Problem 38 - Precalculus Mathematics for Calculus

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° (x-axis), 125° (y-axis), and 114° (z-axis).

Step 1: Determine the components of the vector

The given vector is expressed as \( \mathbf{i} - 2 \mathbf{j} - \mathbf{k} \). This means the vector has components along the x, y, and z axes. The components are \( (1, -2, -1) \) respectively.

Step 2: Calculate the magnitude of the vector

The magnitude \( \|\mathbf{v}\| \) of a vector \( \mathbf{v} = (x, y, z) \) is calculated using the formula \( \sqrt{x^2 + y^2 + z^2} \). Substituting the components, we have:\[ \|\mathbf{v}\| = \sqrt{1^2 + (-2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

Step 3: Calculate the direction cosine for x-axis

The direction cosine \( \cos \alpha \) for the angle with the x-axis is given by:\[ \cos \alpha = \frac{x}{\|\mathbf{v}\|} = \frac{1}{\sqrt{6}} \]Thus, \( \alpha \) is the angle whose cosine is \( \frac{1}{\sqrt{6}} \).

Step 4: Calculate the direction angle for x-axis

Using a calculator to find \( \alpha \), take the inverse cosine:\[ \alpha = \cos^{-1}\left(\frac{1}{\sqrt{6}}\right) \approx 65.91^\circ \]Rounding to the nearest degree, \( \alpha \approx 66^\circ \).

Step 5: Calculate the direction cosine for y-axis

The direction cosine \( \cos \beta \) for the angle with the y-axis is:\[ \cos \beta = \frac{y}{\|\mathbf{v}\|} = \frac{-2}{\sqrt{6}} \]Thus, \( \beta \) is the angle whose cosine is \( \frac{-2}{\sqrt{6}} \).

Step 6: Calculate the direction angle for y-axis

Using a calculator to find \( \beta \), take the inverse cosine:\[ \beta = \cos^{-1}\left(\frac{-2}{\sqrt{6}}\right) \approx 125.26^\circ \]Rounding to the nearest degree, \( \beta \approx 125^\circ \).

Step 7: Calculate the direction cosine for z-axis

The direction cosine \( \cos \gamma \) for the angle with the z-axis is:\[ \cos \gamma = \frac{z}{\|\mathbf{v}\|} = \frac{-1}{\sqrt{6}} \]Thus, \( \gamma \) is the angle whose cosine is \( \frac{-1}{\sqrt{6}} \).

Step 8: Calculate the direction angle for z-axis

Using a calculator to find \( \gamma \), take the inverse cosine:\[ \gamma = \cos^{-1}\left(\frac{-1}{\sqrt{6}}\right) \approx 114.09^\circ \]Rounding to the nearest degree, \( \gamma \approx 114^\circ \).

Key Concepts

Vector Components

Vectors are essential parts of mathematics that describe both direction and magnitude in a space. In the case of a three-dimensional vector like \( \mathbf{i} - 2 \mathbf{j} - \mathbf{k} \), it is represented by three components. These components define the direction along the x, y, and z axes. This is like saying that the vector moves:

• 1 unit in the x direction,
• -2 units in the y direction, and
• -1 unit in the z direction.

Understanding vector components helps break down a vector's movement across each axis, which is fundamental for calculating its magnitude and direction angles.

Vector Magnitude

The vector magnitude is an essential property that measures the length or size of a vector. It indicates how far the vector extends in space regardless of its direction. To calculate the magnitude of a vector \((x, y, z)\), we use the formula \[\|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2}\]For our vector \( \mathbf{i} - 2 \mathbf{j} - \mathbf{k} \), the components are \((1, -2, -1)\). As such, the magnitude calculation is:\[\|\mathbf{v}\| = \sqrt{1^2 + (-2)^2 + (-1)^2} = \sqrt{6}\]This result \(\sqrt{6}\) represents the vector's length.

Direction Cosines

Direction cosines are critical in understanding the orientation of a vector. They are the cosines of the angles the vector makes with each of the coordinate axes. Essentially, direction cosines help indicate the efficiency of the vector components along each axis. For a vector \(\mathbf{v} = (x, y, z)\), direction cosines \(\cos \alpha, \cos \beta, \cos \gamma\) with the x, y, and z axes respectively can be calculated using\[\cos \alpha = \frac{x}{\|\mathbf{v}\|}, \quad \cos \beta = \frac{y}{\|\mathbf{v}\|}, \quad \cos \gamma = \frac{z}{\|\mathbf{v}\|}\]For our example vector,

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

These ratios help determine the direction angles concerning each axis.

Inverse Cosine

Inverse cosine, often expressed as \(\cos^{-1}\), is a trigonometric function used to find angles when the value of cosine is known. It serves as a powerful tool to derive direction angles from direction cosines. Once you have calculated the direction cosines:

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

you use the inverse cosine function to find the actual angles. This process involves a calculator or an inverse cosine table:

• \(\alpha = \cos^{-1}\left(\frac{1}{\sqrt{6}}\right) \approx 66^\circ\)
• \(\beta = \cos^{-1}\left(\frac{-2}{\sqrt{6}}\right) \approx 125^\circ\)
• \(\gamma = \cos^{-1}\left(\frac{-1}{\sqrt{6}}\right) \approx 114^\circ\)

These angles quantify the vector's direction in three-dimensional space.