Problem 32
Question
In Problems , let $$ \mathbf{u}=\left[\begin{array}{l} 3 \\ 4 \end{array}\right], \quad \mathbf{v}=\left[\begin{array}{l} -1 \\ -2 \end{array}\right], \quad \text { and } \quad \mathbf{w}=\left[\begin{array}{r} 1 \\ -2 \end{array}\right] $$ $$ \text { Compute } \mathbf{v}-\frac{1}{2} \mathbf{u} \text { and illustrate the result graphically. } $$
Step-by-Step Solution
Verified Answer
The result of \( \mathbf{v} - \frac{1}{2} \mathbf{u} \) is \( \begin{bmatrix} -\frac{5}{2} \\ -4 \end{bmatrix} \).
1Step 1: Understanding the Problem
We need to calculate the vector \( \mathbf{v} - \frac{1}{2} \mathbf{u} \). This involves scalar multiplication of \( \mathbf{u} \) and vector subtraction of \( \mathbf{v} \) and the resultant vector.
2Step 2: Scalar Multiplication
Multiply the vector \( \mathbf{u} = \begin{bmatrix} 3 \ 4 \end{bmatrix} \) by the scalar \( \frac{1}{2} \). Calculate: \[ \frac{1}{2} \mathbf{u} = \frac{1}{2} \begin{bmatrix} 3 \ 4 \end{bmatrix} = \begin{bmatrix} \frac{3}{2} \ 2 \end{bmatrix} \]
3Step 3: Vector Subtraction
Subtract the result from the previous step from vector \( \mathbf{v} = \begin{bmatrix} -1 \ -2 \end{bmatrix} \). Calculate: \[ \mathbf{v} - \frac{1}{2} \mathbf{u} = \begin{bmatrix} -1 \ -2 \end{bmatrix} - \begin{bmatrix} \frac{3}{2} \ 2 \end{bmatrix} = \begin{bmatrix} -1 - \frac{3}{2} \ -2 - 2 \end{bmatrix} = \begin{bmatrix} -\frac{5}{2} \ -4 \end{bmatrix} \]
4Step 4: Graphical Interpretation
To graphically illustrate \( \mathbf{v} - \frac{1}{2} \mathbf{u} \), plot the vectors on a 2D plane. Start by plotting \( \mathbf{v} = \begin{bmatrix} -1 \ -2 \end{bmatrix} \), then plot \( \frac{1}{2} \mathbf{u} = \begin{bmatrix} \frac{3}{2} \ 2 \end{bmatrix} \). The result of the subtraction \( \begin{bmatrix} -\frac{5}{2} \ -4 \end{bmatrix} \) should be visualized starting at the origin. This shows how the vectors relate in terms of both direction and magnitude.
Key Concepts
Vector SubtractionScalar MultiplicationGraphical Interpretation
Vector Subtraction
In vector calculus, vector subtraction is a fundamental operation that allows us to find a new vector representing the difference between two vectors. To carry out vector subtraction, you simply subtract the components of one vector from the corresponding components of another vector. For example, if we have two vectors \( \mathbf{a} = \begin{bmatrix} a_1 \ a_2 \end{bmatrix} \) and \( \mathbf{b} = \begin{bmatrix} b_1 \ b_2 \end{bmatrix} \), the subtraction \( \mathbf{a} - \mathbf{b} \) gives us the difference vector \( \begin{bmatrix} a_1 - b_1 \ a_2 - b_2 \end{bmatrix} \).
In the given exercise, we performed the operation \( \mathbf{v} - \frac{1}{2} \mathbf{u} \). This involves first finding \( \frac{1}{2} \mathbf{u} \) and then subtracting this resultant vector from \( \mathbf{v} \). The subtraction results in the vector \( \begin{bmatrix} -\frac{5}{2} \ -4 \end{bmatrix} \), which represents the change from the initial vector \( \mathbf{v} \) when the scaled vector \( \frac{1}{2} \mathbf{u} \) is removed.
In the given exercise, we performed the operation \( \mathbf{v} - \frac{1}{2} \mathbf{u} \). This involves first finding \( \frac{1}{2} \mathbf{u} \) and then subtracting this resultant vector from \( \mathbf{v} \). The subtraction results in the vector \( \begin{bmatrix} -\frac{5}{2} \ -4 \end{bmatrix} \), which represents the change from the initial vector \( \mathbf{v} \) when the scaled vector \( \frac{1}{2} \mathbf{u} \) is removed.
Scalar Multiplication
Scalar multiplication in vector calculus is another critical concept, which involves multiplying a vector by a scalar (a real number). This operation scales the vector - stretching or shrinking it based on the scalar's magnitude. If a vector \( \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \) is multiplied by a scalar \( k \), the resulting vector is \( k\mathbf{x} = \begin{bmatrix} kx_1 \ kx_2 \end{bmatrix} \).
For instance, in our example, we scaled the vector \( \mathbf{u} = \begin{bmatrix} 3 \ 4 \end{bmatrix} \) by the scalar \( \frac{1}{2} \). This resulted in a new vector \( \begin{bmatrix} \frac{3}{2} \ 2 \end{bmatrix} \). This shows how scalar multiplication can reduce the length of a vector, maintaining its direction, but altering its magnitude.
For instance, in our example, we scaled the vector \( \mathbf{u} = \begin{bmatrix} 3 \ 4 \end{bmatrix} \) by the scalar \( \frac{1}{2} \). This resulted in a new vector \( \begin{bmatrix} \frac{3}{2} \ 2 \end{bmatrix} \). This shows how scalar multiplication can reduce the length of a vector, maintaining its direction, but altering its magnitude.
Graphical Interpretation
Graphically interpreting vectors helps to visualize the operations performed and their results. On a 2D plane, each vector can be represented as an arrow pointing from the origin to the coordinates described by its components.
To visualize vector subtraction, we first plot the initial vectors starting from the origin: \( \mathbf{v} = \begin{bmatrix} -1 \ -2 \end{bmatrix} \) and \( \frac{1}{2} \mathbf{u} = \begin{bmatrix} \frac{3}{2} \ 2 \end{bmatrix} \). The subtraction of \( \frac{1}{2} \mathbf{u} \) from \( \mathbf{v} \) can be illustrated by placing the tail of the vector \( \frac{1}{2} \mathbf{u} \) at the head of \( \mathbf{v} \) and finding where the resultant vector \( \begin{bmatrix} -\frac{5}{2} \ -4 \end{bmatrix} \) points relative to the origin.
This final vector can also be plotted to show the change in direction and magnitude after the subtraction operation has been completed. Graphical representation helps in understanding both the geometric change and the scale of the resultant vector.
To visualize vector subtraction, we first plot the initial vectors starting from the origin: \( \mathbf{v} = \begin{bmatrix} -1 \ -2 \end{bmatrix} \) and \( \frac{1}{2} \mathbf{u} = \begin{bmatrix} \frac{3}{2} \ 2 \end{bmatrix} \). The subtraction of \( \frac{1}{2} \mathbf{u} \) from \( \mathbf{v} \) can be illustrated by placing the tail of the vector \( \frac{1}{2} \mathbf{u} \) at the head of \( \mathbf{v} \) and finding where the resultant vector \( \begin{bmatrix} -\frac{5}{2} \ -4 \end{bmatrix} \) points relative to the origin.
This final vector can also be plotted to show the change in direction and magnitude after the subtraction operation has been completed. Graphical representation helps in understanding both the geometric change and the scale of the resultant vector.
Other exercises in this chapter
Problem 32
Let \(A=\left[\begin{array}{lll}1 & 4 & -2\end{array}\right]\) and \(B=\left[\begin{array}{r}-1 \\ 2 \\ 3\end{array}\right]\) (a) Compute \(A B\). (b) Compute \
View solution Problem 32
Determine whether each system is overdetermined or underdetermined; then solve each system. $$ \begin{array}{r} 4 y-3 z=6 \\ 2 y+z=1 \\ y+z=0 \end{array} $$
View solution Problem 33
33\. Let $$ A=\left[\begin{array}{rr} 2 & 1 \\ -1 & -3 \end{array}\right] $$ Find \(A^{2}, A^{3}\), and \(A^{4}\).
View solution Problem 33
Determine whether each system is overdetermined or underdetermined; then solve each system. $$ \begin{array}{r} 2 x-7 y+z=2 \\ x+y-2 z=4 \end{array} $$
View solution