Question

Find the magnitude and direction of each of the following vectors, which are given in terms of their \(x\) - and \(y\) -components: \(\vec{A}=(23.0,59.0)\), and \(\vec{B}=(90.0,-150.0)\).

Short answer

Question: Given the vectors \( \vec{A} = (23.0, 59.0) \) and \( \vec{B} = (90.0, -150.0) \), find their magnitudes and directions. Answer: The magnitude and direction of vector \(\vec{A}\) are: \(|\vec{A}| \approx 23.03\) and \(\theta_A \approx 68.92^{\circ}\); The magnitude and direction of vector \(\vec{B}\) are: \(|\vec{B}| \approx 177.49\) and \(\theta_B \approx -59.04^{\circ}\).

Step-by-step solution

Calculate the magnitude of both vectors.

For the magnitude of a vector, we can use the formula \(|\vec{A}| = \sqrt{A_x^2 + A_y^2}\) and \(|\vec{B}| = \sqrt{B_x^2 + B_y^2}\). Now calculate the magnitudes for both vectors: \(|\vec{A}| = \sqrt{(23.0)^2 + (59.0)^2}\) \(|\vec{A}| = \sqrt{530} \approx 23.03\) \(|\vec{B}| = \sqrt{(90.0)^2 + (-150.0)^2}\) \(|\vec{B}| = \sqrt{31,500} \approx 177.49\)

Calculate the direction of both vectors.

For the direction of a vector, we can use the arctangent formula \(\theta_A = \text{tan}^{-1}(\frac{A_y}{A_x})\) and \(\theta_B = \text{tan}^{-1}(\frac{B_y}{B_x})\). Now calculate the directions for both vectors (in degrees): \(\theta_A = \text{tan}^{-1}(\frac{59.0}{23.0})\) \(\theta_A = \text{tan}^{-1}(2.56) \approx 68.92^{\circ}\) \(\theta_B = \text{tan}^{-1}(\frac{-150.0}{90.0})\) \(\theta_B = \text{tan}^{-1}(-1.67) \approx -59.04^{\circ}\) Note: The negative angle for \(\vec{B}\) means its direction is below the \(x\)-axis. Now we have the magnitude and direction for both given vectors: • The magnitude and direction of vector \(\vec{A}\) are: \(|\vec{A}| \approx 23.03\) and \(\theta_A \approx 68.92^{\circ}\) • The magnitude and direction of vector \(\vec{B}\) are: \(|\vec{B}| \approx 177.49\) and \(\theta_B \approx -59.04^{\circ}\).

Key concepts

Magnitude of a Vector

Understanding the magnitude of a vector is an essential concept in physics and mathematics. It refers to the size or length of a vector and is a measure of how far the vector 'stretches'. This single number captures the vector's overall impact without considering its direction.

To calculate the magnitude, we use the Pythagorean Theorem applied to the vector’s components. If a vector \(\vec{V}\) has components along the x-axis and y-axis—denoted as \(V_x\) and \(V_y\), respectively—the magnitude \(|\vec{V}|\) is found by taking the square root of the sum of the squares of its components:
\[|\vec{V}| = \sqrt{V_x^2 + V_y^2}\].

This is analogous to finding the hypotenuse of a right triangle whose sides are the vector components. For positive or negative component values, the magnitude remains positive, affirming that it simply denotes size, unrelated to direction.

Direction of a Vector

Whereas magnitude measures size, the direction of a vector tells us the orientation of the vector in space. Direction is crucial because it provides us with the information on where the vector is pointing. It is often described using angles. In a two-dimensional space, this is an angle from a reference axis (typically the positive x-axis) to the vector and is measured counterclockwise in mathematical contexts.

The direction can be calculated using trigonometry, especially when the vector's components along the x- and y-axis are known. The tangent of the angle \(\theta\) that the vector makes with the x-axis is equal to the ratio of the y-component to the x-component of the vector. Hence, we take the arctangent of this ratio to find \(\theta\).

It is worth noting that the direction can be interpreted as an angle ranging from 0 to 360 degrees or from 0 to \(2\pi\) radians, depending on the context. Positive angles usually indicate measurements counterclockwise from the reference axis, and negative values indicate a clockwise direction.

Vector Components

Vector components serve as the building blocks of a vector. They represent the vector's influence in each dimension of the space it occupies. In two dimensions, a vector can be divided into two perpendicular components: one in the direction of the x-axis (horizontal) and one in the direction of the y-axis (vertical).

The components \(V_x\) and \(V_y\) of a vector \(\vec{V}\) can be understood as projections onto the axes. If you imagine a right triangle with the vector as the hypotenuse, \(V_x\) would be the adjacent side to the angle, and \(V_y\) would be the opposite side.

These components are fundamental because they allow us to apply vector operations like addition, subtraction, and determining the resultant in a straightforward, component-wise manner. They also simplify calculations of many other properties, such as dot products, cross products, and even physical quantities like velocity, force, and displacement.

Arctangent Formula

The arctangent formula is a crucial trigonometric function for determining the angle whose tangent is a given value. It is the inverse operation of the tangent function and is often written as \(\text{tan}^{-1}\) or \(\text{arctan}\).

In the context of vectors, the arctangent formula is used to find the direction angle from the vector's components. By taking the arctangent of the ratio of the y-component to the x-component, we can resolve the angle in radians or degrees:
\[\theta = \text{tan}^{-1}\left(\frac{V_y}{V_x}\right)\].

Importantly, because the tangent function has a period of \(\pi\) radians (180 degrees), it can produce the same value for two different angles within one cycle. Therefore, one must consider the signs of the vector components to determine the correct quadrant for the angle. The outcome from a calculator or computer might only provide an angle between -90 and +90 degrees (-\(\pi/2\) and +\(\pi/2\) radians), which is why additional consideration of the vector’s quadrant is necessary to find the true direction.