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Determine whether \(\mathbf{u}\) and \(\mathbf{v}\) are parallel in each of the following cases. a. \(\mathbf{u}=\left[\begin{array}{r}-3 \\ -6 \\ 3\end{array}\right] ; \mathbf{v}=\left[\begin{array}{r}5 \\ 10 \\ -5\end{array}\right]\) b. \(\mathbf{u}=\left[\begin{array}{r}3 \\ -6 \\ 3\end{array}\right] ; \mathbf{v}=\left[\begin{array}{r}-1 \\ 2 \\ -1\end{array}\right]\) c. \(\mathbf{u}=\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right] ; \mathbf{v}=\left[\begin{array}{r}-1 \\ 0 \\ 1\end{array}\right]\) d. \(\mathbf{u}=\left[\begin{array}{r}2 \\ 0 \\ -1\end{array}\right] ; \mathbf{v}=\left[\begin{array}{r}-8 \\ 0 \\ 4\end{array}\right]\)

Short Answer

Expert verified
a, b, and d are parallel; c is not.

Step by step solution

01

Understanding Parallel Vectors

Two vectors \(\mathbf{u}\) and \(\mathbf{v}\) are parallel if there exists a scalar \(k\) such that \(\mathbf{u} = k\mathbf{v}\). Thus, to check if they are parallel, we need each component of \(\mathbf{u}\) to be a multiple of the corresponding component in \(\mathbf{v}\).
02

Checking Parallelism for Part a

Let's start with the vectors \(\mathbf{u} = \begin{bmatrix} -3 \ -6 \ 3 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} 5 \ 10 \ -5 \end{bmatrix}\). Check if there is a scalar \(k\):- If \(-3 = 5k\), then \(k = -\frac{3}{5}\).- If \(-6 = 10k\), then \(k = -\frac{3}{5}\). This is consistent.- If \(3 = -5k\), then \(k = -\frac{3}{5}\). This is consistent.All components yield \(k = -\frac{3}{5}\), so \(\mathbf{u}\) and \(\mathbf{v}\) are parallel.
03

Checking Parallelism for Part b

Consider vectors \(\mathbf{u} = \begin{bmatrix} 3 \ -6 \ 3 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} -1 \ 2 \ -1 \end{bmatrix}\).- If \(3 = -1k\), then \(k = -3\).- If \(-6 = 2k\), then \(k = -3\).- If \(3 = -1k\), then \(k = -3\).All components are consistent with \(k = -3\), indicating the vectors are parallel.
04

Checking Parallelism for Part c

Now address \(\mathbf{u} = \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} -1 \ 0 \ 1 \end{bmatrix}\).- For the components \(1\) and \(-1\), if \(1 = -1k\), then \(k = -1\).- For the zero components, there's no restriction.- For the components \(1\) and \(1\), we also have \(k = 1\).Since the signs are not consistent, \(\mathbf{u}\) and \(\mathbf{v}\) are not parallel.
05

Checking Parallelism for Part d

For the vectors \(\mathbf{u} = \begin{bmatrix} 2 \ 0 \ -1 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} -8 \ 0 \ 4 \end{bmatrix}\):- If \(2 = -8k\), then \(k = -\frac{1}{4}\).- The zero components don't affect equality.- If \(-1 = 4k\), then \(k = -\frac{1}{4}\).All components are consistent, thus \(\mathbf{u}\) and \(\mathbf{v}\) are parallel.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Mathematics
Vectors are central to mathematics and play a crucial role in fields such as physics, engineering, and computer science. A vector is a mathematical object that has both magnitude and direction. In simpler terms, it's like an arrow pointing from one place to another.
Vectors can represent different quantities like force, velocity, or displacement. For example, if you are moving from point A to point B, the vector would provide both the information about how far you are moving and in which direction.
In mathematics, vectors are often expressed in terms of their components along each axis. For instance, a vector in a two-dimensional space might be represented as \( \begin{bmatrix} 1 \ 2 \end{bmatrix} \), whereas a three-dimensional vector might look like \( \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} \). Understanding vectors is key to exploring numerous mathematical concepts, including parallelism.
Scalar Multiplication
Scalar multiplication is a fundamental operation in vector mathematics. It involves multiplying a vector by a scalar value, which is a single number.
When you perform scalar multiplication, each component of the vector is multiplied by the scalar. For example, if you have a vector \( \mathbf{u} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} \) and you multiply it by a scalar of 2, the result would be \( \begin{bmatrix} 2 \ 4 \ 6 \end{bmatrix} \).
This operation effectively changes the magnitude of the vector without altering its direction—unless multiplied by a negative scalar, which reverses its direction. Scalar multiplication is a key concept in determining if two vectors are parallel because parallel vectors are scalar multiples of each other.
Linear Algebra
Linear algebra is a branch of mathematics concerned with vectors, vector spaces, and linear mappings between these spaces. It includes the study of lines, planes, and subspaces, but extends these concepts to higher dimensions.
One of the core operations in linear algebra is determining the relationship between vectors, such as whether they are parallel or orthogonal. Parallel vectors, as shown in the exercise, are scalar multiples of each other. This makes it easy to determine parallelism using tools from linear algebra, like solving systems of linear equations
Linear algebra provides powerful techniques for solving mathematical problems involving vectors. It forms the foundation for many applications, from computer graphics to engineering and machine learning.
3D Vectors
3D vectors are vectors with three components, typically represented in the form \( \begin{bmatrix} x \ y \ z \end{bmatrix} \). Each component corresponds to a dimension in space: x, y, and z.
These vectors are frequently used in applications dealing with physical spaces, such as physics and computer graphics, where it’s essential to account for three-dimensional spatial relationships.
Analyzing 3D vectors involves understanding operations like addition, scalar multiplication, and dot products. When two 3D vectors are parallel, each component of one vector is a scalar multiple of the corresponding component of the other vector.
This consistency across components simplifies determining parallelism, as demonstrated in the step-by-step solution of the exercise. Understanding 3D vectors is essential for comprehending many real-world phenomena and complex systems.

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Most popular questions from this chapter

Let \(A\) be a \(2 \times 3\) matrix of rank 2 with rows \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\). Show that $$P=\\{X A \mid X=[x y] ; x, y \text { arbitrary }\\}$$ is the plane through the origin with normal \(\mathbf{r}_{1} \times \mathbf{r}_{2}\).

Show that points \(A, B,\) and \(C\) are all on one line if and only if \(\overrightarrow{A B} \times \overrightarrow{A C}=0\).

In each case, find a vector equation of the line. a. Passing through \(P(3,-1,4)\) and perpendicular to the plane \(3 x-2 y-z=0\) b. Passing through \(P(2,-1,3)\) and perpendicular to the plane \(2 x+y=1\) c. Passing through \(P(0,0,0)\) and perpendicular $$ \begin{array}{l} \text { to the lines }\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right]+t\left[\begin{array}{r} 2 \\ 0 \\ -1 \end{array}\right] \text { and } \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 2 \\ 1 \\ -3 \end{array}\right]+t\left[\begin{array}{r} 1 \\ -1 \\ 5 \end{array}\right]} \end{array} $$ d. Passing through \(P(1,1,-1)\), and perpendicular to the lines $$ \begin{array}{l} {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 1 \end{array}\right]+t\left[\begin{array}{r} 1 \\ 1 \\ -2 \end{array}\right] \text { an }} \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 5 \\ 5 \\ -2 \end{array}\right]+t\left[\begin{array}{r} 1 \\ 2 \\ -3 \end{array}\right]} \end{array} $$ e. Passing through \(P(2,1,-1)\), intersecting the line \(\left[\begin{array}{l}x \\\ y \\ z\end{array}\right]=\left[\begin{array}{r}1 \\ 2 \\\ -1\end{array}\right]+t\left[\begin{array}{l}3 \\ 0 \\ 1\end{array}\right],\) and perpendicular to that line. f. Passing through \(P(1,1,2)\), intersecting the line \(\left[\begin{array}{l}x \\\ y \\ z\end{array}\right]=\left[\begin{array}{l}2 \\ 1 \\\ 0\end{array}\right]+t\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right],\) and perpendicular to line.

In each case, find the point \(Q:\) a. \(\overrightarrow{P Q}=\left[\begin{array}{r}2 \\ 0 \\\ -3\end{array}\right]\) and \(P=P(2,-3,1)\) b. \(\overrightarrow{P Q}=\left[\begin{array}{r}-1 \\ 4 \\\ 7\end{array}\right]\) and \(P=P(1,3,-4)\)

Show that if the diagonals of a parallelogram are perpendicular, it is necessarily a rhombus.

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