Problem 27 - Advanced Engineering Mathematics

Question

Find the direction cosines and direction angles of the given vector. $$ \mathbf{a}=\langle 1,0,-\sqrt{3}\rangle $$

Step-by-Step Solution

Answer

Direction cosines are \((\frac{1}{2}, 0, -\frac{\sqrt{3}}{2})\). Direction angles are \(60^\circ, 90^\circ, 150^\circ\).

Step 1: Find the Magnitude of the Vector

First, we need to find the magnitude of the vector \( \mathbf{a} = \langle 1, 0, -\sqrt{3} \rangle \). The magnitude \( ||\mathbf{a}|| \) is calculated as \( \sqrt{1^2 + 0^2 + (-\sqrt{3})^2} \). Simplifying that, we get \( \sqrt{1 + 0 + 3} = \sqrt{4} = 2 \).

Step 2: Determine the Direction Cosines

The direction cosines \( (l, m, n) \) are given by the formulas \( l = \frac{x}{||\mathbf{a}||} \), \( m = \frac{y}{||\mathbf{a}||} \), \( n = \frac{z}{||\mathbf{a}||} \). Substitute the values: \( l = \frac{1}{2} \), \( m = \frac{0}{2} \), \( n = \frac{-\sqrt{3}}{2} \). So, the direction cosines are \( (\frac{1}{2}, 0, -\frac{\sqrt{3}}{2}) \).

Step 3: Calculate the Direction Angles

The direction angles \( \alpha, \beta, \gamma \) are calculated using \( \cos \alpha = l \), \( \cos \beta = m \), \( \cos \gamma = n \). Using inverse cosine function: \( \alpha = \cos^{-1}(\frac{1}{2}) = 60^\circ \), \( \beta = \cos^{-1}(0) = 90^\circ \), \( \gamma = \cos^{-1}(-\frac{\sqrt{3}}{2}) = 150^\circ \).

Key Concepts

Understanding Vector Magnitude

Finding the magnitude of a vector is an essential step in many vector calculations. It helps to understand the size or length of the vector in the vector space.
A vector is represented as \( \mathbf{a} = \langle x, y, z \rangle \). The magnitude of this vector, often denoted as \( ||\mathbf{a}|| \), is calculated using the formula:

\[||\mathbf{a}|| = \sqrt{x^2 + y^2 + z^2}\]
This formula is derived from the Pythagorean theorem, considering the vector as the hypotenuse of a right-angled triangle in three-dimensional space.

For instance, if you have a vector \( \langle 1, 0, -\sqrt{3} \rangle \), you substitute the values into the formula:

\[||\mathbf{a}|| = \sqrt{1^2 + 0^2 + (-\sqrt{3})^2} = \sqrt{1 + 0 + 3} = \sqrt{4} = 2\]
This magnitude represents the length of the vector in three-dimensional space, across all its components.

Trigonometric Identities - The Building Blocks

Trigonometric identities are fundamental in understanding the relationship between angles and side lengths in triangles, which helps in vector analysis, especially when we're dealing with direction cosines.
These identities serve as the rules that govern the behavior of trigonometric functions like sine, cosine, and tangent.

Common trigonometric identities include:

• The Pythagorean identity: \( \sin^2\theta + \cos^2\theta = 1 \)
• Sum and difference identities: \( \cos(a \pm b) = \cos(a)\cos(b) \mp \sin(a)\sin(b) \)
• Double angle formulas: \( \cos(2a) = \cos^2(a) - \sin^2(a) \)

In the context of vectors, we often use trigonometric identities to transform or simplify expressions involving direction cosines \( l, m, n \), where each represents the cosine of the angle the vector forms with the coordinate axes.

Understanding the Inverse Cosine Function

The inverse cosine function, often written as \( \cos^{-1}(x) \), is used to determine the angle whose cosine is \( x \). This function is vital when calculating direction angles from direction cosines.
The direction cosine \( l \), for instance, corresponds to \( \cos(\alpha) \), meaning the angle \( \alpha \) can be found by:

\[\alpha = \cos^{-1}(l)\]
The range for the inverse cosine function is between 0 and 180 degrees.
When applying this to find direction angles \( \alpha, \beta, \gamma \), we do the following:

• \( \alpha = \cos^{-1}(\frac{1}{2}) = 60^\circ \)
• \( \beta = \cos^{-1}(0) = 90^\circ \)
• \( \gamma = \cos^{-1}(-\frac{\sqrt{3}}{2}) = 150^\circ \)

Understanding how to use \( \cos^{-1} \) effectively is crucial for calculating angles from cosines.
It ensures accuracy in determining the orientation of the vector in relation to the axes.