Problem 34 - Algebra and Trigonometry

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°.

Step 1: Understand the Vector Components

The vector \( \mathbf{v} = \langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \rangle \) has components \( v_x = -\frac{\sqrt{2}}{2} \) and \( v_y = -\frac{\sqrt{2}}{2} \). These components describe the vector's position in the coordinate plane.

Step 2: Calculate the Magnitude of the Vector

The magnitude of a vector \( \mathbf{v} = \langle a, b \rangle \) is given by the formula \[ \| \mathbf{v} \| = \sqrt{a^2 + b^2} \]Substitute \( a = -\frac{\sqrt{2}}{2} \) and \( b = -\frac{\sqrt{2}}{2} \) into the formula:\[ \| \mathbf{v} \| = \sqrt{\left(-\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2} \]\[ = \sqrt{\frac{2}{4} + \frac{2}{4}} \]\[ = \sqrt{\frac{4}{4}} = \sqrt{1} = 1 \]Thus, the magnitude is 1.

Step 3: Determine the Direction Angle of the Vector

The direction angle \( \theta \) of a vector is calculated using the formula\[ \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) \]Substitute \( v_x = -\frac{\sqrt{2}}{2} \) and \( v_y = -\frac{\sqrt{2}}{2} \):\[ \theta = \tan^{-1}\left(\frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}\right) = \tan^{-1}(1) \]\[ \tan^{-1}(1) = 45^\circ \]Since both components are negative, the vector is in the third quadrant. Therefore, the angle should be adjusted to the third quadrant by adding 180°: \[ \theta = 45^\circ + 180^\circ = 225^\circ \]Thus, the direction angle is 225°.

Key Concepts

Understanding Vector Components

A vector is typically represented by its components in a coordinate system, often written as \( \langle a, b \rangle \). These components tell us the direction and magnitude of the vector in each of the axis directions. In the context of the given vector \( \mathbf{v} = \langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \rangle \), the components are:

• \( v_x = -\frac{\sqrt{2}}{2} \), which is the horizontal component
• \( v_y = -\frac{\sqrt{2}}{2} \), which is the vertical component

These negative values indicate that the vector is positioned in the third quadrant of the Cartesian plane, meaning it points both to the left and downward. This helps us understand which direction in space the vector points to before further analysis.
Recognizing components is an integral first step to evaluating magnitude and direction.

Applying the Vector Magnitude Formula

Calculating the magnitude gives us the length of the vector irrespective of its direction. The magnitude formula for a vector \( \mathbf{v} = \langle a, b \rangle \) is:\[\| \mathbf{v} \| = \sqrt{a^2 + b^2}\]Substituting in our given components for the calculation:\[\| \mathbf{v} \| = \sqrt{\left(-\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2}\]Breaking it down:

• Each component squared is \( \frac{2}{4} = \frac{1}{2} \)
• The summed value is \( \frac{2}{2} = 1 \)
• The square root of 1 equals 1

Thus, the magnitude of the vector is 1. This result shows us that the vector, despite its direction, has a unit length in space, simplifying many further calculations as the vector remains on the unit circle.

Calculating the Direction Angle

Determining a vector's direction angle involves utilizing trigonometry, specifically the tangent inverse function. The formula used is:\[\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)\]For our vector, substituting the component values leads to:\[\theta = \tan^{-1}\left(\frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}\right) = \tan^{-1}(1)\]Mathematically, \( \tan^{-1}(1) \) equates to \( 45^\circ \). However, because both components are negative, the vector resides within the third quadrant.

• To adjust for quadrant placement, 180° is added to 45°, resulting in 225°

Hence, the direction angle of the vector is \( 225^\circ \). The angle accentuates not just the direction but the quadrant orientation of the vector, crucial for understanding its full spatial dynamics.