Question

If angle between two unit vectors \(\underline{a} \& \underline{b}\) is \(\alpha\), then \(|\underline{\mathrm{a}}-\underline{\mathrm{b}} \cos \alpha|=0<\alpha<(\pi / 2)\) (A) \(\sin \alpha\) (B) \(\sin (\alpha / 2)\) (C) \(\sin 2 \alpha\) (D) \(\sin ^{2}(\alpha / 2)\)

Short answer

The short answer is: \(\bold{(A)} \; \sin \alpha\).

Step-by-step solution

Find the magnitude of the difference vector

The difference vector is \(\underline{a}-\underline{b} \cos \alpha\). First, we find the magnitude of this vector using the properties of the dot product: \(|\underline{a}-\underline{b} \cos \alpha|^2 = (\underline{a}-\underline{b} \cos \alpha) \cdot (\underline{a}-\underline{b} \cos \alpha)\) Expanding the dot product: \(|\underline{a}-\underline{b} \cos \alpha|^2 = \underline{a} \cdot \underline{a} - 2(\underline{a} \cdot \underline{b})\cos\alpha +(\underline{b} \cdot \underline{b})\cos^2 \alpha\)

Simplify the expression

Since \(\underline{a}\) and \(\underline{b}\) are unit vectors, we have \(\underline{a} \cdot \underline{a} = \underline{b} \cdot \underline{b} = 1\). And we know that \(\underline{a} \cdot \underline{b} = |\underline{a}||\underline{b}|\cos(\alpha) = 1\cdot 1 \cdot \cos(\alpha)=\cos(\alpha)\). Now, substitute these values into the expression: \(|\underline{a}-\underline{b} \cos \alpha|^2 = 1 - 2(\cos \alpha)^2 + (\cos \alpha)^2\) Combining terms, we get: \(|\underline{a}-\underline{b} \cos \alpha|^2 = 1-\cos^2\alpha\)

Determine the correct answer

Recall the Pythagorean identity: \(1-\cos^2\alpha = \sin^2\alpha\). Hence, we have: \(|\underline{a}-\underline{b} \cos \alpha|^2 = \sin^2\alpha\) Taking the square root of both sides to obtain the magnitude: \(|\underline{a}-\underline{b} \cos \alpha| = \sin\alpha\) By comparing with the provided options, we can conclude that the correct expression is: \(\bold{(A)} \; \sin \alpha\)

Key concepts

Vector Arithmetic

Vector arithmetic involves operations such as addition, subtraction, and multiplication by a scalar. These operations follow specific rules that are different from regular arithmetic with numbers. For instance, when subtracting two vectors, like in the example \(\underline{a}-\underline{b} \cos \alpha\), it's not as simple as subtracting numerical values. The operation takes into account both the magnitude and the direction of the vectors involved.

Here's a practical way to understand vector subtraction: if you were to walk from point A to B, represented by vector \(\underline{a}\), and then from point B to C, represented by \(\underline{b} \cos \alpha\), your overall displacement (which is another vector) would be the vector \(\underline{a}-\underline{b} \cos \alpha\). Taking the magnitude of this difference vector gives us the 'straight-line' distance of your journey from start to finish.

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is proper when you’re working on problems that require projection or finding an angle between two vectors, as seen in \(\underline{a} \cdot \underline{a}\) or \(\underline{a} \cdot \underline{b}\).

The dot product can be calculated in two ways: algebraically or geometrically. Algebraically, it's the sum of the products of the corresponding entries of the two sequences of numbers. Geometrically, it expresses the multiplication of the magnitudes of the two vectors and the cosine of the angle between them, as in \(\underline{a} \cdot \underline{b} = |\underline{a}||\underline{b}|\cos(\alpha)\). Understanding the dot product is vital for calculating the magnitude of the difference vector effectively.

Unit Vectors

Unit vectors are vectors with a magnitude of 1. They are the building blocks of vector analysis and are especially important in providing direction. A unit vector retains the direction of the original vector but normalizes its length to 1. In vector arithmetic, such as when you deal with \(\underline{a}\) and \(\underline{b}\), knowing that these are unit vectors simplifies computations.

In the context of our problem, unit vectors simplify the calculations of the dot product; since their magnitudes are 1, \(\underline{a} \cdot \underline{a}\) and \(\underline{b} \cdot \underline{b}\) merely evaluate to 1. This property reduces potential complexity and makes it easier to focus on the angle \(\alpha\) when working through the problem.

Pythagorean Trigonometric Identity

The Pythagorean trigonometric identity is an essential principle in trigonometry, stated as \( \sin^2 \theta + \cos^2 \theta = 1 \). It expresses the inherent relationship between the sine and cosine of an angle. In the context of vector calculations, particularly when working with unit vectors and angles between them, this identity helps simplify expressions and solve for unknowns.

In our exercise, after applying the dot product and unit vector properties, the problem was reduced to \( 1 - \cos^2\alpha \), which then invokes the Pythagorean identity to become \( \sin^2\alpha \). Recognizing this opportunity to use the Pythagorean trigonometric identity can be a powerful tool for students to simplify and solve vector problems, especially when dealing with magnitudes and angles.