Problem 10 - Algebra and Trigonometry

Question

If \(\mathbf{v}\) is a nonzero vector with direction angle \(\alpha, 0^{\circ} \leq \alpha<360^{\circ},\) between \(\mathbf{v}\) and \(\mathbf{i},\) then \(\mathbf{v}\) equals which of the following? (a) \(\|\mathbf{v}\|(\cos \alpha \mathbf{i}-\sin \alpha \mathbf{j})\) (b) \(\|\mathbf{v}\|(\cos \alpha \mathbf{i}+\sin \alpha \mathbf{j})\) (c) \(\|\mathbf{v}\|(\sin \alpha \mathbf{i}-\cos \alpha \mathbf{j})\) (d) \(\|\mathbf{v}\|(\sin \alpha \mathbf{i}+\cos \alpha \mathbf{j})\)

Step-by-Step Solution

Answer

Choice (b).

Step 1: Understanding vector form

A vector in the plane can be expressed in terms of its magnitude and direction angle as \[ \mathbf{v} = \| \mathbf{v}\| \begin{bmatrix} \cos(\alpha) \ \sin(\alpha) \end{bmatrix} \] where \(\alpha\) is the direction angle with the positive x-axis, and \(\mathbf{i}\) and \(\mathbf{j}\) are the unit vectors in the x and y directions, respectively.

Step 2: Expressing the vector

Replacing the components with \( \mathbf{i} \) and \( \mathbf{j} \), the vector \( \mathbf{v} \) becomes \[ \mathbf{v} = \| \mathbf{v}\|(\cos \alpha \mathbf{i} + \sin \alpha \mathbf{j}) \]

Step 3: Comparing with the given choices

Compare the derived expression of the vector \(\mathbf{v}\) with the given choices:- (a) \(\| \mathbf{v}\|(\cos \alpha \mathbf{i} - \sin \alpha \mathbf{j})\) - (b) \(\| \mathbf{v}\|(\cos \alpha \mathbf{i} + \sin \alpha \mathbf{j})\) - (c) \(\| \mathbf{v}\|(\sin \alpha \mathbf{i} - \cos \alpha \mathbf{j})\)- (d) \(\| \mathbf{v}\|(\sin \alpha \mathbf{i} + \cos \alpha \mathbf{j})\).The expression matches with choice (b).

Key Concepts

Vector Magnitude

To understand vectors, it's essential to grasp the concept of vector magnitude. The magnitude of a vector indicates how long the vector is, or its length. This is represented as \(\backslash| \backslashmathbf{v} \backslash|\). For a vector in two-dimensional plane with components \(x\) and \(y\), the magnitude is calculated using the Pythagorean theorem: \(\backslash| \backslashmathbf{v} \backslash| = \backslashsqrt{x^2 + y^2}\). This formula helps in determining how far away a vector lies from the origin point (0,0).

Direction Angles

Direction angles are key to understanding the orientation of a vector. The direction angle \(\backslasha \backslashalpha \backslash)\) is defined as the angle a vector makes with the positive x-axis. It ranges between \(0^{\circ} \leq \backslashalpha \lt 360^\backslashcirc\). Using trigonometric functions, we can break down the vector into its components: cosine for the x-direction and sine for the y-direction. The vector can then be expressed in terms of its magnitude and direction angle as \( \backslashmathbf{v} = \backslash| \backslashmathbf{v} \backslash| ( \backslashcos \backslasha \backslashalpha \backslashmathbf{i} + \backslashsin \backslasha \backslashalpha \backslashmathbf{j} )\). This form is very helpful in transforming vectors into manageable parts with clear interpretations.

Unit Vectors

Unit vectors are vectors with a magnitude of 1. They are particularly useful for indicating direction without affecting magnitude. Common unit vectors are \(\backslashmathbf{i}\) in the x-direction and \(\backslashmathbf{j}\) in the y-direction. Any vector can be represented as a combination of these unit vectors. For instance, a vector \( \backslashmathbf{v} \) with direction angle \(\backslasha \backslashalpha \backslash)\) can be written as \( \backslash| \backslashmathbf{v} \backslash| ( \backslashcos \backslasha \backslashalpha \backslashmathbf{i} + \backslashsin \backslasha \backslashalpha \backslashmathbf{j} )\). Using unit vectors simplifies calculations and provides a clear geometric interpretation of vector components.