Problem 37 - Algebra and Trigonometry

Question

Find the direction angles of the given vector, rounded to the nearest degree. $$ 3 \mathbf{i}+4 \mathbf{j}+5 \mathbf{k} $$

Step-by-Step Solution

Answer

The direction angles are approximately \( 68^\circ \), \( 57^\circ \), and \( 45^\circ \).

Step 1: Identify the vector components

The given vector can be expressed with components as \( \mathbf{v} = \langle 3, 4, 5 \rangle \), where the component along \( \mathbf{i} \) is 3, along \( \mathbf{j} \) is 4, and along \( \mathbf{k} \) is 5.

Step 2: Calculate the magnitude of the vector

The magnitude \( \| \mathbf{v} \| \) of the vector \( \mathbf{v} = \langle 3, 4, 5 \rangle \) is calculated using the formula: \[\| \mathbf{v} \| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}.\]

Step 3: Find angles with each axis

To find the direction angles \( \alpha \), \( \beta \), and \( \gamma \) with the \( x \), \( y \), and \( z \) axes respectively, use the formulas:- \( \cos(\alpha) = \frac{3}{5\sqrt{2}} \)- \( \cos(\beta) = \frac{4}{5\sqrt{2}} \)- \( \cos(\gamma) = \frac{5}{5\sqrt{2}} \),and calculate each angle by taking the inverse cosine.

Step 4: Calculate direction angles

Evaluate each inverse cosine to find the direction angles:- \( \alpha = \cos^{-1}\left(\frac{3}{5\sqrt{2}}\right) \approx 68^\circ \)- \( \beta = \cos^{-1}\left(\frac{4}{5\sqrt{2}}\right) \approx 57^\circ \)- \( \gamma = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \approx 45^\circ \).

Step 5: Round angles to nearest degrees

After calculating, ensure each angle is rounded to the nearest whole degree, yielding: \( \alpha \approx 68^\circ \), \( \beta \approx 57^\circ \), \( \gamma \approx 45^\circ \).

Key Concepts

Vector Components

When trying to understand a vector in three-dimensional space, it's crucial to break it down into its components. A vector can be described by how much it stretches along each of the three main axes: the x-axis, y-axis, and z-axis. For example, if we have a vector \(3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\), it means that:

• The vector stretches 3 units along the x-axis, represented by the component \(3\) of \(\mathbf{i}\).
• It stretches 4 units along the y-axis, represented by the component \(4\) of \(\mathbf{j}\).
• It stretches 5 units along the z-axis, represented by the component \(5\) of \(\mathbf{k}\).

Breaking down a vector into its components makes it easier to understand its behavior and use in calculations. This breakdown provides a clear way to examine the vector's direction and distance in space.

Magnitude of a Vector

The magnitude of a vector is like the "length" of the vector, showing how far the vector extends in space from its initial point to its terminal point. To find the magnitude, you use the formula:\[\| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2}\]This formula effectively applies the Pythagorean theorem to three dimensions. For our example vector \(\langle 3, 4, 5 \rangle\), the magnitude is calculated as:

• First, square each component: \(3^2 = 9\), \(4^2 = 16\), \(5^2 = 25\).
• Add these squared values: \(9 + 16 + 25 = 50\).
• Finally, take the square root: \(\sqrt{50} = 5\sqrt{2}\).

This magnitude tells you the vector's "size" and is crucial for understanding things like direction cosines.

Direction Cosines

Direction cosines connect a vector's components with its magnitude, helping define the angles between the vector and each axis. For any vector \(\mathbf{v} = \langle x, y, z \rangle\), the direction cosines are defined as:

• \(\cos(\alpha) = \frac{x}{\|\mathbf{v}\|}\)
• \(\cos(\beta) = \frac{y}{\|\mathbf{v}\|}\)
• \(\cos(\gamma) = \frac{z}{\|\mathbf{v}\|}\)

In our case:

• \(\cos(\alpha) = \frac{3}{5\sqrt{2}}\)
• \(\cos(\beta) = \frac{4}{5\sqrt{2}}\)
• \(\cos(\gamma) = \frac{5}{5\sqrt{2}}\)

These cosines help you find the angles between the vector and the axes, which are essential for validating directional movements or rotations in physics and engineering.

Inverse Cosine Function

The inverse cosine function, or \(\cos^{-1}\), allows you to find an angle when you know the cosine value of that angle. It is extremely useful when dealing with vectors, as it provides the specific direction angles needed. To perform this calculation effectively:

• Calculate \(\cos^{-1}\left(\frac{3}{5\sqrt{2}}\right)\) to find angle \(\alpha\), yielding approximately \(68^\circ\).
• Calculate \(\cos^{-1}\left(\frac{4}{5\sqrt{2}}\right)\) for angle \(\beta\), yielding approximately \(57^\circ\).
• Compute \(\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)\) for angle \(\gamma\), yielding approximately \(45^\circ\).

Rounding each result to the nearest whole number gives us the direction angles necessary to completely describe the vector's orientation. These calculations are pivotal in fields such as navigation, robotics, and computer graphics, where understanding precise angles is critical.