All lines are in the \((x, y)\) plane. Write the equation of the straight line through (2,-3) with slope \(3 / 4,\) in the parametric form \(\mathbf{r}=\mathbf{r}_{0}+\mathbf{A} t\)

Short Answer

Expert verified
\mathbf{r} = \begin{pmatrix} 2 \ -3 \end{pmatrix} + t \begin{pmatrix} 4 \ 3 \end{pmatrix}\.

Step by step solution

01

Identify components

Identify the given components: the point \(2, -3\) as the position vector \mathbf{r}_{0}\ and the slope \ \frac{3}{4} \ which will help define the direction vector.
02

Form the position vector

The position vector \ \mathbf{r}_{0} \ is given by \ \[ \mathbf{r}_{0} = \begin{pmatrix} 2 \ -3 \end{pmatrix} \] .
03

Find the direction vector

The slope \frac{3}{4}\ means for every 4 units moved in the x-direction, the y-direction changes by 3. Hence, the direction vector \mathbf{A}\ is \[ \mathbf{A} = \begin{pmatrix} 4 \ 3 \end{pmatrix} \].
04

Write the parametric equation

Using the formula \mathbf{r} = \mathbf{r}_{0} + \mathbf{A}t\, substitute the position vector and direction vector: \ \[ \mathbf{r} = \begin{pmatrix} 2 \ -3 \end{pmatrix} + t \begin{pmatrix} 4 \ 3 \end{pmatrix} \].

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Key Concepts

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

Position Vector
A position vector is a vector that points from the origin of the coordinate system to a specific point. It essentially indicates the location of that point in space. In the context of our exercise, the position vector \(\backslash\text{\( \mathbf{r}_{0} \)}\) is given by the point \((2, -3)\). This means \(\backslash\text{\( \mathbf{r}_{0} = \begin{pmatrix} 2 & -3 \end{pmatrix} \)}\). This vector tells us exactly where on the x-y plane our line passes through.
Understanding the position vector is crucial because it represents a fixed point that our line will always go through.
In the parametric equation of a line, \(\backslash\text{\( \mathbf{r} = \mathbf{r}_{0} + \mathbf{A}t \)}\), the term \(\backslash\text{\( \mathbf{r}_{0} \)}\) anchors the line to a specific location in the coordinate system.
Direction Vector
The direction vector indicates the direction in which a line moves from any given point. For our line, the direction vector \(\backslash\text{\( \mathbf{A} \)}\) can be derived from the slope. If the slope given is \(\backslashfrac{3}{4} \), it tells us that for every 4 units we move in the x-direction, the y-direction changes by 3 units. Therefore, the direction vector is \(\backslash\text{\( \mathbf{A} = \begin{pmatrix} 4 & 3 \end{pmatrix} \)}\). This vector points in the direction our line travels.
When dealing with lines, the direction vector is important because it defines the 'path' or 'route' of the line. It guides the line's slope and ensures it follows the correct trajectory. In the parametric equation \(\backslash\text{\( \mathbf{r} = \mathbf{r}_{0} + \mathbf{A}t \)}\), \(\backslash\text{\( \mathbf{A} \)}\) dictates how the line extends from the fixed point (the position vector).
Slope
The slope of a line describes how steep the line is. It indicates the ratio of the vertical change to the horizontal change between two points on a line. Given the slope \(\backslashfrac{3}{4} \), it means that for every 4 units moved horizontally (in the x-direction), the vertical movement (in the y-direction) is 3 units.
Slopes are fundamental in trigonometry and geometry as they provide a clear measure of a line's steepness and direction. When we convert slope into a direction vector like \(\backslash\text{\( \mathbf{A} = \begin{pmatrix} 4 & 3 \end{pmatrix} \)}\), it helps us define the line's trajectory in a more tangible form.
In our parametric equation \(\backslash\text{\( \mathbf{r} = \mathbf{r}_{0} + \mathbf{A}t \)}\), incorporating the slope as part of the direction vector ensures that the line constructed has the correct incline relative to the coordinate system.

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Most popular questions from this chapter

(a) Write the equations of a straight line through the points (2,7,-1) and (5,7,3) (b) Find the equation of the plane determined by the two lines \(\mathbf{r}=(\mathbf{i}-2 \mathbf{j}+\mathbf{k}) t\) and \(\mathbf{r}=(6 \mathbf{i}-3 \mathbf{j}+2 \mathbf{k}) t\) (c) Find the angle which the line in (a) makes with the plane in (b). (d) Find the distance from (1,1,1) to the plane in \((\mathrm{b})\) (e) Find the distance from (1,6,-3) to the line in (a).

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