Question

if the difference of two unit vectors is unit then angle between two vectors = (a) \((\pi / 2)\) (b) \((\pi / 3)\) (c) \((\pi / 4)\) (d) \((2 \pi / 3)\)

Short answer

To find the angle between the two given unit vectors \(\vec{A}\) and \(\vec{B}\), where the difference between these unit vectors also results in a unit vector, we can use the dot product formula. After calculating the dot product and rearranging the equation, we find that \(\cos{\theta} = \dfrac{1}{2}\). Thus, \(\theta = \boxed{\dfrac{\pi}{3}}\), which corresponds to option (b).

Step-by-step solution

Define the given_unit_vectors

Let the two given unit vectors be \(\vec{A}\) and \(\vec{B}\). Given that the difference of two unit vectors is a unit vector, so \(\vec{C}=\vec{A}-\vec{B}\) is also a unit vector.

Use the dot product formula

The dot product formula is: \(\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos{\theta}\), where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\). Since both \(\vec{A}\) and \(\vec{B}\) are unit vectors, their magnitudes are equal to 1. Thus, the dot product formula becomes: \(\vec{A} \cdot \vec{B}=\cos{\theta}\).

Calculate the magnitude of the difference vector

We know that \(\vec{C} = \vec{A} - \vec{B}\). The square of the magnitude of \(\vec{C}\) is given by: \(|\vec{C}|^2 =(\vec{A} - \vec{B})\cdot(\vec{A} - \vec{B})\) We can expand this expression: \(|\vec{C}|^2 =\vec{A}\cdot\vec{A} - \vec{A}\cdot\vec{B} - \vec{B}\cdot\vec{A} + \vec{B}\cdot\vec{B}\) Since both \(\vec{A}\) and \(\vec{B}\) are unit vectors, their magnitudes are equal to 1: \(|\vec{C}|^2 = 1 - (\vec{A}\cdot\vec{B}) - (\vec{B}\cdot\vec{A}) + 1\) \(|\vec{C}|^2 = 2 - 2 (\vec{A}\cdot\vec{B})\) Since \(\vec{C}\) is a unit vector, we know that \(|\vec{C}|^2=1\). So, substituting this onto the above equation: \(1 = 2 - 2(\vec{A}\cdot\vec{B})\)

Solve for the dot product

We can rearrange the equation from step 3 to find the dot product: \(2 (\vec{A}\cdot\vec{B}) = 1\) \(\vec{A}\cdot\vec{B} = \dfrac{1}{2}\)

Find the angle

Substituting the dot product back into the dot product formula: \(\cos{\theta} = \dfrac{1}{2}\) Thus, \(\theta = \boxed{\dfrac{\pi}{3}}\), which corresponds to option (b).

Key concepts

Unit Vector

When studying vectors, one term that often comes up is the 'unit vector'. A unit vector has a magnitude (also known as length) of exactly one unit and points in a specific direction. It's a standard way to represent direction without considering magnitude.

For a given vector \( \vec{v} \), you can find its unit vector \( \hat{v} \) by dividing the vector by its magnitude: \( \hat{v} = \frac{\vec{v}}{|\vec{v}|} \). Take note that the magnitude of any unit vector is always one, which simplifies various vector calculations, as seen in the exercise where the difference of two unit vectors results in another unit vector.

Dot Product

The dot product, also called scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation combines two vectors and outputs a scalar.

The dot product of vectors \( \vec{A} \) and \( \vec{B} \) is calculated as \( \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos{\theta} \), where \( \theta \) is the angle between the two vectors, and \( |\vec{A}| \) and \( |\vec{B}| \) are their magnitudes. In the context of our exercise, the vectors are unit vectors meaning their magnitudes are 1, which simplifies the formula considerably.

Vector Magnitude

The magnitude of a vector is essentially its 'length' and is a measure of how far the vector extends in space. The magnitude is always a non-negative number.

For a vector \( \vec{A} \) with components \( \vec{A} = (x, y, z) \), the magnitude is calculated using the Pythagorean theorem: \( |\vec{A}| = \sqrt{x^2 + y^2 + z^2} \). In two dimensions, it'll just be \( |\vec{A}| = \sqrt{x^2 + y^2} \). In our exercise scenario, since we're working with unit vectors, their magnitudes are conveniently already equal to one.

Angle Between Two Vectors

The angle between two vectors is a measure of their orientation with respect to each other and can be determined using the dot product formula.

If you have two vectors, \( \vec{A} \) and \( \vec{B} \) with a dot product \( \vec{A} \cdot \vec{B} \) and magnitudes \( |\vec{A}| \) and \( |\vec{B}| \) respectively, the cosine of the angle \( \theta \) between them is given by \( \cos{\theta} = \frac{\vec{A} \cdot \vec{B}}{{|\vec{A}| |\vec{B}|}} \). In the case where the vectors are unit vectors (magnitude of 1), this equation becomes even simpler. This concept is pivotal in understanding the solution to the exercise, as it directly correlates the dot product to the angle between vectors.