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

Answer: The magnitude of \(\vec{A}\) is \(\sqrt{5186}\) and its direction is approximately \(68.56^\circ\). The magnitude of \(\vec{B}\) is \(\sqrt{31500}\) and its direction is approximately \(-59.04^\circ\).

Step-by-step solution

Find the magnitude of \(\vec{A}\)

To find the magnitude of \(\vec{A}=(23.0, 59.0)\), we use the formula, $$ |\vec{A}| = \sqrt{A_x^2 + A_y^2} $$ Where \(A_x = 23.0\) and \(A_y = 59.0\). Insert the values into the formula: $$ |\vec{A}| = \sqrt{(23.0)^2 + (59.0)^2} $$ Calculate the magnitude: $$ |\vec{A}| = \sqrt{5186} $$ The magnitude of \(\vec{A}\) is \(\sqrt{5186}\).

Find the direction of \(\vec{A}\)

To find the direction of \(\vec{A}\), we use the arctangent function. $$ \theta_A = \arctan \left( \frac{A_y}{A_x} \right) $$ Insert the values of components \(A_x\) and \(A_y\): $$ \theta_A = \arctan \left( \frac{59.0}{23.0} \right) $$ Calculate the direction: $$ \theta_A \approx 68.56^\circ $$ The direction of vector \(\vec{A}\) is approximately \(68.56^\circ\). Now, we will find the magnitude and direction of the vector \(\vec{B}\).

Find the magnitude of \(\vec{B}\)

To find the magnitude of \(\vec{B}=(90.0, -150.0)\), we again use the formula, $$ |\vec{B}| = \sqrt{B_x^2 + B_y^2} $$ Where \(B_x = 90.0\) and \(B_y = -150.0\). Insert the values into the formula: $$ |\vec{B}| = \sqrt{(90.0)^2 + (-150.0)^2} $$ Calculate the magnitude: $$ |\vec{B}| = \sqrt{31500} $$ The magnitude of \(\vec{B}\) is \(\sqrt{31500}\).

Find the direction of \(\vec{B}\)

To find the direction of \(\vec{B}\), we again use the arctangent function. $$ \theta_B = \arctan \left( \frac{B_y}{B_x} \right) $$ Insert the values of components \(B_x\) and \(B_y\): $$ \theta_B = \arctan \left( \frac{-150.0}{90.0} \right) $$ Calculate the direction: $$ \theta_B \approx -59.04^\circ $$ The direction of vector \(\vec{B}\) is approximately \(-59.04^\circ\). In conclusion, the magnitude and direction of vector \(\vec{A}\) are \(\sqrt{5186}\) and \(68.56^\circ\), while the magnitude and direction of vector \(\vec{B}\) are \(\sqrt{31500}\) and \(-59.04^\circ\).

Key concepts

Magnitude of Vectors

The magnitude of a vector represents its length or size. Think about it as how long a vector is on a graph. When working with vectors in the plane, often represented by their \(x\) and \(y\) components as \( (x, y) \), we can use the Pythagorean theorem to find the magnitude. The formula for the magnitude \(|\vec{V}|\) of a vector \(\vec{V}\) with components \(V_x\) and \(V_y\) is:
\[|\vec{V}| = \sqrt{V_x^2 + V_y^2}\]
Here's the process in simple steps:

• Square each component \(V_x\) and \(V_y\).
• Add the squares together.
• Take the square root of the sum.

For example, given a vector \(\vec{A} = (23.0, 59.0)\), we calculate its magnitude by substituting into the formula:

• \(23.0^2 = 529.0\)
• \(59.0^2 = 3481.0\)
• Add them to get \(4010.0\)
• Then compute \(\sqrt{4010.0}\) for the magnitude, which gives the length of \(\vec{A}\) in its graphical representation.

Understanding vector magnitude is crucial as it helps describe the strength or effect of vectors, similar to how one would describe the intensity of a force.

Direction of Vectors

The direction of a vector is like the arrow on a compass; it tells us where the vector is pointing. To determine this direction, we use the angle \(\theta\) that the vector makes with the positive \(x\)-axis. Calculating this angle helps us understand the orientation of the vector in the plane.
The direction is often measured using the arctangent function of the ratio of the \(y\) component over the \(x\) component:
\[\theta = \arctan \left( \frac{V_y}{V_x} \right)\]
For example, for vector \(\vec{A} = (23.0, 59.0)\), the direction can be calculated as follows:

• Find the ratio \(\frac{59.0}{23.0}\).
• Compute \(\arctan(\frac{59.0}{23.0})\) to find the angle \(\theta\).

This calculation results in an angle \(\theta_A \approx 68.56^\circ\), showing that \(\vec{A}\) points upward and to the right.
Understanding vector direction is essential in fields like physics where knowing an object's path and orientation is crucial.

Arctangent Function

The arctangent function, denoted as \(\arctan\), is a trigonometric function that helps you find the angle whose tangent is a given number. In simpler terms, if you know the ratio of the opposite side to the adjacent side in a right triangle, \(\arctan\) helps you find the angle.
When working with vectors, this is useful for determining the vector's direction relative to the \(x\)-axis. For any vector with \(y\) and \(x\) components, the arctangent calculates the angle \(\theta\) through:
\[\theta = \arctan \left( \frac{y}{x} \right)\]
Here are key points about the \(\arctan\) function:

• It returns the angle in radians, but you can convert to degrees if needed (\(1\) radian \(\approx 57.3^\circ\)).
• The function considers angle directions, so depending on the sign of \(y\) and \(x\), \(\theta\) might need adjustments (like adding 180 degrees for vectors in certain quadrants).

For example, for the vector \(\vec{B} = (90.0, -150.0)\), we calculated

• \(\theta_B = \arctan\left(\frac{-150.0}{90.0}\right) \approx -59.04^\circ\)

This negative angle indicates a direction below the \(x\)-axis, which may need adjusting based on context.