicular to every line lying on a plane. Therefore, it serves as a perfect tool to understand the orientation of the plane.
The plane equation \( x - 2y + z = 6 \) allows us to directly find the normal vector. This vector is made up of the coefficients of \( x \), \( y \), and \( z \) from the plane's equation. Hence, the normal vector \( \mathbf{n} \) is \( \langle 1, -2, 1 \rangle \).
This normal vector tells us the direction extending out from the surface of the plane and can be used to test how other vectors, such as the direction vector of a line, relate in terms of parallelism or perpendicularity.
Dot Product
The dot product is a very informative mathematical operation that helps determine the angle between two vectors. Specifically, it can tell us if vectors are perpendicular. This is handy when comparing a line's direction vector with a plane's normal vector to check for parallelism.
The dot product of two vectors, say \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), is computed as \( \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \).
In our example, finding the dot product of the direction vector \( \mathbf{d} = \langle 2, 3, 4 \rangle \) and the normal vector \( \mathbf{n} = \langle 1, -2, 1 \rangle \) yields 0: \( \mathbf{d} \cdot \mathbf{n} = 2 \cdot 1 + 3 \cdot (-2) + 4 \cdot 1 = 0 \).
When the dot product is zero, it confirms that the direction vector and the normal vector are perpendicular. Thus, the line is parallel to the plane, as the direction of the line (represented by its direction vector) does not "pierce" through the plane.

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Most popular questions from this chapter

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