Problem 38 - Algebra and Trigonometry

Question

Find the direction angles of the given vector, rounded to the nearest degree. $$ \mathbf{i}-2 \mathbf{j}-\mathbf{k} $$

Step-by-Step Solution

Answer

The direction angles are approximately \(66^{\circ}\), \(126^{\circ}\), and \(116^{\circ}\).

Step 1: Write the Vector's Components

The vector given is \( \mathbf{i} - 2 \mathbf{j} - \mathbf{k} \). Identify its components as: \( a = 1 \), \( b = -2 \), and \( c = -1 \).

Step 2: Calculate the Magnitude of the Vector

The magnitude of the vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) is given by the formula: \[ |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \]. Substitute the vector's components to get:\[|\mathbf{v}| = \sqrt{1^2 + (-2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}.\]

Step 3: Calculate the Direction Cosines

The direction cosines are calculated using the components of the vector and its magnitude:\[\cos \alpha = \frac{a}{|\mathbf{v}|} = \frac{1}{\sqrt{6}}, \\cos \beta = \frac{b}{|\mathbf{v}|} = \frac{-2}{\sqrt{6}}, \\cos \gamma = \frac{c}{|\mathbf{v}|} = \frac{-1}{\sqrt{6}}.\]

Step 4: Calculate the Direction Angles

Find the angles using the inverse cosine function:\[\alpha = \cos^{-1}\left(\frac{1}{\sqrt{6}}\right), \\beta = \cos^{-1}\left(\frac{-2}{\sqrt{6}}\right), \\gamma = \cos^{-1}\left(\frac{-1}{\sqrt{6}}\right).\]

Step 5: Convert to Degrees and Round

Using a calculator, convert each angle from radians to degrees and round to the nearest degree:- \( \alpha \approx 66^{\circ} \)- \( \beta \approx 126^{\circ} \)- \( \gamma \approx 116^{\circ} \)

Key Concepts

Understanding Vector Components

Vector components are the building blocks of any vector. When we have a vector expressed in terms of the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), its components denote how far and in what direction the vector moves along the axes of a 3D coordinate system. In the example, the vector \( \mathbf{i} - 2 \mathbf{j} - \mathbf{k} \) has components \( a = 1 \), \( b = -2 \), and \( c = -1 \). Each represents a signed magnitude in their respective directions, indicating:

• 1 unit in the positive x-direction
• 2 units in the negative y-direction
• 1 unit in the negative z-direction

Understanding these components helps us visualize the vector's orientation in space. Breaking vectors down into components is essential for performing more complex operations, like calculating angles and magnitude.

Calculating the Magnitude of a Vector

The magnitude of a vector, also known as its 'length' or 'norm', gives us the measure from the vector's tail to its tip in a straight line. The formula \( |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \) allows us to find this magnitude by summing the squares of its components and then taking the square root. For the vector \( \mathbf{i} - 2 \mathbf{j} - \mathbf{k} \), we compute:

• Square the components: \( 1^2 = 1 \), \((-2)^2 = 4 \), and \((-1)^2 = 1 \)
• Add these values: \( 1 + 4 + 1 = 6 \)
• Take the square root: \( \sqrt{6} \)

The magnitude is thus \( \sqrt{6} \). Knowing the magnitude is crucial before finding direction angles, as it acts as a reference for vector comparisons.

Significance of Direction Cosines

Direction cosines are cosines of the angles between the vector and the coordinate axes. Calculating these requires both the vector components and its magnitude. For any vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \):

• \( \cos \alpha = \frac{a}{|\mathbf{v}|} \)
• \( \cos \beta = \frac{b}{|\mathbf{v}|} \)
• \( \cos \gamma = \frac{c}{|\mathbf{v}|} \)

These formulas give us ratios that express how much of each component contributes to the overall orientation of the vector. In our case:

• \( \cos \alpha = \frac{1}{\sqrt{6}} \)
• \( \cos \beta = \frac{-2}{\sqrt{6}} \)
• \( \cos \gamma = \frac{-1}{\sqrt{6}} \)

These values show how inclined the vector is to each principal axis, helping us understand its spatial direction.

Understanding the Inverse Cosine Function

The inverse cosine function, denoted as \( \cos^{-1} \), provides the angle whose cosine is a given number, a critical step in determining the orientation of a vector. To find the direction angles of a vector, you apply \( \cos^{-1} \) to the direction cosines computed previously.The angles \( \alpha \), \( \beta \), and \( \gamma \) represent the angles between the vector and the x, y, and z axes, respectively. Let's look at how it's applied:

• \( \alpha = \cos^{-1}\left(\frac{1}{\sqrt{6}}\right) \approx 66^\circ \)
• \( \beta = \cos^{-1}\left(\frac{-2}{\sqrt{6}}\right) \approx 126^\circ \)
• \( \gamma = \cos^{-1}\left(\frac{-1}{\sqrt{6}}\right) \approx 116^\circ \)

Each angle is rounded to the nearest degree, making it easier to interpret. Using a calculator is typically helpful here, as dealing with inverse trigonometric functions can be cumbersome without one.