Problem 25 - Advanced Engineering Mathematics

Question

In Problems \(25-28\), find the direction cosines and direction angles of the given vector. $$ \mathbf{a}=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k} $$

Step-by-Step Solution

Answer

Direction cosines: \(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\). Direction angles: \(\alpha, \beta, \gamma\).

Step 1: Understanding the Vector

We are given the vector \( \mathbf{a} = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k} \). This vector can be represented with its components as \( \mathbf{a} = (1, 2, 3) \), where \( 1 \) is the component in the \( x \)-direction, \( 2 \) is in the \( y \)-direction, and \( 3 \) is in the \( z \)-direction.

Step 2: Compute the Magnitude of the Vector

The magnitude \( \|\mathbf{a}\| \) of the vector \( \mathbf{a} \) is calculated using the formula \( \|\mathbf{a}\| = \sqrt{x^2 + y^2 + z^2} \). Substituting the values for \( \mathbf{a} = (1, 2, 3) \), we get \( \|\mathbf{a}\| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14} \).

Step 3: Calculate the Direction Cosines

The direction cosines of the vector \( \mathbf{a} \) are given by \( \cos \alpha = \frac{x}{\|\mathbf{a}\|} \), \( \cos \beta = \frac{y}{\|\mathbf{a}\|} \), and \( \cos \gamma = \frac{z}{\|\mathbf{a}\|} \). For \( \mathbf{a} = (1, 2, 3) \) and \( \|\mathbf{a}\| = \sqrt{14} \), the direction cosines are: \( \cos \alpha = \frac{1}{\sqrt{14}} \), \( \cos \beta = \frac{2}{\sqrt{14}} \), and \( \cos \gamma = \frac{3}{\sqrt{14}} \).

Step 4: Find the Direction Angles

The direction angles \( \alpha \), \( \beta \), and \( \gamma \) can be found by taking the inverse cosine of the direction cosines. Thus, \( \alpha = \cos^{-1}\left(\frac{1}{\sqrt{14}}\right) \), \( \beta = \cos^{-1}\left(\frac{2}{\sqrt{14}}\right) \), and \( \gamma = \cos^{-1}\left(\frac{3}{\sqrt{14}}\right) \).

Key Concepts

Vector Mathematics

Vector mathematics involves the study of quantities that have both magnitude and direction. Unlike scalar quantities, which only have magnitude, vectors are critical in fields like physics, engineering, and even computer science due to their ability to represent various multi-dimensional phenomena. In our exercise, the vector \( \mathbf{a} = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k} \) serves as a perfect example of a 3D vector. Here, the vector is defined in terms of unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), which are vectors of length one in the directions of the x, y, and z axes respectively.
\( \mathbf{a} \)'s components \((1, 2, 3)\) tell us how much the vector stretches along each axis, and these components are essential for calculating properties like magnitude and direction. Understanding these basics of vector representation allows you to solve more complex mathematical problems involving vectors.

3D Vectors

In three-dimensional space, 3D vectors are vital for representing positions, forces, velocities, and more. A vector in 3D space is often written as \( (x, y, z) \), where each component represents the vector's projection on each of the three axes. For the vector \( (1, 2, 3) \), these numbers mean you're moving 1 unit along the x-axis, 2 units along the y-axis, and 3 units along the z-axis.
To find its magnitude, or the vector's length, we use the formula \( \|\mathbf{a}\| = \sqrt{x^2 + y^2 + z^2} \). By substituting in our values, \( \|\mathbf{a}\| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14} \). This provides a measure of how "far" the vector reaches in space, an essential step before delving into direction cosines and angles.
3D vectors are widely used in various applications such as computer graphics, where they help simulate realistic movements and transformations.

Mathematical Problems

Solving mathematical problems involving vectors in 3D often includes finding direction cosines and angles. Direction cosines are the cosines of the angles between the vector and the positive x, y, and z axes, respectively. They are calculated by dividing each component by the vector's magnitude. For instance, the direction cosines for the vector \( (1, 2, 3) \) with magnitude \( \sqrt{14} \) are:

• \( \cos \alpha = \frac{1}{\sqrt{14}} \)
• \( \cos \beta = \frac{2}{\sqrt{14}} \)
• \( \cos \gamma = \frac{3}{\sqrt{14}} \)

Once we have these, the direction angles \( \alpha, \beta, \gamma \) can be found by taking the inverse cosine (arccos) of the direction cosines:

• \( \alpha = \cos^{-1}\left(\frac{1}{\sqrt{14}}\right) \)
• \( \beta = \cos^{-1}\left(\frac{2}{\sqrt{14}}\right) \)
• \( \gamma = \cos^{-1}\left(\frac{3}{\sqrt{14}}\right) \)

Understanding how to compute these values allows you to tackle a wide array of mathematical problems, providing critical insights into the vector’s orientation in space.