Question

Two vectors \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{s}}\) lie in the \(x y\) plane. Their magnitudes are \(r=4.5\) units and \(s=7.3\) units. Their directions are, respectively, \(320^{\circ}\) and \(85^{\circ}\) measured counterclockwise from the positive \(x\) axis. Find the magnitude and the direction of \(\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{s}}\)

Short answer

The magnitude of vector \(\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{s}}\) is calculated as \(4.5 \times 7.3 \times |\sin((320 - 85) \times \frac{\pi}{180})|\). The direction of this cross product vector in the \(xy\)-plane is not defined as it is perpendicular to this plane.

Step-by-step solution

Convert angles to radians

Angular measures are converted from degrees to radians. For \(\overrightarrow{\mathbf{r}}\) this is \(\theta_r = 320^{\circ} \times \frac{\pi}{180} = 320 \times \frac{\pi}{180}\) radians and for \(\overrightarrow{\mathbf{s}}\) we get \(\theta_s = 85^{\circ} \times \frac{\pi}{180} = 85 \times \frac{\pi}{180}\) radians.

Find the x and y components of \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{s}}\)

The x and y components of the vectors can be found using their magnitudes and the cosine and sine of their angles. For \(\overrightarrow{\mathbf{r}}\), \(r_x = r\cos(\theta_r) = 4.5\cos(320 \times \frac{\pi}{180})\) and \(r_y = r\sin(\theta_r) = 4.5\sin(320 \times \frac{\pi}{180})\). Similarly, for \(\overrightarrow{\mathbf{s}}\), \(s_x = s\cos(\theta_s) = 7.3\cos(85 \times \frac{\pi}{180})\) and \(s_y = s\sin(\theta_s) = 7.3\sin(85 \times \frac{\pi}{180})\).

Calculate \(\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{s}}\) and its magnitude

The cross product of two vectors \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{s}}\), denoted by \(\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{s}}\), is given by \(|\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{s}}| = rs\sin(\theta_s - \theta_r) = 4.5 \times 7.3 \times |\sin((320 - 85) \times \frac{\pi}{180})|\).

Calculate the direction

The direction \(\theta_{rs}\) of this cross product vector is obtained by taking the arctangent of the ratio of its y and x components. Since the cross product of two vectors in the \(xy\)-plane is a vector which is perpendicular to this plane and thus, doesn't lie in the same plane of \(x\) and \(y\), its direction in the \(xy\)-plane is not defined.

Key concepts

Angular Conversion from Degrees to Radians

Understanding the conversion between degrees and radians is vital in various branches of mathematics and physics where rotational or cyclic phenomena are involved. Degrees and radians are both units used to measure angles. One complete revolution around a circle is equal to 360 degrees or \(2\pi\) radians. To convert from degrees to radians, we multiply the angle in degrees by \(\frac{\pi}{180}\).For instance, a degree measurement of \(320^\circ\) is converted to radians as \(320^\circ \times \frac{\pi}{180}\), resulting in approximately \(5.59\) radians. This conversion forms the cornerstone for utilizing trigonometric functions in physics, as most scientific calculations require angles to be in radians.
Let's put this into practice with the provided exercise. For vectors \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{s}}\), with directions of \(320^\circ\) and \(85^\circ\) respectively, the angular conversion becomes crucial for determining their components in a plane, a step that's necessary before we can calculate a cross product.

Vector Components in a Plane

Vectors are mathematical entities with both magnitude and direction, and they are essential in representing physical quantities like force, velocity, and displacement. In a two-dimensional plane, a vector \(\overrightarrow{\mathbf{v}}\) with magnitude \(v\) and angle \(\theta\) from the positive \(x\)-axis can be resolved into its horizontal (\(x\)-component) and vertical (\(y\)-component) components using trigonometry.
The \(x\)-component is found using \(v_x = v\cos(\theta)\) and the \(y\)-component using \(v_y = v\sin(\theta)\). In the context of our exercise, finding the components of \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{s}}\) using their magnitudes and the angles in radians is the second step after converting the angles from degrees to radians. This process is critical for further operations, such as finding the cross product.

Magnitude of a Cross Product

The cross product of two vectors in three-dimensional space is a vector that is perpendicular to the plane formed by the original vectors. In a plane, the magnitude of the cross product can tell us about the area of the parallelogram spanned by the two vectors. The formula to compute the magnitude of the cross product \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}\) is \( |\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}| = ab\sin(\theta)\), where \(a\) and \(b\) are the magnitudes of the vectors and \(\theta\) is the angle between them.

Application in the Exercise

In our example, the cross product magnitude for vectors \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{s}}\) is calculated using their magnitudes and the sine of the difference between their angles. This value provides not only an insight into the relationship between vectors but also into physical properties such as torque whenever two forces are applied at an angle.

Trigonometry in Physics

Trigonometry's role in physics cannot be overstated—it is fundamental in describing the relationships between the angles and sides of triangles and is deeply embedded in the study of waves, oscillations, and circular motion.
In our case, using trigonometric functions like sine and cosine to determine the components of vectors lays the groundwork for calculating the cross product in a plane. Moreover, trigonometric principles help in analyzing the resultant direction of forces, the motion of particles on a circular path, and much more. While the concept of direction for the cross product does not apply in the same way it does for vectors lying in the plane, recognizing the application of trigonometry in physics helps students understand how a cross product relates to rotational motion and other three-dimensional phenomena.