Problem 49 - Algebra and Trigonometry

Question

Parallel Vectors Two nonzero vectors are parallel if they point in the same direction or in opposite directions. This means that if two vectors are parallel, one must be a scalar multiple of the other. Determine whether the given vectors \(\mathbf{u}\) and \(\mathbf{v}\) are parallel. If they are, express \(\mathbf{v}\) as a scalar multiple of \(\mathbf{u} .\) (a) \(\mathbf{u}=\langle 3,-2,4\rangle, \mathbf{v}=\langle- 6,4,-8\rangle\) (b) \(\mathbf{u}=\langle- 9,-6,12\rangle, \mathbf{v}=\langle 12,8,-16\rangle\) (c) \(\mathbf{u}=\mathbf{i}+\mathbf{j}+\mathbf{k}, \quad \mathbf{v}=2 \mathbf{i}+2 \mathbf{j}-2 \mathbf{k}\)

Step-by-Step Solution

Answer

(a) Yes, \(\mathbf{v} = -2\mathbf{u}\); (b) Yes, \(\mathbf{v} = -\frac{4}{3}\mathbf{u}\); (c) No.

Step 1: Verify Parallelism for Part (a)

To determine if vectors \(\mathbf{u} = \langle 3, -2, 4 \rangle\) and \(\mathbf{v} = \langle -6, 4, -8 \rangle\) are parallel, check if there exists a scalar \(k\) such that \(v_i = k \times u_i\) for each component. Calculate \(k\) for each component: 1. For \(3\) and \(-6\), \(-6 = k \times 3\) implies \(k = -2\). 2. For \(-2\) and \(4\), \(4 = k \times (-2)\) implies \(k = -2\).3. For \(4\) and \(-8\), \(-8 = k \times 4\) implies \(k = -2\).Since the same scalar \(k = -2\) satisfies all components, \(\mathbf{v}\) is a scalar multiple of \(\mathbf{u}\). Thus, the vectors are parallel, and \(\mathbf{v} = -2 \mathbf{u}\).

Step 2: Verify Parallelism for Part (b)

For vectors \( \mathbf{u} = \langle -9, -6, 12 \rangle \) and \( \mathbf{v} = \langle 12, 8, -16 \rangle \), find scalar \( k \):1. From \(-9\) to \(12\), \(12 = k \times (-9)\) gives \(k = -\frac{4}{3}\).2. From \(-6\) to \(8\), \(8 = k \times (-6)\) gives \(k = -\frac{4}{3}\).3. From \(12\) to \(-16\), \(-16 = k \times 12\) gives \(k = -\frac{4}{3}\).All components yield the same \(k\), confirming \(\mathbf{v} = -\frac{4}{3} \mathbf{u}\). Thus, the vectors are parallel.

Step 3: Verify Parallelism for Part (c)

Given \( \mathbf{u} = \langle 1, 1, 1 \rangle \) and \( \mathbf{v} = \langle 2, 2, -2 \rangle \), check for parallelism:1. From \(1\) to \(2\), \(2 = k \times 1\) implies \(k = 2\).2. From \(1\) to \(2\), \(2 = k \times 1\) implies \(k = 2\).3. From \(1\) to \(-2\), \(-2 = k \times 1\) implies \(k = -2\).The inconsistent values of \(k\) (2 and -2) mean these vectors are not parallel.

Key Concepts

scalar multiple

To determine if two vectors are parallel, we need to check if one is a scalar multiple of the other. This simply means that you can multiply all components of one vector by a constant scalar to get the other vector. For example, if we have vector \ \( \mathbf{u} = \langle 3, -2, 4 \rangle \ \) and vector \ \( \mathbf{v} = \langle -6, 4, -8 \rangle \ \), we can see that each component of \( \mathbf{v} \) can be expressed as a multiple of the corresponding component of \( \mathbf{u} \): \

• \( -6 \) is \( -2 \times 3 \)
• \( 4 \) is \( -2 \times -2 \)
• \( -8 \) is \( -2 \times 4 \)

The scalar here is -2, and since it's consistent across all components, vector \( \mathbf{v} \) is indeed a scalar multiple of vector \( \mathbf{u} \), making them parallel.
Scalar multiples indicate two vectors either point in the same or opposite directions. This concept is pivotal in understanding vector parallelism.

vector components

Vectors in three-dimensional space are represented by their components, often in the form \( \langle x, y, z \rangle \). These components tell us how much a vector extends in each of the three-dimensional axes: x, y, and z. For instance, the vector \( \mathbf{u} = \langle 3, -2, 4 \rangle \) means:

• It stretches 3 units along the x-axis,
• moves -2 units along the y-axis (opposite the positive y direction),
• and extends 4 units along the z-axis.

Understanding vector components is essential for performing operations like addition, subtraction, and finding scalar multiples. By examining each component, one can easily manipulate and analyze vectors, identify parallelism, and solve various vector algebra problems.

vector algebra

Vector algebra allows us to perform mathematical operations on vectors such as addition, subtraction, and scaling by a scalar. These operations are crucial for solving real-world problems involving directions and magnitudes, especially in physics and engineering.
One key operation is scaling, where each component of a vector is multiplied by a scalar, adjusting its magnitude without changing its direction. For example, multiplying \( \mathbf{u} = \langle 3, -2, 4 \rangle \) by the scalar \(-2\), we get \( \mathbf{v} = \langle -6, 4, -8 \rangle \), indicating a change in direction but maintaining parallelism.
In vector algebra, parallel vectors remain one of the primary characteristics we often look to identify, providing insights into vector relationships and enabling us to solve various practical and theoretical problems efficiently.