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Chapter 17: Problem 73

Find parametric equations for the line. The line through the point (-4,2,3) and parallel to a line in the \(y z\) -plane which makes a \(45^{\circ}\) angle with the positive \(y\) -axis and the positive \(z\) -axis.

Short Answer

Expert verified
Parametric equations: \( x = -4 \), \( y = 2 + \frac{\sqrt{2}}{2}t \), \( z = 3 + \frac{\sqrt{2}}{2}t \).

Step by step solution

01

Understand the Direction in the yz-plane

Since the line makes a 45掳 angle with both the positive y and z axes, we construct a direction vector in the yz-plane. Recall that the cosine of 45掳 is \( \frac{\sqrt{2}}{2} \). The direction vector parallel to the yz-plane is \( \mathbf{d}_{yz} = (0, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) \).
02

Apply the Direction to 3D

Since the direction of the line should be parallel to the yz-plane with no x-component, the full direction vector in 3D is \( \mathbf{d} = (0, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) \).
03

Set an Initial Point and Construct Parametric Equations

Given point on the line is P(-4, 2, 3). The parametric equations of the line use this point and the direction vector:\[ x = -4 \]\[ y = 2 + \frac{\sqrt{2}}{2}t \]\[ z = 3 + \frac{\sqrt{2}}{2}t \] where \( t \) is a parameter.

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Key Concepts

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

Direction Vectors
Direction vectors are essential in defining the orientation of a line or a line segment in space. They are often used in geometry and physics to describe motion and orientation. A direction vector essentially tells us which way the line is pointing.

In our scenario, a line is specified to be parallel to the yz-plane and also makes a 45掳 angle with both the positive y-axis and z-axis. To develop a direction vector, we analyze the components along each axis:
  • If a line is parallel to the yz-plane, its x-component is zero. This tells us that there is no movement along the x-axis.
  • Both the y- and z-components are equal when measured at a 45掳 angle to their respective axes, due to the symmetry of this angle. The cosine of 45掳 is \( \frac{\sqrt{2}}{2} \), giving us a vector of \((0, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\).
This direction vector describes the orientation of our line in 3D space with clarity and precision.
Angles in Geometry
Angles are a fundamental concept in geometry, defining the rotation needed to align one direction with another. They are crucial in establishing orientation and constructing direction vectors.

When a problem states a line makes a specific angle with the axes, it implies a particular direction of orienting the vector components. In our exercise, the line forms a 45掳 angle with both the y-axis and the z-axis. This specific angle helps in understanding the transformation of coordinate components into vectors:
  • A 45掳 angle means the y- and z-components of the vector must be equivalent due to equal cosine values, i.e., \( \cos(45掳) = \frac{\sqrt{2}}{2} \).
  • Such angles lead us to create symmetric and balanced vector components when orienting our direction vector in the yz-plane.
These geometric concepts of angles provide the foundation for creating precisely oriented lines in any dimensional space.
3D Coordinate Geometry
In 3D coordinate geometry, every point and line has a clear position in space through coordinates and direction vectors. It's a mathematical language used to describe points and lines in space spanning across three dimensions (x, y, and z).

Our exercise requested the construction of a line moving through a specific point, \((-4, 2, 3)\). Incorporating the direction from the yz-plane, we define its path using parametric equations:
  • The starting point is denoted by our coordinates \((-4, 2, 3)\). This acts as a reference or anchor from which the direction vector projects.
  • The direction vector \((0, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\) advances us along the line constantly as we increase or decrease the parameter \(t\).
Thus, we form the parametric equations:
- \( x = -4 \)
- \( y = 2 + \frac{\sqrt{2}}{2}t \)
- \( z = 3 + \frac{\sqrt{2}}{2}t \)
Each equation describes movement along its respective axis, based on the direction taken from a given point in 3D space.

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Most popular questions from this chapter

Find \(c\) so that one revolution about the \(z\) axis of the helix gives an increase of \(\Delta z\) in the \(z\) -coordinate. $$x=2 \cos t, y=2 \sin t, z=c t, \Delta z=15$$

Explain what is wrong with the statement. When a particle moves around a circle its velocity and acceleration are always orthogonal.

Suppose \(\vec{r}(t)=\cos t \vec{i}+\sin t \vec{j}+2 t \vec{k}\) represents the position of a particle on a helix, where \(z\) is the height of the particle above the ground. (a) Is the particle ever moving downward? When? (b) When does the particle reach a point 10 units above the ground? (c) What is the velocity of the particle when it is 10 units above the ground? (d) When it is 10 units above the ground, the particle leaves the helix and moves along the tangent. Find parametric equations for this tangent line.

Are the statements true or false? Give reasons for your answer. The equation \(\vec{r}(t)=3 t \vec{i}+(6 t+1) \vec{j}\) parameterizes a line.

Sketch the vector field and its flow. $$\vec{v}=3 \vec{i}-2 \vec{j}$$
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