Two vectors are given in component form in the following way: v (4, –3) s (3, 2) Which of the following vectors represents 30-25 in component form а. (6, —13) b. (18,–5) c. (13,6) S 121

Algebra and Trigonometry (6th Edition)
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ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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## Vectors in Component Form

Two vectors are given in component form in the following way:
\[ \vec{v} \langle 4, -3 \rangle \]
\[ \vec{s} \langle 3, 2 \rangle \]

### Question:
Which of the following vectors represents \(3\vec{v} - 2\vec{s}\) in component form?
a. \(\langle 6, -13 \rangle\)
b. \(\langle 18, -5 \rangle\)
c. \(\langle 13, 6 \rangle\)
d. \(\langle 6, 12 \rangle\)

To solve this, we need to calculate \(3\vec{v} - 2\vec{s}\):

1. **Multiply each component of \(\vec{v}\) by 3:**
   \[
   3\vec{v} = 3 \cdot \langle 4, -3 \rangle = \langle 12, -9 \rangle
   \]

2. **Multiply each component of \(\vec{s}\) by 2:**
   \[
   2\vec{s} = 2 \cdot \langle 3, 2 \rangle = \langle 6, 4 \rangle
   \]

3. **Subtract the corresponding components of \(2\vec{s}\) from \(3\vec{v}\):**
   \[
   3\vec{v} - 2\vec{s} = \langle 12, -9 \rangle - \langle 6, 4 \rangle = \langle 12 - 6, -9 - 4 \rangle = \langle 6, -13 \rangle
   \]

So, the correct answer is:
a. \(\langle 6, -13 \rangle\)
Transcribed Image Text:## Vectors in Component Form Two vectors are given in component form in the following way: \[ \vec{v} \langle 4, -3 \rangle \] \[ \vec{s} \langle 3, 2 \rangle \] ### Question: Which of the following vectors represents \(3\vec{v} - 2\vec{s}\) in component form? a. \(\langle 6, -13 \rangle\) b. \(\langle 18, -5 \rangle\) c. \(\langle 13, 6 \rangle\) d. \(\langle 6, 12 \rangle\) To solve this, we need to calculate \(3\vec{v} - 2\vec{s}\): 1. **Multiply each component of \(\vec{v}\) by 3:** \[ 3\vec{v} = 3 \cdot \langle 4, -3 \rangle = \langle 12, -9 \rangle \] 2. **Multiply each component of \(\vec{s}\) by 2:** \[ 2\vec{s} = 2 \cdot \langle 3, 2 \rangle = \langle 6, 4 \rangle \] 3. **Subtract the corresponding components of \(2\vec{s}\) from \(3\vec{v}\):** \[ 3\vec{v} - 2\vec{s} = \langle 12, -9 \rangle - \langle 6, 4 \rangle = \langle 12 - 6, -9 - 4 \rangle = \langle 6, -13 \rangle \] So, the correct answer is: a. \(\langle 6, -13 \rangle\)
**Understanding Rotation Transformation Matrices**

In linear algebra, transformation matrices are used to perform linear transformations on vectors. One such common transformation is rotation. 

**Problem Statement:**

What is the transformation matrix for rotating 270 degrees clockwise about the origin?

**Options:**

a. \[ \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \]

b. \[ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \]

c. \[ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \]

d. \[ \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \]

e. \[ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \]

**Explanation:**

To determine the correct transformation matrix, we must recall the standard form of a rotation matrix for a counterclockwise rotation by an angle θ:

\[ \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \]

For a clockwise rotation, we modify the signs:

\[ \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \]

For rotating 270 degrees clockwise, we use θ = 270° (which is equivalent to θ = -90° counterclockwise):

\[
\cos(270^\circ) = 0, \quad \sin(270^\circ) = -1 \\
\begin{pmatrix}
 \cos 270^\circ & \sin 270^\circ \\
 -\sin 270^\circ & \cos 270^\circ
 \end{pmatrix} = 
 \begin{pmatrix}
 0 & 1 \\
 -1 & 0
 \end{pmatrix}
\]

Thus, the correct transformation matrix for a 270° clockwise rotation about the origin is:

**b.** \[ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \]
Transcribed Image Text:**Understanding Rotation Transformation Matrices** In linear algebra, transformation matrices are used to perform linear transformations on vectors. One such common transformation is rotation. **Problem Statement:** What is the transformation matrix for rotating 270 degrees clockwise about the origin? **Options:** a. \[ \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \] b. \[ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \] c. \[ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \] d. \[ \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \] e. \[ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \] **Explanation:** To determine the correct transformation matrix, we must recall the standard form of a rotation matrix for a counterclockwise rotation by an angle θ: \[ \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \] For a clockwise rotation, we modify the signs: \[ \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \] For rotating 270 degrees clockwise, we use θ = 270° (which is equivalent to θ = -90° counterclockwise): \[ \cos(270^\circ) = 0, \quad \sin(270^\circ) = -1 \\ \begin{pmatrix} \cos 270^\circ & \sin 270^\circ \\ -\sin 270^\circ & \cos 270^\circ \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \] Thus, the correct transformation matrix for a 270° clockwise rotation about the origin is: **b.** \[ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \]
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