Problem 66 - Precalculus with Limits

Question

In Exercises 63-66, find the magnitude and direction angle of the vector \(\mathbf{v}\). \(\small{\mathbf{v} = 8(\cos 135^{\circ} \mathbf{i} + \sin 135^{\circ} \mathbf{j})}\)

Step-by-Step Solution

Answer

The magnitude of the vector \( \mathbf{v} \) is 8, and its direction angle is 135°.

Step 1: Decompose the Vector

The given vector \( \mathbf{v} \) can be decomposed into its component forms: \( \mathbf{v} = 8(\cos 135^{\circ} \mathbf{i} + \sin 135^{\circ} \mathbf{j}) = 8*(-0.7071\mathbf{i} + 0.7071\mathbf{j}) = -5.65\mathbf{i} + 5.65\mathbf{j} \)

Step 2: Find the Magnitude of the Vector

The magnitude of a vector is given by the Pythagorean theorem in two dimensions: \( || \mathbf{v} || = \sqrt{(-5.65)^2 + 5.65^2} = 8 \)

Step 3: Find the Direction Angle of the Vector

The direction angle of a vector is found by taking the inverse tangent of the ratio of its y-component to its x-component: \( \theta = \arctan( \frac{-5.65}{5.65}) = 135^\circ \)

Key Concepts

Magnitude

The magnitude of a vector is much like the length of a line segment. It tells us how long the vector is, regardless of its direction. For a two-dimensional vector, the magnitude can be found by using the Pythagorean theorem. This is because we can think of the vector as the hypotenuse of a right triangle, whose legs are the components of the vector.

In the given example, the vector \(\mathbf{v} = -5.65\mathbf{i} + 5.65\mathbf{j}\) is decomposed into its x and y components as -5.65 and 5.65, respectively. To find its magnitude:

• Square the x-component: \((-5.65)^2 = 31.9225\).
• Square the y-component: \((5.65)^2 = 31.9225\).
• Add these squares: \(31.9225 + 31.9225 = 63.845\).
• Take the square root of the sum to get the magnitude: \(\sqrt{63.845} = 8\).

This process gives us the vector's magnitude of 8, which is consistent with what the problem states: the vector is scaled by 8.

Direction Angle

The direction angle of a vector provides information about the angle it makes with the positive x-axis. This is important because two vectors can have the same magnitude, but completely different directions, telling us how they're oriented in space.

In our exercise, the original vector is given with a direction that involves the trigonometric functions cosine and sine with an angle of \(135^{\circ}\). This angle was found using the inverse tangent function, or arctan, which calculates the angle based on the ratio of the vector's y-component to its x-component.

• The formula used is: \(\theta = \arctan(\frac{y}{x})\).
• Plug in the y-component (5.65) and x-component (-5.65): \(\theta = \arctan\left(\frac{-5.65}{5.65}\right)\).
• This results in \(\theta = 135^{\circ}\), which represents the vector's orientation in standard position on the coordinate plane.

The angle of \(135^{\circ}\) is further confirmed by the vector's placement in the second quadrant, where both cosine and sine values indicate a direction towards the top-left of the Cartesian plane.

Component Forms

Component forms of a vector describe it in terms of its x and y directions. This is particularly useful for operations like addition or subtraction, which rely on the ability to separately manipulate each component of the vector.

The given vector \(\mathbf{v}\) is initially represented as \(8 (\cos 135^{\circ} \mathbf{i} + \sin 135^{\circ} \mathbf{j})\). This illustrates the vector as a combination of its horizontal (x) and vertical (y) influences:

• Substitute `\(\cos 135^{\circ}\)` and `\(\sin 135^{\circ}\)` with their approximate values.
• \(\cos 135^{\circ} \approx -0.7071\) and \(\sin 135^{\circ} \approx 0.7071\)
• The vector components multiply by 8 to create: \(-5.65\mathbf{i} + 5.65\mathbf{j}\).

This breakdown allows one to clearly see how much of the vector points in each direction. The algebra here reflects how vectors can be visualized and calculated based on their real-world implications, such as moving left and up in this case.