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Chapter 10: Problem 10

In point form (as in Exercise 7 ). find an equation for the line through the point \((4,1,-3)\) and parallel to a line that has the direction vector \((-1,2,5)\)

Short Answer

Expert verified
The line equation is \( \vec{r} = (4 - t)\hat{i} + (1 + 2t)\hat{j} + (-3 + 5t)\hat{k} \).

Step by step solution

01

Understand the Line Equation

The equation for a line in three-dimensional space based on a point \((x_0, y_0, z_0)\) and a direction vector \((a, b, c)\) can be expressed as: \( \vec{r} = \vec{r_0} + t\vec{d} \). Here, \( \vec{r_0} \) is the position vector of the given point, \( \vec{d} \) is the direction vector, and \( t \) is a parameter.
02

Identify the Given Elements

From the problem, we are given a point \((4,1,-3)\) through which the line passes. We are also given a direction vector \((-1,2,5)\) to which the desired line is parallel.
03

Write the Position Vector

The position vector \(\vec{r_0}\) for the point \((4,1,-3)\) is \(4\hat{i} + 1\hat{j} - 3\hat{k}\).
04

Formulate the Line Equation

Plug the position vector and direction vector into the line equation: \[ \vec{r} = (4\hat{i} + 1\hat{j} - 3\hat{k}) + t(-1\hat{i} + 2\hat{j} + 5\hat{k}). \] Expand this to \[ \vec{r} = (4 - t)\hat{i} + (1 + 2t)\hat{j} + (-3 + 5t)\hat{k}. \]
05

Verify the Components

Ensure each component form follows the standard line equation form: \( x = x_0 + at, y = y_0 + bt, z = z_0 + ct \). Here, they are \( x = 4 - t \), \( y = 1 + 2t \), \( z = -3 + 5t \).

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Key Concepts

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

Direction Vector
In three-dimensional space, a direction vector defines the direction a line moves across the spatial plane. It's essentially the vector that helps us understand in which direction our line is heading. Given a direction vector, \((-1, 2, 5)\), for this example, it signifies how much the line moves in each of the x, y, and z components, respectively.
  • The first component, \(-1\), means the line moves one unit backwards or negatively along the x-axis.
  • The second component, \(+2\), indicates that the line moves two units positively along the y-axis.
  • The third component, \(+5\), shows a movement of five units forward along the z-axis.

Remember, the direction vector is the consistent line guide. It remains unchanged along the line's entirety, hence why it's called the direction vector.
Position Vector
The position vector is your starting point. It determines the exact point in space from which the line begins. It's like placing the tip of your pencil at a specific spot on a piece of paper, from which you will draw your line. For example, in the given problem, the position vector is derived from the point \(4, 1, -3)\).

This means that:
  • The line starts at \(x = 4\) on the x-axis.
  • It begins at \(y = 1\) on the y-axis.
  • And it marks its starting point at \(z = -3\) on the z-axis.
When writing it as a vector, it appears as \(4\hat{i} + 1\hat{j} - 3\hat{k}\). This vector pinpoints exactly where we begin to draw our line.
Parametric Equation of a Line
A parametric equation of a line in three dimensions is a powerful tool to describe every point on the line as a function of a parameter, usually denoted as \t\. Using both the direction vector and position vector, we can formulate this equation.
The formula is given by \( \vec{r} = \vec{r_0} + t\vec{d} \)
  • \( \vec{r_0} \) is your starting position vector (where the line starts). In our example, it's \(4\hat{i} + 1\hat{j} - 3\hat{k} \).
  • \( t \) is a parameter that can change (think of it as a time variable).
  • \( \vec{d} \) is the direction vector \((-1\hat{i} + 2\hat{j} + 5\hat{k}) \), which tells us the direction of the line.
When these elements are assembled into the parametric equation, they allow us to find any point on the line by simply changing the value of \t\. For example, substituting into the parametric equation yields component equations like:
  • \( x = 4 - t \)
  • \( y = 1 + 2t \)
  • \( z = -3 + 5t \)
These equations help you find each component of every point along the line as you vary \t\.

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Most popular questions from this chapter

Refer to formulas for Chanter 9 Find the exact volume of the solid that results when the triangular region with vertices at \((0,0),(5,0),\) and \((0,9)\) is rotated about the a) \(x\) axis. b) \(y\) axis.

Use the Distance Formula to show that the equation of the parabola with focus \(F(0,1)\) and directrix \(y=-1\) is $$y=\frac{1}{4} x^{2}$$

Use algebra to find the point of intersection of the two lines whose equations are provided. Use Example 7 as a guide. \(2 x+y=11\) and \(3 x+2 y=16\)

"Two intersecting lines are perpendicular if their direction vectors \(v_{1}=(a, b, c)\) and \(v_{2}=(d, e, f)\) satisfy the condition that ad \(+b e+c f=0\). Are these lines perpendicular? \(\ell_{1}:(x, y, z)=(1,-2,5)+n(4,1,-3)\) and \(\ell_{2}:(x, y, z)=(1,-2,5)+r(-3,6,-2)\)

The planes \(x+2 y+z=9\) and \(3 x+y-2 z=2\) intersect in the line \(\ell\) whose equation is \((x, y, z)=(-1,5,0)+n(1,-1,1)\) Find the point in both planes for which: a) \(x=7\) b) \(y=16\)
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