Understanding Unit Vectors and Magnitude Calculation

This problem involves unit vectors and their magnitudes when added and subtracted. A unit vector is simply a vector with a magnitude of 1. Let's denote the two unit vectors as \(\vec{A}\) and \(\vec{B}\). Since they are unit vectors, their magnitudes are given by:

• \(|\vec{A}| = 1\)
• \(|\vec{B}| = 1\)

We are given that the magnitude of the difference between these two unit vectors is \(\sqrt 3\). This means:

\(|\vec{A} - \vec{B}| = \sqrt 3\)

We need to find the magnitude of the sum of these two vectors, which is \(|\vec{A} + \vec{B}|\).

Vector Magnitude Formulas

For any two vectors \(\vec{A}\) and \(\vec{B}\) with an angle \(\theta\) between them, the magnitudes of their sum and difference are given by the following formulas:

• Magnitude of Sum: \(|\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta\)
• Magnitude of Difference: \(|\vec{A} - \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|\cos\theta\)

Step-by-Step Calculation

First, let's use the given information about the magnitude of the difference to find the angle \(\theta\) between the unit vectors. We substitute the known values into the magnitude of difference formula:

\(|\vec{A} - \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|\cos\theta\)

We know \(|\vec{A} - \vec{B}| = \sqrt 3\), \(|\vec{A}| = 1\), and \(|\vec{B}| = 1\). Plugging these in:

\((\sqrt 3)^2 = (1)^2 + (1)^2 - 2(1)(1)\cos\theta\)

\(3 = 1 + 1 - 2\cos\theta\)

\(3 = 2 - 2\cos\theta\)

Now, we solve for \(\cos\theta\):

\(3 - 2 = -2\cos\theta\)

\(1 = -2\cos\theta\)

\(\cos\theta = -\frac{1}{2}\)

The angle \(\theta\) between the two unit vectors is the angle whose cosine is \(-\frac{1}{2}\). This angle is \(\theta = 120^\circ\) or \(\frac{2\pi}{3}\) radians.

Now that we have the angle between the vectors, we can find the magnitude of their sum using the magnitude of sum formula:

\(|\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta\)

Substitute the values \(|\vec{A}| = 1\), \(|\vec{B}| = 1\), and \(\cos\theta = -\frac{1}{2}\):

\(|\vec{A} + \vec{B}|^2 = (1)^2 + (1)^2 + 2(1)(1)(-\frac{1}{2})\)

\(|\vec{A} + \vec{B}|^2 = 1 + 1 - 1\)

\(|\vec{A} + \vec{B}|^2 = 1\)

To find \(|\vec{A} + \vec{B}|\), we take the square root of both sides:

\(|\vec{A} + \vec{B}| = \sqrt{1}\)

\(|\vec{A} + \vec{B}| = 1\)

The magnitude of the sum of the two unit vectors is 1 unit.

Result Summary

Given two unit vectors \(\vec{A}\) and \(\vec{B}\):

The magnitude of the sum of the two unit vectors is 1 unit.

Revision Table: Key Vector Concepts

Additional Information: Vector Calculations and Unit Vectors

Understanding unit vectors is fundamental in physics and mathematics. They provide a convenient way to express direction without affecting the magnitude of other quantities. For example, a velocity vector can be written as the product of its speed (magnitude) and a unit vector in the direction of motion.

The formulas for the magnitude of the sum and difference of two vectors are derived from the law of cosines applied to the triangle formed by the two vectors and their resultant (sum or difference). When dealing with unit vectors, these formulas simplify because \(|\vec{A}|\) and \(|\vec{B}|\) are both 1.

In this specific problem, the fact that the magnitude of the difference is \(\sqrt 3\) is a strong hint about the angle between the unit vectors. The case where two unit vectors have a difference magnitude of \(\sqrt 3\) corresponds to an angle of \(120^\circ\) between them. Similarly, if the difference magnitude were 1, the angle would be \(60^\circ\).

Knowing the angle allows us to calculate the magnitude of the sum. In this particular scenario (\(\theta = 120^\circ\) for difference \(\sqrt{3}\)), the magnitude of the sum happens to be 1, which means the sum vector is also a unit vector. This is not a general rule; the magnitude of the sum of two unit vectors can range from 0 (if they are opposite, \(\theta = 180^\circ\)) to 2 (if they are in the same direction, \(\theta = 0^\circ\)).