Problem 4 - Applied Mathematics

Question

Two vectors inclined at an angle \(\theta\) have magnitudes \(3 \mathrm{~N}\) and \(5 \mathrm{~N}\) and their resultant is of magnitude \(4 \mathrm{~N}\). The angle \(\theta\) is: (a) \(90^{\circ}\) (b) \(\arccos \frac{4}{5}\) (c) \(\arccos \frac{3}{5}\) (d) \(\arccos \frac{-3}{5}\) (e) \(60^{\circ}\)

Step-by-Step Solution

Answer

\(\theta = \arccos\left(-\frac{3}{5}\right)\)

Step 1: Understand the resultant vector formula

The magnitude of the resultant vector \textbf{R} of two vectors \textbf{A} and \textbf{B} inclined at an angle \theta can be found using the formula: \[ R = \sqrt{A^2 + B^2 + 2AB\cos\theta} \]

Step 2: Substitute the given values into the equation

The magnitudes of the vectors are given as \(3 \mathrm{~N}\) and \(5 \mathrm{~N}\), and the magnitude of the resultant is given as \(4 \mathrm{~N}\). Substitute these values into the formula: \[ 4 = \sqrt{3^2 + 5^2 + 2 \cdot 3 \cdot 5 \cdot \cos\theta} \]

Step 3: Simplify the equation

First, calculate the squares and the product: \[ 4 = \sqrt{9 + 25 + 30\cos\theta} \] \[ 4 = \sqrt{34 + 30\cos\theta} \]

Step 4: Solve for \(\cos\theta\)

Square both sides of the equation to remove the square root: \[ 16 = 34 + 30\cos\theta \] Solve the equation for \(\cos\theta\): \[ 16 - 34 = 30\cos\theta \] \[ -18 = 30\cos\theta \] \[ \cos\theta = -\frac{18}{30} \] \[ \cos\theta = -\frac{3}{5} \]

Step 5: Determine the angle \(\theta\)

Using the cosine inverse function, find \(\theta\): \[ \theta = \arccos\left(-\frac{3}{5}\right) \]

Key Concepts

vector magnitude

The magnitude of a vector represents its length. It is a scalar quantity that indicates how large or small the vector is, without considering its direction.
For any vector \textbf{A} represented by its components along the x, y, and z axes as \textbf{A} = [Ax, Ay, Az], the magnitude can be calculated using the formula: \[\|\textbf{A}\| = \sqrt{A_x^2 + A_y^2 + A_z^2}\]
This formula is derived from the Pythagorean theorem and can be applied in two or three-dimensional space. For instance, if we have a vector \textbf{A} with components 3N and 4N in the x and y directions respectively, its magnitude would be: \[\|\textbf{A}\| = \sqrt{3^2 + 4^2} = 5N\]
In the exercise, vectors are given directly as magnitudes of 3N and 5N. So, these values are used directly in further calculations.

angle between vectors

The angle between two vectors, when they are inclined to each other, determines how they interact to form a resultant vector. This angle, often denoted as \(\theta\), plays a crucial role.
To find the resultant vector \textbf{R}, we use the law of cosines for vectors: \[\|\textbf{R}\| = \sqrt{A^2 + B^2 + 2AB\cos\theta}\]
Here, \(\cos\theta\) bridges the relationship between the magnitudes of the vectors and the angle between them.
In our problem, substituting the magnitudes we get: \[4 = \sqrt{3^2 + 5^2 + 2 \cdot 3 \cdot 5 \cdot \cos\theta} \]
This simplifies further to find the angle between the vectors. The angle \(\theta\) directly affects the resultant vector's magnitude.

cosine inverse

The cosine inverse function \(\arccos\) is used to find the angle whose cosine value is known.
For any given value of \(\cos\theta\), the angle \(\theta\) is: \[\theta = \arccos(\cos\theta) \]
In our problem, after solving the equation for \(\cos\theta\), we get: \[\theta = \arccos\left(-\frac{3}{5}\right) \]
Using the cosine inverse function, we can determine \(\theta\) precisely. The value \(-\frac{3}{5}\) provides the cosine of the angle directly. Remember, the function \(\arccos\) gives the principal value of the angle in the range \([0,\pi] \) radians or \([0^{\circ}, 180^{\circ}]\).
The calculation and understanding of \(\cos\theta\) and \(\arccos\) are fundamental in finding the precise angle between vectors.