Problem 66 - Precalculus with Limits

Question

Finding the Direction Angle of a Vector In Exercises \(63-66\) , find the magnitude and direction angle of the vector v. $$\mathbf{v}=8\left(\cos 135^{\circ} \mathbf{i}+\sin 135^{\circ} \mathbf{j}\right)$$

Step-by-Step Solution

Answer

The magnitude of the vector \(v = 8(\cos 135^{\circ} \mathbf{i}+\sin 135^{\circ} \mathbf{j})\) is 8 and the direction angle is \(315^\circ\).

Step 1: Calculate the magnitude of the vector

The magnitude of the vector \(v = 8(\cos 135^{\circ} \mathbf{i}+\sin 135^{\circ} \mathbf{j})\) is given by multiplying the scalar with the magnitude of unit vector which is 1. Thus magnitude \(|v| = 8 * 1 = 8\)

Step 2: Determine the direction angle

The given vector's x-component is \(8\cos 135^{\circ}\) and y-component is \(8\sin 135^{\circ}\). Since \(\cos 135^{\circ} = -1/\sqrt{2}\) and \(\sin 135^{\circ} = 1/\sqrt{2}\), the x and y-components equate to \(-8/\sqrt{2}\) and \(8/\sqrt{2}\) respectively. The direction angle \(\theta\) in standard position can be found by taking \(\tan^{-1}(y/x)\) where y is the vector's y component and x is its x component. So, \(\theta = \tan^{-1}[(8/\sqrt{2}) /(-8/\sqrt{2})]\), which simplifies to \(\theta = \tan^{-1}(-1) = -45^{\circ}\) in standard position. As this angle is negative, it's measured clockwise from the positive x-axis, but direction angles are typically measured counterclockwise. Therefore, we add \(360^\circ\) to \(-45^\circ\) to get the direction angle as \(315^\circ\).

Step 3: Finalize the results

After calculating, we have found the magnitude of the vector to be 8 and its direction angle to be \(315^\circ\). Thus, the magnitude and the direction angle of vector \(v\) are 8 and \(315^\circ\) respectively.

Key Concepts

Magnitude of a Vector

The magnitude of a vector is a measure of its length. For a two-dimensional vector, imagine that the vector represents a point in the plane relative to the origin (0,0). The magnitude is simply the distance from the origin to that point.

When given a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \), where \(\mathbf{i}\) and \(\mathbf{j}\) are the unit vectors along the x and y axes respectively, the magnitude can be found using the Pythagorean theorem giving us \( |\mathbf{v}| = \sqrt{a^2 + b^2} \). In the example \( \mathbf{v}=8(\cos 135^\circ \mathbf{i}+\sin 135^\circ \mathbf{j})\), the magnitude is calculated to be 8, as it is a scaled version of a unit vector.

Trigonometric Functions

Trigonometric functions are a cornerstone of geometry and physics, as they relate the angles of a triangle to its sides. The main functions are sine (sin), cosine (cos), and tangent (tan). These are functions of an angle, often represented by the Greek letter \( \theta \).

For a right-angled triangle with an angle \( \theta \), the sine function gives the ratio of the length of the opposite side to the length of the hypotenuse, while the cosine gives the ratio of the adjacent side to the hypotenuse. The tangent represents the ratio of the sine to the cosine of that angle. Each function has a range of applications, from calculating the components of vectors to modeling periodic phenomena.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as arc sine, arc cosine, and arc tangent (often written as \(\tan^{-1}\)), are the reverse operations used to find the angle when the value of the trigonometric function is known.

In the context of vectors, these inverse functions are particularly important when determining the direction angle. If you know the ratio of the y-component to the x-component of the vector, you can use the arc tangent function to find the original angle \(\theta\). For example, the direction angle of the vector calculated using \(\tan^{-1}\) resulted in \(\theta = \tan^{-1}(-1) = -45^\circ\). These inverse functions often require adjustments to be made for the angle's quadrant, which was done by adding \(360^\circ\) in this case to find the positive direction angle.

Vector Components

Vector components are projections of a vector onto the axes of the coordinate system. For a vector \(\mathbf{v}\) in two dimensions, it's made up of two components: an x-component along the horizontal axis and a y-component along the vertical axis. These components can be expressed as \(\mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j}\), where \(v_x\) and \(v_y\) are the magnitudes of the x and y components, respectively.

In the exercise, the components \(8\cos 135^\circ\) and \(8\sin 135^\circ\) were multiplied by 8 to account for the vector's magnitude. Consequently, these components gave us \(v_x = -8/\sqrt{2}\) and \(v_y = 8/\sqrt{2}\) since \(\cos 135^\circ\) and \(\sin 135^\circ\) are the ratios relevant to the unit circle associated with that angle.