Question

A position vector has components \(x=34.6 \mathrm{~m}\) and \(y=-53.5 \mathrm{~m} .\) Find the vector's length and angle with the \(x\) -axis.

Short answer

Answer: The magnitude of the position vector is approximately 63.7 m, and its angle with the x-axis is approximately 303.04 degrees.

Step-by-step solution

Calculate the Magnitude of the Vector

To find the length (magnitude) of the position vector, we can use the Pythagorean theorem, which relates the magnitudes of the components \((x, y)\) to the magnitude of the vector \(R\) by the formula: \(R^2 = x^2 + y^2\). In our case, \(x = 34.6\) m, and \(y = -53.5\) m, so we need to plug these values into the equation and find the square root of the result to get the magnitude of the vector. \(R = \sqrt{x^2 + y^2} = \sqrt{34.6^2 + (-53.5)^2}\) \(R \approx \sqrt{(34.6)^2 + (-53.5)^2}\) \(R \approx \sqrt{1197.96 + 2862.25}\) \(R \approx \sqrt{4060.21}\) \(R \approx 63.7\) m So, the magnitude of the position vector is approximately \(63.7\) m.

Calculate the Angle with the x-axis

Now, we need to find the angle \(\theta\) between the position vector and the x-axis. To find this angle, we can use the arctangent function: \(\theta = \arctan(\frac{y}{x})\). It is important to note that we need to consider in which quadrant the vector is located. In our case, \(x = 34.6\) and \(y = -53.5\) (which is in the 4th quadrant). So, we should add \(360^\circ\) or \(2\pi\) radians (depending on units) to the arctangent to get the correct angle. \(\theta = \arctan(\frac{-53.5}{34.6})\) \(\theta \approx -56.96^\circ\) Since the calculated angle is negative, we can add \(360^\circ\) to get the angle in the range of \(0^\circ \leq \theta < 360^\circ\): \(\theta \approx 360^\circ - 56.96^\circ\) \(\theta \approx 303.04^\circ\) So, the angle with the x-axis is approximately \(303.04^\circ\).

Final Answer

The length of the position vector is approximately \(63.7\) m, and its angle with the x-axis is approximately \(303.04^\circ\).

Key concepts

Position Vector

A position vector is a directed line segment that points from one specified point to another. In the coordinate system, it describes how far and in which direction an object is from the origin. For example, if a vector is represented with coordinates \((x, y)\), it means the tip of the vector is located at these coordinates.

• In our original exercise, the position vector has the components \(x=34.6\) m and \(y=-53.5\) m.
• This means the vector starts at the origin \((0, 0)\) and ends at the point \((34.6, -53.5)\).

Understanding the components helps in visualizing the position and direction of the vector.

Magnitude Calculation

The magnitude of a vector represents its length and is a key component in vector analysis.
To calculate it, we use the Pythagorean theorem. This theorem applies to right triangles and relates the length of the hypotenuse to the lengths of the other two sides.
For a vector with components \((x, y)\), the formula is:

• \(R = \sqrt{x^2 + y^2}\)

In our case, substitute \(x = 34.6\) m and \(y = -53.5\) m,
we calculate:
\[R = \sqrt{(34.6)^2 + (-53.5)^2} \approx 63.7\text{ m}\]
Thus, the magnitude or length of the vector is approximately 63.7 meters.

Angle Determination

Finding the angle between the position vector and the x-axis helps us understand the vector's direction. The direction is important in determining how the force or movement is distributed across different axes.
We use the arctangent function for this purpose.
First, calculate the angle \(\theta\) as:

• \(\theta = \arctan\left(\frac{y}{x}\right)\)

Apply the values from our vector:
\[\theta = \arctan\left(\frac{-53.5}{34.6}\right)\approx -56.96^\circ\]
This angle is initially negative because the vector is in the fourth quadrant. We need to adjust it:
\[\theta \approx 360^\circ - 56.96^\circ \approx 303.04^\circ\]
This gives us a clearer idea of the vector's direction in standard position angle.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry that relates to right triangles. In the context of vectors, it allows us to determine the magnitude by treating the vector's components as the two shorter sides of a right triangle.
When you have a vector given by \((x, y)\) components:

• The vector's magnitude \(R\) is the hypotenuse of the triangle formed by the components.
• We use\(R^2 = x^2 + y^2\) to find this length.

This application of the theorem is crucial in breaking down the vector into manageable parts for calculations.

Arctangent Function

The arctangent function, often written as \(\arctan\), helps find an angle when given an opposite side and an adjacent side in right triangle geometry.
It's the inverse function of the tangent function.
How we use it in vector analysis:

• Take the ratio \(\frac{y}{x}\) of the vector's components.
• Apply \(\theta = \arctan\left(\frac{y}{x}\right)\) to find the angle with respect to the x-axis.

This function is critical in converting between Cartesian coordinates and direction angles. Remember, adjusting the angle based on the vector's quadrant is essential to ensure the angle correctly reflects the vector's orientation in the coordinate plane.