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Chapter 1: Problem 29

For the following exercises, point \(P\) and vector \(\mathbf{v}\) are given. Let \(L\) be the line passing through point \(P\) with direction \(\mathbf{v}\). Find parametric equations of line \(L\). Find symmetric equations of line \(L\). Find the intersection of the line with the \(x y\) -plane. \(P(1,-2,3), \mathbf{v}=\langle 1,2,3\rangle\)

Short Answer

Expert verified
Parametric equations: \(x = 1 + t\), \(y = -2 + 2t\), \(z = 3 + 3t\). Symmetric equations: \(x - 1 = \frac{y + 2}{2} = \frac{z - 3}{3}\). Intersection with \(xy\)-plane: \((0, -4, 0)\).

Step by step solution

01

Identify the Components

We have a point \(P(1,-2,3)\) and a direction vector \(\mathbf{v}=\langle 1,2,3\rangle\). To define the line, we'll use these components to form parametric equations.
02

Formulate the Parametric Equations

The parametric equations of a line involve using a parameter \(t\) such that \(x = x_0 + at\), \(y = y_0 + bt\), and \(z = z_0 + ct\), where \((x_0, y_0, z_0)\) is the point and \(\langle a, b, c \rangle\) is the direction vector. For point \(P(1,-2,3)\) and vector \(\mathbf{v}=\langle 1,2,3\rangle\), the equations are: \[x = 1 + t,\quad y = -2 + 2t,\quad z = 3 + 3t\].
03

Convert to Symmetric Equations

The symmetric equations remove the parameter \(t\). From the parametric equations, we solve each for \(t\): \(t = x - 1 = \frac{y + 2}{2} = \frac{z - 3}{3}\). So, the symmetric equations are: \[x - 1 = \frac{y + 2}{2} = \frac{z - 3}{3}\].
04

Find Intersection with the xy-plane

The intersection with the \(xy\)-plane occurs when \(z = 0\). Set \(3 + 3t = 0\) to solve for \(t\): \[3t = -3 \quad \Rightarrow \quad t = -1\]. Substituting \(t = -1\) into the parametric equations, we find \(x = 1 + (-1) = 0\) and \(y = -2 + 2(-1) = -4\). Thus, the intersection point is \((0, -4, 0)\).

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Key Concepts

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

Symmetric Equations
Symmetric equations provide a concise way to express a line by eliminating the parameter \( t \). By removing \( t \), we can directly relate the coordinates \( x, y, \) and \( z \) to each other. This is incredibly useful when we want a clear, direct relationship between the three variables.
  • The symmetric form usually arises by expressing \( t \) from each parametric equation: solving for \( t \) in each case.
  • For this exercise, the parametric equations are \( x = 1 + t \), \( y = -2 + 2t \), and \( z = 3 + 3t \).
  • Solving for \( t \) in these equations, we get: \( t = x - 1 \), \( t = \frac{y + 2}{2} \), \( t = \frac{z - 3}{3} \).
  • We then set these equations equal to each other: \[ x - 1 = \frac{y + 2}{2} = \frac{z - 3}{3} \]
This format of the equation offers a straightforward method to trace the point of intersection or alignment of the coordinates in three-dimensional space.
Intersection with the xy-plane
Finding the intersection of a line with the \(xy\)-plane allows us to understand where the line crosses this typical two-dimensional plane within the three-dimensional coordinate system.
  • The \(xy\)-plane is defined by all points where \( z = 0 \).
  • The parametric equation \( z = 3 + 3t \) will help us set \( z = 0 \) and then solve for \( t \).
  • Solving \( 3 + 3t = 0 \) results in \( t = -1 \).
  • This value of \( t \) is then substituted back into the remaining parametric equations: \( x = 1 + (-1) = 0 \) and \( y = -2 + 2(-1) = -4 \).
  • Hence, these calculations reveal the intersection point of the line with the \(xy\)-plane: \( (0, -4, 0) \).
Understanding intersections plays a crucial role in visualizing how geometric entities interact within plane boundaries.
Direction Vector
A direction vector is an essential component in vector equations, guiding the path along which a line travels. This vector indicates the line's direction in three-dimensional space.
  • The direction vector here is \( \mathbf{v}=\langle 1, 2, 3 \rangle \).
  • Each component of the vector \( \langle a, b, c \rangle \) informs the directional movement in the \( x \)-, \( y \)-, and \( z \)-directions, respectively.
  • For our line, \( 1 \) unit in the \( x \)-direction, \( 2 \) units in the \( y \)-direction, and \( 3 \) units in the \( z \)-direction orchestrate the line's path.
  • This vector works in tandem with the point \( P(1, -2, 3) \) to create the parametric equations that define the line.
Direction vectors are fundamental as they determine the precise way a line makes its journey through space, impacting everything from collision detection to path tracing in computational geometry applications.

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Most popular questions from this chapter

The equation of a quadric surface is given. Use the method of completing the square to write the equation in standard form. Identify the surface. $$ 4 x^{2}-y^{2}+z^{2}-8 x+2 y+2 z+3=0 $$

Find the equation of the quadric surface with points \(P(x, y, z)\) that are equidistant from point \(Q(0,-1,0)\) and plane of equation \(y=1\). Identify the surface.

The intersection between cylinder \((x-1)^{2}+y^{2}=1\) and sphere \(x^{2}+y^{2}+z^{2}=4\) is called a Viviani curve. a. Solve the system consisting of the equations of the surfaces to find the equation of the intersection curve. (Hint: Find \(x\) and \(y\) in terms of \(z .)\) b. Use a computer algebra system (CAS) to visualize the intersection curve on sphere \(x^{2}+y^{2}+z^{2}=4\).

For the following exercises, find the traces for the surfaces in planes \(x=k, y=k\), and \(z=k .\) Then, describe and draw the surfaces\(x^{2}=y^{2}+z^{2}\)

Consider \(\mathbf{r}(t)=\langle\cos t, \sin t, 2 t\rangle\) the position vector of a particle at time \(t \in[0,30]\), where the components of are expressed in centimeters and time in seconds. Let \(\overrightarrow{O P}\) be the position vector of the particle after \(1 \mathrm{sec}\). a. Show that all vectors \(\overrightarrow{P Q}\), where \(Q(x, y, z)\) is an arbitrary point, orthogonal to the instantaneous velocity vector \(\mathbf{v}(1)\) of the particle after \(1 \mathrm{sec}\), can be expressed as \(\overrightarrow{P Q}=\langle x-\cos 1, y-\sin 1, z-2\rangle\), where \(x \sin 1-y \cos 1-2 z+4=0 .\) The set of point \(Q\) describes a plane called the normal plane to the path of the particle at point \(P\) b. Use a CAS to visualize the instantaneous velocity vector and the normal plane at point \(P\) along with the path of the particle.
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