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Find a point \(P\) on the line and a vector \(\mathbf{v}\) parallel to the line by inspection. $$ \begin{array}{l}{\text { (a) } x \mathbf{i}+y \mathbf{j}=(2 \mathbf{i}-\mathbf{j})+t(4 \mathbf{i}-\mathbf{j})} \\ {\text { (b) }\langle x, y, z\rangle=\langle- 1,2,4\rangle+ t\langle 5,7,-8\rangle}\end{array} $$

Short Answer

Expert verified
(a) Point P: (2, -1), Vector v: (4, -1); (b) Point P: (-1, 2, 4), Vector v: (5, 7, -8).

Step by step solution

01

Identify the point P and vector v for part (a)

For part (a), the given line equation is \( x \mathbf{i} + y \mathbf{j} = (2 \mathbf{i} - \mathbf{j}) + t(4 \mathbf{i} - \mathbf{j}) \).From the equation, the point \( P \) on the line is the constant part \((2 \mathbf{i} - \mathbf{j})\), which corresponds to the point \((2, -1)\).The vector \( \mathbf{v} \) parallel to the line is the coefficient of \( t \), which is \((4 \mathbf{i} - \mathbf{j})\), translating to \((4, -1)\).
02

Identify the point P and vector v for part (b)

For part (b), the given line equation is \( \langle x, y, z \rangle = \langle -1, 2, 4 \rangle + t\langle 5, 7, -8 \rangle \).The point \( P \) on the line is the constant vector \( \langle -1, 2, 4 \rangle \), which corresponds to the point \((-1, 2, 4)\).The vector \( \mathbf{v} \) parallel to the line is the coefficient of \( t \), which is \( \langle 5, 7, -8 \rangle \).

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Key Concepts

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

Points on a Line
When working with vector equations of lines, identifying points on the line is an essential step that helps visualize how the line behaves in space. A point on a line in vector form is typically represented as a fixed part of the equation, often seen as a constant or starting vector. For example, in the vector equation \( x \mathbf{i} + y \mathbf{j} = (2 \mathbf{i} - \mathbf{j}) + t(4 \mathbf{i} - \mathbf{j}) \), the point on the line is indicated by the constant portion \( (2 \mathbf{i} - \mathbf{j}) \), which can be rewritten as the coordinate \( (2, -1) \) in two-dimensional space.

In part (b) of the exercise, the three-dimensional vector equation \( \langle x, y, z \rangle = \langle -1, 2, 4 \rangle + t\langle 5, 7, -8 \rangle \) shows that the point on the line is \( \langle -1, 2, 4 \rangle \). This tells you that before any movement determined by the parameter \( t \), the line passes through the point \( (-1, 2, 4) \) in three-dimensional space.

Points like these are crucial since they anchor the line and give a specific location in space from where the line extends. Recognizing this point helps in mapping out the line's path efficiently.
Direction Vectors
Direction vectors play a key role in defining lines in vector equations. They instruct us on the path and orientation of the line without specifying a length because the parameter \( t \) can stretch it indefinitely.

In vector form, these are the vectors that multiply the parameter \( t \), indicating direction. For instance, in part (a) of the exercise:
  • The given vector equation is \( x \mathbf{i} + y \mathbf{j} = (2 \mathbf{i} - \mathbf{j}) + t(4 \mathbf{i} - \mathbf{j}) \).
  • The vector \( (4 \mathbf{i} - \mathbf{j}) \) acts as the direction vector, suggesting that the line moves 4 units along \( \mathbf{i} \) and -1 unit along \( \mathbf{j} \) as \( t \) varies.
Similarly, in part (b) with the equation \( \langle x, y, z \rangle = \langle -1, 2, 4 \rangle + t \langle 5, 7, -8 \rangle \), the direction vector \( \langle 5, 7, -8 \rangle \) signifies the directions in all three axes.

Direction vectors are valuable as they provide not only an orientation but also a potential infinite extension of the line as \( t \) changes. This concept allows us to navigate and understand vector spaces more intuitively.
Three-Dimensional Vectors
Switching from two to three dimensions in vector equations adds depth to the representation of lines in space. In three-dimensional vector equations, the variables represent coordinates along the \( x \), \( y \), and \( z \) axes.

The vector form \( \langle x, y, z \rangle = \langle a, b, c \rangle + t \langle d, e, f \rangle \) is quite similar to its two-dimensional counterpart but involves an additional dimension. It combines a fixed point \( \langle a, b, c \rangle \), indicating its position in 3D space, and a direction vector \( \langle d, e, f \rangle \) that shows the line's path.

In part (b) of the exercise, the vector equation \( \langle x, y, z \rangle = \langle -1, 2, 4 \rangle + t \langle 5, 7, -8 \rangle \) can be well-understood if you think of the line as stretching through 3D space starting at \( (-1, 2, 4) \) and moving in the direction specified by the vector \( (5, 7, -8) \).

Three-dimensional vectors empower us to model and work within real-world scenarios that involve height, width, and depth, allowing for a complete spatial comprehension. Understanding these vectors is fundamental for studying physics, engineering, and any field that operates in multidimensional spaces.

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