Question

The \(x\) and \(y\) components of a vector \(\overrightarrow{\mathbf{r}}\) are \(r_{x}=14 \mathrm{~m}\) and \(r_{y}=-9.5 \mathrm{~m}\), respectively. Find (a) the direction and (b) the magnitude of the vector \(\overrightarrow{\mathbf{r}}\). (c) If both \(r_{x}\) and \(r_{y}\) are doubled, how do your answers to parts (a) and (b) change?

Short answer

The magnitude of \( \overrightarrow{\mathbf{r}} \) is approximately 16.92 m, and its direction is \(-34.38^\circ\). Doubling the components doubles the magnitude to approximately 33.83 m without changing the direction.

Step-by-step solution

Calculate Magnitude

The magnitude of a vector \( \overrightarrow{\mathbf{r}} \) with components \( r_x \) and \( r_y \) can be calculated using the Pythagorean theorem: \[ |\overrightarrow{\mathbf{r}}| = \sqrt{r_x^2 + r_y^2} \]Substitute \( r_x = 14 \; m \) and \( r_y = -9.5 \; m \):\[ |\overrightarrow{\mathbf{r}}| = \sqrt{14^2 + (-9.5)^2} = \sqrt{196 + 90.25} = \sqrt{286.25} \approx 16.92 \; m \]

Calculate Direction

The direction of the vector \( \overrightarrow{\mathbf{r}} \) is given by the angle \( \theta \) with the positive x-axis. The angle can be calculated using the formula:\[ \theta = \tan^{-1}\left(\frac{r_y}{r_x}\right) \]Substitute \( r_x = 14 \; m \) and \( r_y = -9.5 \; m \):\[ \theta = \tan^{-1}\left(\frac{-9.5}{14}\right) \approx \tan^{-1}(-0.6786) \approx -34.38^\circ \] Since angle is negative, it indicates the vector is below the x-axis.

Doubling the Components

When both \( r_x \) and \( r_y \) are doubled, the new components become \( 2r_x = 28 \; m \) and \( 2r_y = -19 \; m \).

Calculate New Magnitude

The new magnitude \(|\overrightarrow{\mathbf{r'}}| \) is calculated as:\[ |\overrightarrow{\mathbf{r'}}| = \sqrt{(2r_x)^2 + (2r_y)^2} = \sqrt{28^2 + (-19)^2} = \sqrt{784 + 361} = \sqrt{1145} \approx 33.83 \; m \]Notice this is twice the original magnitude.

Calculate New Direction

The direction with the components doubled uses the same formula for angle:\[ \theta' = \tan^{-1}\left(\frac{2r_y}{2r_x}\right) \]Since both the numerator and denominator are doubled, the argument of the tangent function doesn't change:\[ \theta' = \tan^{-1}\left(\frac{-19}{28}\right) \approx \tan^{-1}(-0.6786) \approx -34.38^\circ \]The angle remains the same.

Key concepts

Vector Magnitude

The magnitude of a vector essentially gives you an idea of how long the vector is; it is a measure of its size or length. To determine the magnitude we use the Pythagorean theorem, which is rooted in basic geometry. For a two-dimensional vector with components \(r_x\) and \(r_y\), the formula to find the magnitude is:

• \( |\overrightarrow{\mathbf{r}}| = \sqrt{r_x^2 + r_y^2} \)

This formula effectively measures the hypotenuse of a right-angled triangle, where \(r_x\) and \(r_y\) are perpendicular to each other.

In the original exercise, substituting \(r_x = 14 \, m\) and \(r_y = -9.5 \, m\) gives us \( \sqrt{286.25} \), resulting in a magnitude of approximately \(16.92 \, m\). When both components are doubled, the magnitude also doubles, as each side of the 'triangle' grows proportionately. Hence, the new magnitude becomes approximately \(33.83 \, m\).

Vector Direction

The direction of a vector is as crucial as its magnitude; it tells us where the vector is pointing. This is represented by the angle \(\theta\) the vector makes with the positive x-axis.

To find this angle, we utilize the tangent function from trigonometry, formulated as:

• \( \theta = \tan^{-1}\left(\frac{r_y}{r_x}\right) \)

Here, the angle is determined by the ratio of the vector's y-component to its x-component.

In the exercise, plugging in the values \(r_x = 14 \, m\) and \(r_y = -9.5 \, m\) results in an angle of \(-34.38^\circ\), indicating the vector points below the x-axis. Upon doubling both components, the ratio remains unchanged, hence the direction or angle \(\theta\) stays the same.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in mathematics that describes the relationship between the sides of a right triangle. It states that for a right triangle:

• The square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Mathematically, this is expressed as \(c^2 = a^2 + b^2\).

In the context of vectors, the theorem helps us determine the magnitude by treating the vector's components as perpendicular sides of a right triangle. For the given vector \(\overrightarrow{\mathbf{r}}\), this theorem allows us to calculate its magnitude through the components' squares, ensuring the formulas in vector mathematics are rooted firmly in geometric principles.

Trigonometry in Vectors

Trigonometry is heavily utilized in the study of vectors to understand their directions and magnitudes in different dimensions. It involves using sine, cosine, and tangent functions based on angle measures.

In vector mathematics, particularly for direction calculation, the tangent function \( \tan(\theta) \) is commonly utilized to express the angle relative to the x-axis. By solving \( \theta = \tan^{-1}\left(\frac{r_y}{r_x}\right) \), one can determine the vector's direction.

Trigonometry also gets applied in converting vectors between rectangular components (x and y) and polar form (magnitude and angle). Understanding these conversions is criticalto breaking down complex physics and engineering problems, ensuring clarity in vector analysis.