Problem 60 - Trigonometry

Question

Find the magnitude of each vector and the angle \(\theta, 0^{\circ} \leq \theta<360^{\circ}\), that the vector makes with the positive \(x\)-axis. $$\mathbf{V}=\langle 5,-5\rangle$$

Step-by-Step Solution

Answer

The magnitude is \(5\sqrt{2}\), and the angle is \(315^{\circ}\).

Step 1: Calculate Vector Magnitude

The magnitude of a vector \( \mathbf{V} = \langle a, b \rangle \) is given by the formula \( \|\mathbf{V}\| = \sqrt{a^2 + b^2} \). For \( \mathbf{V} = \langle 5, -5 \rangle \), plug in the values: \( \|\mathbf{V}\| = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \).

Step 2: Calculate Angle with Positive X-Axis

To find the angle \( \theta \) a vector makes with the positive \( x \)-axis, use the formula \( \tan(\theta) = \frac{b}{a} \). For \( \mathbf{V} = \langle 5, -5 \rangle \), \( a = 5 \) and \( b = -5 \), giving \( \tan(\theta) = \frac{-5}{5} = -1 \). The angle whose tangent is \(-1\) is \( 135^{\circ} \) in the fourth quadrant as the vector points southwest.

Key Concepts

Angle with Positive X-Axis

Understanding the angle a vector makes with the positive x-axis is essential in vector analysis. This angle helps determine the vector's direction in a two-dimensional plane. To find this angle, trigonometric relationships between the components of the vector are used.

In our current exercise, we deal with a vector \( \mathbf{V} = \langle 5, -5 \rangle \). The angle \( \theta \) is calculated using the tangent function. A vector with components \( (a, b) \) forms an angle \( \theta \) with the x-axis such that:

• \( \tan(\theta) = \frac{b}{a} \)

For \( \mathbf{V} = \langle 5, -5 \rangle \), this results in \( \tan(\theta) = \frac{-5}{5} = -1 \). Therefore, the angle corresponding to a tangent value of \(-1\) needs careful interpretation of which quadrant it resides in, as this determines the correct angle to measure from the positive x-axis.

Tangent of Angle

The tangent of an angle in a vector context helps to find the direction of the vector relative to the x-axis. This is crucial in determining where the vector points in the coordinate plane.

A tangent function relates the vertical component \( b \) to the horizontal component \( a \) of a vector. It is calculated as:

• \( \tan(\theta) = \frac{b}{a} \)

Understanding that \( \tan(\theta) = -1 \) means the direction is opposite to that formed by \( \tan(\theta) = 1 \) is essential. This is why angles of \( 45^{\circ} \) or \( 225^{\circ} \) result in a \( \tan(\theta) = 1 \), and angles of \( 135^{\circ} \) or \( 315^{\circ} \) result in \( \tan(\theta) = -1 \). Recognizing these values aids in identifying the quadrant location of the vector. In this case, considering the sign of \( b \), the appropriate angle is \( 315^{\circ} \), placing it in the fourth quadrant.

Quadrants in Trigonometry

When analyzing vectors and the angles they form with axes, understanding the four quadrants of the Cartesian plane is vital. Each quadrant indicates where the vector is headed and influences the sign of trigonometric values.

The four quadrants are defined as follows:

• First Quadrant: Both \( x \) and \( y \) are positive. Angles range from \( 0^{\circ} \) to \( 90^{\circ} \).
• Second Quadrant: \( x \) is negative, \( y \) is positive. Angles range from \( 90^{\circ} \) to \( 180^{\circ} \).
• Third Quadrant: Both \( x \) and \( y \) are negative, and angles range from \( 180^{\circ} \) to \( 270^{\circ} \).
• Fourth Quadrant: \( x \) is positive, \( y \) is negative. Angles range from \( 270^{\circ} \) to \( 360^{\circ} \).

Identifying the correct quadrant helps in determining the correct angle measure corresponding to the vector. For \( \mathbf{V} = \langle 5, -5 \rangle \), the vector is in the fourth quadrant where the x-component is positive and the y-component is negative, which results in an angle of \( 315^{\circ} \). This is essential for both determining the direction of the vector and in applications using this information.