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Chapter 8: Problem 4

A line passes through the points \(A(2,1)\) and \(B(-3,5) .\) Write two different vector equations for this line.

Short Answer

Expert verified
Two vector equations are: \(\mathbf{r} = \langle 2, 1 \rangle + t\langle -5, 4 \rangle\) and \(\mathbf{r} = \langle -3, 5 \rangle + t\langle -5, 4 \rangle\).

Step by step solution

01

Understanding the vector equation format

The vector equation of a line passing through a point \(P(x_0, y_0)\) with direction vector \(\mathbf{d} = \langle a, b \rangle\) is given by \(\mathbf{r} = \mathbf{r}_0 + t\mathbf{d}\), where \(\mathbf{r}_0\) is the position vector of point \(P\), and \(t\) is a scalar.
02

Finding the direction vector of the line

To find the direction vector \(\mathbf{d}\), subtract the coordinates of point \(A(2,1)\) from point \(B(-3,5)\). The direction vector is \(\mathbf{d} = \langle -3 - 2, 5 - 1 \rangle = \langle -5, 4 \rangle\).
03

Writing the first vector equation using point A

Using point \(A(2,1)\) and the direction vector \(\langle -5, 4 \rangle\), the vector equation is \(\mathbf{r} = \langle 2, 1 \rangle + t\langle -5, 4 \rangle\).
04

Writing the second vector equation using point B

Alternatively, using point \(B(-3,5)\) and the same direction vector \(\langle -5, 4 \rangle\), the vector equation is \(\mathbf{r} = \langle -3, 5 \rangle + t\langle -5, 4 \rangle\).

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Key Concepts

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

Direction Vector
The direction vector is a crucial concept when dealing with vector equations of a line. It essentially tells you the direction in which the line moves. Consider it a set of instructions for moving along the line, starting from any point on that line. In mathematical terms, the direction vector \( \mathbf{d} = \langle a, b \rangle \) is derived by subtracting the coordinates of one point on the line from another. This vector shows how much the line goes in the horizontal and vertical directions.
  • For the line passing through points \( A(2,1) \) and \( B(-3,5) \), the direction vector was found by \( \langle -3 - 2, 5 - 1 \rangle = \langle -5, 4 \rangle \).
  • This means to move from any point on the line, you move 5 units left horizontally and 4 units up vertically.
Hence, the direction vector helps you define how the line traverses through the plane.
Position Vector
A position vector is used to denote the position of a point relative to the origin of a coordinate plane. It's like pointing from the center of the universe (origin) to where you stand (a point on the line). This vector is typically represented as \( \mathbf{r}_0 = \langle x_0, y_0 \rangle \), where \( x_0 \) and \( y_0 \) are the coordinates of the point you're referencing.
For example, in our exercises:
  • Using point A(2,1), the position vector is \( \mathbf{r}_0 = \langle 2, 1 \rangle \).
  • Or using point B(-3,5), it becomes \( \mathbf{r}_0 = \langle -3, 5 \rangle \).
Position vectors ensure that we have a starting point for moving along the direction vector to trace the entire line.
Scalar
A scalar is a simple number that stands alone, without any direction. In the context of vector equations, it acts as a multiplier. When you see \( t \) in a vector equation, that is your scalar. Here's how it works: the scalar multiplies the direction vector to give you different points on the line.
  • The scalar \( t \) can be any real number, allowing the line to stretch infinitely in both directions along the direction vector.
  • For example, for \( \mathbf{r} = \mathbf{r}_0 + t\mathbf{d} \), changing \( t \) allows you to compute every possible point on the line.
  • A unique \( t \) for every point ensures that no matter how far you extend it, you'll find the position on the line.
Think of \( t \) as the variable time; as it increases or decreases, it propels the point along the line indefinitely.
Vector Equation of a Line
The vector equation of a line combines all our learned concepts: the position vector, direction vector, and scalar. It's similar to a recipe鈥攜ou need all the ingredients to make the whole dish. This equation is expressed as \( \mathbf{r} = \mathbf{r}_0 + t\mathbf{d} \), where:
  • \( \mathbf{r} \) is any point on the line.
  • \( \mathbf{r}_0 \) is your starting position vector.
  • \( t \) is the scalar that adjusts how far you go along the line.
  • \( \mathbf{d} \) is the direction vector, guiding your movement.
By substituting different values into this equation, you can describe every point the line passes through.
Using points A and B in our example resulted in two vector equations:
  • \( \mathbf{r} = \langle 2, 1 \rangle + t\langle -5, 4 \rangle \)
  • \( \mathbf{r} = \langle -3, 5 \rangle + t\langle -5, 4 \rangle \)
Each original point used creates a backbone for the infinite possibilities traced by the line.

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Most popular questions from this chapter

For the line \(L: \vec{r}=(1,-5)+s(3,5), s \in \mathbf{R},\) determine the following: a. an equation for the line perpendicular to \(L,\) passing through \(P(2,0)\) b. the point at which the line in part a. intersects the \(y\) -axis

Determine the Cartesian equation of the plane that passes through the points (1,4,5) and (3,2,1) and is perpendicular to the plane \(2 x-y+z-1=0\)

State the coordinates of a point on each of the given lines. a. \(\vec{r}=(-3,1,8)+s(-1,1,9), s \in \mathbf{R}\) b. \(\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-1}\) c. \(x=-2+3 t, y=1+(-4 t), z=3-t, t \in \mathbf{R}\) d. \(\frac{x+2}{-1}=\frac{z-1}{2}, y=-3\) e. \(x=3, y=-2, z=-1+2 k, k \in \mathbf{R}\) f. \(\frac{x-\frac{1}{3}}{\frac{1}{2}}=\frac{y+\frac{3}{4}}{\frac{-1}{4}}=\frac{z-\frac{2}{5}}{\frac{1}{2}}\)

A plane has vector equation \(\vec{r}=(2,1,3)+s\left(\frac{1}{3},-2, \frac{3}{4}\right)+t(6,-12,30)\) \(s, t \in \mathbf{R}\) a. Express the first direction vector with only integers. b. Reduce the second direction vector. c. Write a new equation for the plane using the calculations from parts a. and b.

A line passes through the points \(A(-1,5,-4)\) and \(B(2,5,-4)\). a. Write a vector equation for the line containing these points. b. Write parametric equations corresponding to the vector equation you wrote in part a. c. Explain why there are no symmetric equations for this line.
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