Problem 48 - Algebra and Trigonometry

Question

\(47-52\) . Find the magnitude and direction (in degrees) of the vector. $$ \mathbf{v}=\left\langle-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right\rangle $$

Step-by-Step Solution

Answer

The magnitude is 1, and the direction is 225°.

Step 1: Calculate Vector Magnitude

To find the magnitude of the vector \( \mathbf{v}=\left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle \), use the formula for the magnitude of a vector: \[ ||\mathbf{v}|| = \sqrt{x^2 + y^2} \] where \( x = -\frac{\sqrt{2}}{2} \) and \( y = -\frac{\sqrt{2}}{2} \). Calculating, we get:\[ ||\mathbf{v}|| = \sqrt{ \left(-\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2 } = \sqrt{ \frac{2}{4} + \frac{2}{4} } = \sqrt{1} = 1. \]

Step 2: Determine the Vector Direction

To find the direction angle \( \theta \) (in degrees) of the vector \( \mathbf{v} \), use the formula: \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \] With \( x = -\frac{\sqrt{2}}{2} \) and \( y = -\frac{\sqrt{2}}{2} \) for vector \( \mathbf{v} \), we have:\[ \theta = \tan^{-1}\left(\frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}\right) = \tan^{-1}(1). \] This yields \( \theta = 45^\circ \), but since both components are negative, the vector is in the third quadrant. Thus, the actual direction angle is:\[ \theta = 225^\circ. \]

Key Concepts

Vector Magnitude

When we talk about the magnitude of a vector, we are referring to its length. To understand vector magnitude, imagine a line stretching from the origin to the point represented by the vector coordinates. This length is calculated using a formula inspired by the Pythagorean theorem.

For a vector \[\mathbf{v} = \langle x, y \rangle,\]the magnitude is given by\[||\mathbf{v}|| = \sqrt{x^2 + y^2}.\]The formula gives us a measure of how "long" the vector is, regardless of its direction.

In our problem with the vector \[\mathbf{v} = \left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle,\]substitute the values into the formula:

• Calculate \((-\frac{\sqrt{2}}{2})^2 + (-\frac{\sqrt{2}}{2})^2\) to get \[ \frac{2}{4} + \frac{2}{4} = 1. \]
• Then take the square root to find \[||\mathbf{v}|| = \sqrt{1} = 1.\]

The result is that the vector has a magnitude of 1, which makes it a unit vector. A unit vector has a length of one and is often used to represent directions.

Vector Direction

The direction of a vector is indicated by the angle it forms with the positive x-axis. Understanding vector direction involves trigonometric concepts, notably the inverse tangent function (\(\tan^{-1}\)). This function helps in finding the angle associated with the vector's slope.

To find the direction angle \(\theta\) of a vector \[\mathbf{v} = \langle x, y \rangle,\]we use:\[\theta = \tan^{-1}\left(\frac{y}{x}\right).\]In our example, substituting \[ x = -\frac{\sqrt{2}}{2} \quad \text{and} \quad y = -\frac{\sqrt{2}}{2},\]we calculate:

• \( \theta = \tan^{-1}\left(\frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}\right) = \tan^{-1}(1). \)
• This initially gives \(45^\circ.\)

However, given both components are negative, placing \(\mathbf{v}\) in the third quadrant, adjust the angle:

• Vector directions in the third quadrant start from \(180^\circ, \) add \(45^\circ\) to reach \(225^\circ.\)

Hence, the vector direction is \(225^\circ,\) providing a clear path in a negative x and y direction.

Trigonometry

Trigonometry is crucial for understanding vectors because it deals with angles and relationships within triangles. By using trigonometry, particularly the tangent function, we calculate angles pertinent to the vector's slope.

The inverse tangent function \(\tan^{-1}\) is used when we want to determine the angle a vector makes concerning the coordinate axes, especially in cases where both x and y values are either negative or positive.

Here's a quick rundown of why trigonometry matters in vector analysis:

• **Relating Angles and Sides:** Trigonometric functions relate angles to the lengths of a right triangle's sides.
• **Inverse Functions:** These determine angles from known side ratios; for vectors, \(\tan^{-1}\) gives us the angle or direction.
• **Quadrant Consideration:** Adjusting angles based on quadrant placement is possible with trigonometry.

In our exercise, trigonometry helps situate the vector into the correct quadrant and adjust the angle for accurate directional representation.Understanding these principles gives a clearer, more detailed understanding of how a vector behaves in a coordinate plane.