Problem 48 - Precalculus Mathematics for Calculus

Question

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

Magnitude: 1, Direction: 225 degrees.

Step 1: Understand the Vector Components

The vector \( \mathbf{v} = \left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle \) has components \( -\frac{\sqrt{2}}{2} \) for both the x and y directions. This means the vector points in the second quadrant of the coordinate plane.

Step 2: Find the Magnitude of the Vector

The magnitude of a vector \( \mathbf{v} = \langle x, y \rangle \) is given by \( |\mathbf{v}| = \sqrt{x^2 + y^2} \). Substituting the vector components: \[ |\mathbf{v}| = \sqrt{\left(-\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1. \] Thus, the magnitude is 1.

Step 3: Determine the Direction of the Vector

The direction \( \theta \) of a vector with components \( (x, y) \) is given by \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \). For \( \mathbf{v} \), \[ \theta = \tan^{-1}\left(\frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}\right) = \tan^{-1}(1). \] This yields \( \theta = 45^\circ \). Since the vector is in the third quadrant and both components are negative, the actual direction from the positive x-axis is \( 180^\circ + 45^\circ = 225^\circ. \)

Key Concepts

Vector Magnitude

Understanding vector magnitude is essential in vector analysis. The magnitude of a vector represents its length or size, and it is always a positive quantity. If you imagine the vector drawn on a coordinate plane, the magnitude is like the distance from the origin (0,0) to the point where the vector ends. It is calculated using the Pythagorean theorem. This involves taking the square root of the sum of the squares of its components.
For example, if a vector is given by \( \mathbf{v} = \langle x, y \rangle \), its magnitude \( |\mathbf{v}| \) is found using the formula:
\[ |\mathbf{v}| = \sqrt{x^2 + y^2} \]

• Square each of the components: \( x^2 \) and \( y^2 \).
• Add these squares together.
• Take the square root of the sum to find the magnitude.

In the exercise given, for the vector \( \mathbf{v} = \left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle \), the magnitude is 1, showing that it forms a unit line in the coordinate system.

Vector Direction

The direction of a vector shows where the vector is pointing compared to the axes. It is usually measured in degrees from the positive x-axis. This concept becomes obvious using trigonometry, typically with the tangent function, which relates the angle to the opposite and adjacent components of a right triangle.
For a vector \( \mathbf{v} = \langle x, y \rangle \), the direction \( \theta \) can be calculated by:
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]

• The "\( \tan^{-1} \)" is the inverse tangent function, which you can find on most calculators.
• This formula gives the angle based on the relative magnitude of the vector's components.
• Remember, angles returned by \( \tan^{-1} \) are based in the first and fourth quadrants. You might need adjustments if the vector components are negative or lie in other quadrants.

For example, for the vector \( \mathbf{v} \) in the exercise, because both components are negative, the direction adjusts from \( 45^\circ \) to \( 225^\circ \), indicating the vector points in the third quadrant on the coordinate plane.

Coordinate Plane

The coordinate plane is a two-dimensional space defined by an x-axis and a y-axis. This is where all points and vectors "live" when you're working on problems like this one. Each point on this plane is expressed by coordinates \((x, y)\), and vectors are typically directed lines stretching from one point to another, often from the origin (0,0) to a point \((x, y)\).
The plane is divided into four quadrants:

• First Quadrant: Top-right, where both x and y are positive.
• Second Quadrant: Top-left, with negative x and positive y.
• Third Quadrant: Bottom-left, both x and y are negative.
• Fourth Quadrant: Bottom-right, positive x and negative y.

Vectors' positions in these quadrants help determine their direction angles. In this exercise, the vector \( \mathbf{v} \) has both components \(-\frac{\sqrt{2}}{2}\), meaning it is situated in the third quadrant. This affects its direction calculation, highlighting the importance of the coordinate plane when analyzing vectors.