In Exercises 47–52, write the vector v in terms of i and j whose magnitude ||v|| and direction angle θ are given.||v|| = 1/2, θ = 113°

Verified step by step guidance

1. insert step 1: Understand that a vector \( \mathbf{v} \) in terms of \( \mathbf{i} \) and \( \mathbf{j} \) can be expressed as \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \), where \( a \) and \( b \) are the components of the vector.
2. insert step 2: Recall that the magnitude of the vector \( ||\mathbf{v}|| \) is given by \( \sqrt{a^2 + b^2} \). Here, \( ||\mathbf{v}|| = \frac{1}{2} \).
3. insert step 3: Use the direction angle \( \theta = 113^\circ \) to find the components \( a \) and \( b \). The component \( a \) is given by \( a = ||\mathbf{v}|| \cos(\theta) \) and \( b \) is given by \( b = ||\mathbf{v}|| \sin(\theta) \).
4. insert step 4: Substitute the given magnitude and direction angle into the formulas: \( a = \frac{1}{2} \cos(113^\circ) \) and \( b = \frac{1}{2} \sin(113^\circ) \).
5. insert step 5: Write the vector \( \mathbf{v} \) in terms of \( \mathbf{i} \) and \( \mathbf{j} \) using the calculated components: \( \mathbf{v} = \left(\frac{1}{2} \cos(113^\circ)\right)\mathbf{i} + \left(\frac{1}{2} \sin(113^\circ)\right)\mathbf{j} \).

Key Concepts

Magnitude of a Vector

The magnitude of a vector represents its length or size, denoted as ||v||. In this case, the magnitude is given as 1/2, indicating that the vector's length is half a unit. Understanding magnitude is essential for determining how far the vector extends in space.

Direction Angle

The direction angle θ of a vector indicates the angle it makes with the positive x-axis, measured in degrees or radians. Here, θ is 113°, which means the vector is oriented 113° counterclockwise from the positive x-axis. This angle is crucial for calculating the vector's components along the x and y axes.

Vector Components

Vector components are the projections of a vector along the coordinate axes, typically represented as i (x-component) and j (y-component). To express the vector v in terms of i and j, we use the formulas v = ||v|| * cos(θ) * i + ||v|| * sin(θ) * j. This allows us to break down the vector into its horizontal and vertical parts based on its magnitude and direction angle.