### Determining Orthogonality of VectorsIn this lesson, we will determine if the following vectors are orthogonal.Given vectors:\[ \mathbf{u} = \begin{bmatrix} 12 \\ 4 \\ -8 \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} 2 \\ -4 \\ 1 \end{bmatrix}\]**Question:**Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or a simplified fraction.)**Options:**- **A.** The vectors $\mathbf{u}$ and $\mathbf{v}$ are not orthogonal because $\mathbf{u} \cdot \mathbf{v} =$ [ ]- **B.** The vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal because $\mathbf{u} \cdot \mathbf{v} =$ [ ]- **C.** The vectors $\mathbf{u}$ and $\mathbf{v}$ are not orthogonal because $\mathbf{u} + \mathbf{v} =$ [ ]- **D.** The vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal because $\mathbf{u} + \mathbf{v} =$ [ ]**Explanation:**Vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if their dot product is zero. The dot product of $\mathbf{u}$ and $\mathbf{v}$ is calculated as follows:\[ \mathbf{u} \cdot \mathbf{v} = (12 \times 2) + (4 \times -4) + (-8 \times 1)\]Calculate each term:\[ (12 \times 2) = 24 \]\[ (4 \times -4) = -16 \]\[ (-8 \times 1) = -8\]Therefore:\[ \mathbf{u} \cdot \mathbf{v} = 24 + (-16) + (-8) = 0\]Since $\mathbf{u} \cdot \mathbf{v} = 0$, the vectors \(\mathbf

We have to solve given problem:

Determine if the following vectors are orthogonal. 12 2 -8 %3D Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or a simplified fraction) O A. The vectors u and v are not orthogonal because u v = O B. The vectors u and v are orthogonal because u•v = O C. The vectors u and v are not orthogonal because u + v = O D. The vectors u and v are orthogonal because u + v =

Determine if the following vectors are orthogonal. 12 2 -8 %3D Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or a simplified fraction) O A. The vectors u and v are not orthogonal because u v = O B. The vectors u and v are orthogonal because u•v = O C. The vectors u and v are not orthogonal because u + v = O D. The vectors u and v are orthogonal because u + v =

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Concept explainers

Magnitude

A vector is an element with both direction and magnitude. A vector is represented using an arrow, where the arrow represents the direction of the vector and its length is represented using the magnitude. The set of vectors in space is known as vector space. The vectors can be added, subtracted, multiplied, and divided in a vector space. The image below shows a vector AB.

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### Determining Orthogonality of Vectors

In this lesson, we will determine if the following vectors are orthogonal.

Given vectors:
\[
\mathbf{u} = \begin{bmatrix} 12 \\ 4 \\ -8 \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} 2 \\ -4 \\ 1 \end{bmatrix}
\]

**Question:**
Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or a simplified fraction.)

**Options:**
- **A.** The vectors $\mathbf{u}$ and $\mathbf{v}$ are not orthogonal because $\mathbf{u} \cdot \mathbf{v} =$ [ ]
- **B.** The vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal because $\mathbf{u} \cdot \mathbf{v} =$ [ ]
- **C.** The vectors $\mathbf{u}$ and $\mathbf{v}$ are not orthogonal because $\mathbf{u} + \mathbf{v} =$ [ ]
- **D.** The vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal because $\mathbf{u} + \mathbf{v} =$ [ ]

**Explanation:**
Vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if their dot product is zero. The dot product of $\mathbf{u}$ and $\mathbf{v}$ is calculated as follows:
\[
\mathbf{u} \cdot \mathbf{v} = (12 \times 2) + (4 \times -4) + (-8 \times 1)
\]

Calculate each term:
\[
(12 \times 2) = 24
\]
\[
(4 \times -4) = -16
\]
\[
(-8 \times 1) = -8
\]

Therefore:
\[
\mathbf{u} \cdot \mathbf{v} = 24 + (-16) + (-8) = 0
\]

Since $\mathbf{u} \cdot \mathbf{v} = 0$, the vectors \(\mathbf

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