Question

(a) What are the components of a vector $\overrightarrow{\mathbf{a}}$ in the $xy$ plane if its direction is ${252}^{\circ }$ counterclockwise from the positive $x$ axis and its magnitude is $7.34$ units? $(b)$ The $x$ component of a certain vector is $-25$ units and the $y$ component is $+43$ units. What are the magnitude of the vector and the angle between its direction and the positive $x$ axis?

Short answer

The components of the first vector are approximately $a_x=-3.67$ and $a_y=-6.67$. The magnitude of the second vector is approximately 49.39 units and its direction/angle from the positive x-axis is approximately 211 degrees (or 3.68 radians) in the counterclockwise direction.

Step-by-step solution

Find vector components

The components of a vector can be calculated using the formula $a_x=r\cos (\theta )$ for the x-component and $a_y=r\sin (\theta )$ for the y-component. Here, $r$ is the magnitude (7.34 units) and $\mathrm{\Theta }$ is the angle from the positive x-axis (252 degrees). However, the angle should be converted to radians first: ${252}^{\circ }=252\times \frac{\pi }{180}=4.4$ rad. Hence, the components are $$a_x=7.34\cos (4.4)$$ $$a_y=7.34\sin (4.4)$$

Calculate the magnitude and direction of vector

The magnitude of a vector is calculated as $r=\sqrt{x^2+y^2}$, where $x$ and $y$ are the x, y components of the vector respectively. Substituting the given values: $$r=\sqrt{(-25{)}^2+(+43{)}^2}$$ Now, the direction or angle $\mathrm{\Theta }$ from positive x-axis can be calculated using the formula $\mathrm{\Theta }={\tan }^{-1}(\frac{y}{x})$. Hence substituting the given values, the direction is $\mathrm{\Theta }={\tan }^{-1}(\frac{43}{-25})$. Please note because x is negative we need to add $\pi$ (or 180 degrees) to the arctan result. This is necessary due to the properties of arctan function.

Key concepts

Vector Magnitude

The magnitude of a vector gives us a measure of how long or strong the vector is. You can think of it as the "length" of the vector in its space. To find the magnitude, use the formula $r=\sqrt{x^2+y^2}$, where $x$ and $y$ are the components of the vector along the x and y axes, respectively.
This formula is derived from the Pythagorean theorem, as we're essentially finding the hypotenuse of a right triangle. In the exercise, if you have a vector with components $-25$ and $+43$, the magnitude would be $r=\sqrt{(-25{)}^2+(43{)}^2}$.
Breaking down the formula:

• Square each component: $(-25{)}^2$ and ${43}^2$
• Add the squares together: $625+1849=2474$
• Take the square root to find the magnitude: $\sqrt{2474}\approx 49.74$

With the magnitude, you gain a better understanding of the vector's extent in space, regardless of its direction.

Vector Direction

Understanding the direction of a vector is crucial as it tells you "which way" the vector points. This direction is typically measured as an angle from a reference axis. In our exercise, we use the positive x-axis as the reference. For any vector with known $x$ and $y$ components, the direction $\mathrm{\Theta }$ can be calculated using the inverse tangent function: $\mathrm{\Theta }={\tan }^{-1}\left(\frac{y}{x}\right)$.
This angle calculation works well when your vector is in the first quadrant. However, as in part (b) of the exercise, when the x-component is negative, it means the vector might be pointing to the left of the y-axis or into the second or third quadrants. In such cases, it is essential to adjust the angle by adding ${180}^{\circ }$ or $\pi$ radians to find the correct.angle
So, for a vector with components $-25$ and $+43$, you get $\mathrm{\Theta }={\tan }^{-1}\left(\frac{43}{-25}\right)$. After using a calculator to find the arctan of $1.72$, you add $\pi$ to move the angle into the correct quadrant.

Trigonometric Functions

Trigonometric functions are handy tools in vector analysis and help in breaking down a vector into its components. Three primary trigonometric functions are significant when dealing with vectors: sine $\sin$, cosine $\cos$, and tangent $\tan$.
To decompose a vector into its components, use:

• $a_x=r\cos (\theta )$: This calculates the x-component.
• $a_y=r\sin (\theta )$: This calculates the y-component.

Here, $r$ is the vector's magnitude, and $\theta$ is its direction (angle from the positive x-axis). For example, if you have a magnitude of $7.34$ units and a direction of ${252}^{\circ }$ (or $4.4$ radians), converting this angle from degrees to radians is necessary since most trigonometric calculators use radians.
In cases like this exercise, you use the formulas: $a_x=7.34\cos (4.4)$ and $a_y=7.34\sin (4.4)$ to get the respective components of the vector in the x and y directions. By mastering these functions, breaking down vectors becomes an easier task, making it a fundamental skill in physics and engineering.