Problem 20
Question
Find \(\mathbf{x}+\mathbf{y}\) for each pair of vectors \(\mathbf{x}\) and \(\mathbf{v}\). Represent \(\mathrm{x}, \mathbf{y}\), and \(\mathrm{x}+\mathrm{y}\) in the plane, and explain graphically how you add \(\mathbf{x}\) and \(\mathbf{y}\). \(\mathbf{x}=\left[\begin{array}{l}-1 \\ -1\end{array}\right]\) and \(\mathbf{y}=\left[\begin{array}{l}1 \\ 2\end{array}\right]\)
Step-by-Step Solution
Verified Answer
The sum of the vectors \( \mathbf{x} + \mathbf{y} \) is \( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \). Draw \( \mathbf{x} \) and \( \mathbf{y} \) and use the tip-to-tail method to visualize the vector addition.
1Step 1: Understand the Vectors
Begin by identifying the given vectors. Here, we have two vectors: \( \mathbf{x} = \begin{bmatrix} -1 \ -1 \end{bmatrix} \) and \( \mathbf{y} = \begin{bmatrix} 1 \ 2 \end{bmatrix} \). Each vector has two components, corresponding to their x and y coordinates.
2Step 2: Add the Vectors Algebraically
To find \( \mathbf{x} + \mathbf{y} \), add the corresponding components of the vectors. So, the x-components are \(-1 + 1 = 0\), and the y-components are \(-1 + 2 = 1\). Therefore, \( \mathbf{x} + \mathbf{y} = \begin{bmatrix} 0 \ 1 \end{bmatrix} \).
3Step 3: Graphical Representation of Vectors
Draw the initial point of \( \mathbf{x} \) at the origin (0,0) and point towards (-1,-1). Then plot \( \mathbf{y} \) starting at the origin as well, pointing towards (1,2).
4Step 4: Visualizing the Vector Addition
Place the tail of \( \mathbf{y} \) at the head of \( \mathbf{x} \), which moves vector \( \mathbf{y} \) to be positioned with origin at (-1,-1). The resultant vector \( \mathbf{x} + \mathbf{y} \) is drawn from the tail of \( \mathbf{x} \) at the origin to the new head of \( \mathbf{y} \), ending at (0,1). This graphical method is known as the "tip-to-tail" method.
Key Concepts
Algebraic Addition of VectorsGraphical Representation of VectorsTip-to-Tail Method
Algebraic Addition of Vectors
Adding vectors algebraically involves straightforward addition of the corresponding components from each vector, like adding integers. Vectors are often represented in component form, \( \mathbf{x} = \begin{bmatrix} x_1 \ y_1 \end{bmatrix} \) and \( \mathbf{y} = \begin{bmatrix} x_2 \ y_2 \end{bmatrix} \), making it simple to compute their sum.
- Step 1: Identify the components of each vector.
- Step 2: Add the x-components of the vectors (i.e., \(x_1 + x_2\)).
- Step 3: Add the y-components of the vectors (i.e., \(y_1 + y_2\)).
Graphical Representation of Vectors
Visualizing vectors on a coordinate plane offers a clear picture of how vectors can be added. When you draw vectors, you represent their magnitude and direction clearly. For \( \mathbf{x} = \begin{bmatrix} -1 \ -1 \end{bmatrix} \), one would draw an arrow starting at the origin (0,0) and ending at the point (-1,-1). Similarly, for \( \mathbf{y} = \begin{bmatrix} 1 \ 2 \end{bmatrix} \), the arrow starts at (0,0) and points to (1,2).
- Consistent Scaling: Ensure each unit on the axes represents the same length for accuracy.
- Arrowheads: Show the direction of the vector.
Tip-to-Tail Method
The "tip-to-tail" method is a fundamental approach for vector addition. It involves placing the tail of one vector at the tip (head) of another. This visual technique aids in understanding how vectors combine to form a resultant vector.
- Step 1: Begin by plotting \( \mathbf{x} \) from the origin.
- Step 2: Move \( \mathbf{y} \) such that its tail begins at the head of \( \mathbf{x} \).
- Step 3: Draw the resultant vector from the origin to the head of \( \mathbf{y} \).
Other exercises in this chapter
Problem 19
In Problems 19-24, solve each system of linear equations. $$ \begin{array}{r} 2 x-3 y+z=-1 \\ x+y-2 z=-3 \\ 3 x-2 y+z=2 \end{array} $$
View solution Problem 20
Use the dot product to compute the length of \([1,1,3]\) '.
View solution Problem 20
Suppose that \(A\) is an \(m \times k\) matrix and \(B\) is a \(k \times n\) matrix. Show that $$ (A B)^{\prime}=B^{\prime} A^{\prime} $$
View solution Problem 20
In Problems 19-24, solve each system of linear equations. $$ \begin{aligned} 5 x-y+2 z &=6 \\ x+2 y-z &=-1 \\ 3 x+2 y-2 z &=1 \end{aligned} $$
View solution