Problem 29
Question
In Problems 29-34, let $$ \mathbf{u}=\left[\begin{array}{l} 3 \\ 4 \end{array}\right], \quad \mathbf{v}=\left[\begin{array}{l} -1 \\ -2 \end{array}\right], \quad \text { and } \quad \mathbf{w}=\left[\begin{array}{r} 1 \\ -2 \end{array}\right] $$ $$ \text { Compute } \mathbf{u}+\mathbf{v} \text { and illustrate the result graphically. } $$
Step-by-Step Solution
Verified Answer
\( \mathbf{u} + \mathbf{v} = \left[ \begin{array}{c} 2 \\ 2 \end{array} \right] \). Graphically, it is a vector from the origin to (2, 2).
1Step 1: Add the Vectors
To find \( \mathbf{u} + \mathbf{v} \), we add the corresponding components of \( \mathbf{u} \) and \( \mathbf{v} \). That means, add the first component of \( \mathbf{u} \) to the first component of \( \mathbf{v} \), and then add the second component of \( \mathbf{u} \) to the second component of \( \mathbf{v} \). Mathematically:\[\mathbf{u} + \mathbf{v} = \left[\begin{array}{c} 3 \ 4 \end{array}\right] + \left[\begin{array}{c} -1 \ -2 \end{array}\right] = \left[\begin{array}{c} 3 + (-1) \ 4 + (-2) \end{array}\right] = \left[\begin{array}{c} 2 \ 2 \end{array}\right]\]
2Step 2: Graphical Representation
To illustrate the result graphically, plot the vectors \( \mathbf{u} \), \( \mathbf{v} \), and their resultant \( \mathbf{u} + \mathbf{v} \) on an XY plane. Start \( \mathbf{u} \) at the origin and draw a vector to \((3, 4)\). Then, starting at \((3, 4)\), draw the vector \( \mathbf{v} \) falling to \((2, 2)\). The resultant vector \( \mathbf{u} + \mathbf{v} \) connects the origin directly to \((2, 2)\). This shows the tail-to-tip method and confirms that \( \mathbf{u} + \mathbf{v} = \left[ \begin{array}{c} 2 \ 2 \end{array} \right] \).
Key Concepts
Graphical Representation of VectorsTail-to-Tip MethodComponent-Wise Addition
Graphical Representation of Vectors
When working with vectors, visualizing them graphically can greatly enhance understanding. Imagine the XY plane as a blank canvas where every vector can be drawn as an arrow. The origin, point (0,0), is the starting point for our arrows.
For vector \( \mathbf{u} \), represented by \( [3, 4] \), we start drawing an arrow from the origin to the point (3,4). This arrow illustrates the direction and magnitude of \( \mathbf{u} \). Similarly, vector \( \mathbf{v} \) begins at the origin and points towards (-1, -2). However, when performing vector addition, we don't place every vector at the origin.
Instead, we utilize the tail-to-tip method. After plotting \( \mathbf{u} \), move to the tip of \( \mathbf{u} \) at (3,4) and start \( \mathbf{v} \) from there. This stepwise connection of vectors provides clear visual evidence of the operation and helps conclude the resultant vector \( \mathbf{u} + \mathbf{v} \) with ease, which in this exercise is \( [2, 2] \). By using graphical techniques alongside arithmetic solutions, we reinforce understanding with every vector operation.
For vector \( \mathbf{u} \), represented by \( [3, 4] \), we start drawing an arrow from the origin to the point (3,4). This arrow illustrates the direction and magnitude of \( \mathbf{u} \). Similarly, vector \( \mathbf{v} \) begins at the origin and points towards (-1, -2). However, when performing vector addition, we don't place every vector at the origin.
Instead, we utilize the tail-to-tip method. After plotting \( \mathbf{u} \), move to the tip of \( \mathbf{u} \) at (3,4) and start \( \mathbf{v} \) from there. This stepwise connection of vectors provides clear visual evidence of the operation and helps conclude the resultant vector \( \mathbf{u} + \mathbf{v} \) with ease, which in this exercise is \( [2, 2] \). By using graphical techniques alongside arithmetic solutions, we reinforce understanding with every vector operation.
Tail-to-Tip Method
The tail-to-tip method is a straightforward and valuable way to add vectors visually. It emphasizes the continuous path from one vector to the next, by connecting them in order.
To apply this method, start with the first vector, drawing it from the origin to its specific coordinates. When the first vector ends, the second vector begins, with its tail at the tip of the first.
Consider \( \mathbf{u} + \mathbf{v} \). Draw \( \mathbf{u} \) from (0,0) to (3,4), then draw \( \mathbf{v} \) from (3,4) to (2,2). This continuous line from the origin to \( (2, 2) \) represents the resultant vector. This method not only aids in visualizing results but also provides a geometric understanding of vector operations.
To apply this method, start with the first vector, drawing it from the origin to its specific coordinates. When the first vector ends, the second vector begins, with its tail at the tip of the first.
Consider \( \mathbf{u} + \mathbf{v} \). Draw \( \mathbf{u} \) from (0,0) to (3,4), then draw \( \mathbf{v} \) from (3,4) to (2,2). This continuous line from the origin to \( (2, 2) \) represents the resultant vector. This method not only aids in visualizing results but also provides a geometric understanding of vector operations.
Component-Wise Addition
Component-wise addition is a basic but essential skill in vector mathematics. It involves adding corresponding components of two vectors, thus simplifying vector addition.
For vectors \( \mathbf{u} = [3, 4] \) and \( \mathbf{v} = [-1, -2] \), the process involves simple arithmetic:
For vectors \( \mathbf{u} = [3, 4] \) and \( \mathbf{v} = [-1, -2] \), the process involves simple arithmetic:
- First component: \( 3 + (-1) = 2 \)
- Second component: \( 4 + (-2) = 2 \)
Other exercises in this chapter
Problem 29
Suppose that \(A\) is a \(4 \times 3\) matrix, \(B\) is a \(1 \times 3\) matrix, \(C\) is a \(3 \times 1\) matrix, and \(D\) is a \(4 \times 3\) matrix. Which o
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In Problems 29-34, determine whether each system is overdetermined or underdetermined; then solve each system. $$ \begin{array}{r} x-2 y+z=3 \\ 2 x-3 y+z=8 \end
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Let \(\mathbf{x}=[2,0,-1]^{\prime}\). Find \(\mathbf{y}\) so that \(\mathbf{x}\) and \(\mathbf{y}\) are perpendicular.
View solution Problem 30
In Problems , let $$ \mathbf{u}=\left[\begin{array}{l} 3 \\ 4 \end{array}\right], \quad \mathbf{v}=\left[\begin{array}{l} -1 \\ -2 \end{array}\right], \quad \te
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