Problem 63 - Precalculus with Limits

Question

Finding the Direction Angle of a Vector In Exercises \(63-66\) , find the magnitude and direction angle of the vector v. $$\mathbf{v}=6 \mathbf{i}-6 \mathbf{j}$$

Step-by-Step Solution

Answer

The magnitude of the vector is \( 6\sqrt{2} \) and the direction angle is \( 135° \).

Step 1: Calculate the magnitude

The magnitude is calculated as follows: \( |\mathbf{v}| = \sqrt{(6)^2 + ( -6 )^2} = 6\sqrt{2}\).

Step 2: Calculate the direction angle

The direction angle is calculated by first determining the tangent of the angle from the components of the vector: \( \tan \theta = \frac{-6}{6} = -1 \). Then, we take the arctangent of this value to get the angle in radians, \( \theta = \arctan(-1) \). This value is in the third quadrant since i-component is positive and j-component is negative, thus we need to add 180 to the radians value to correctly represent the direction of the vector into the third quadrant.

Step 3: Convert the direction angle to degrees

Multiplying by \( \frac{180}{\pi} \), we have that \( \theta = \arctan(-1) \times \frac{180}{\pi} = -45° \). Adding 180 to this value for quadrant adjustment, we arrive at \( \theta = 135° \).

Key Concepts

Vector Magnitude

The concept of vector magnitude is essential for determining the length of a vector. Understanding this can help us grasp the scale of the physical quantities a vector represents. The magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} \) is given by the formula:

• \( |\mathbf{v}| = \sqrt{a^2 + b^2} \)

This formula is analogous to the Pythagorean theorem, where \(a\) and \(b\) are the components of the vector along the x (horizontal) and y (vertical) axes respectively.
For the vector \(\mathbf{v}=6 \mathbf{i}-6 \mathbf{j}\), we calculate:

• \( |\mathbf{v}| = \sqrt{6^2 + (-6)^2} = \sqrt{36 + 36} = 6\sqrt{2} \)

This result tells us that the actual length of the vector from its initial point to its endpoint is \(6\sqrt{2}\). Magnitude provides a scalar measure of the vector’s size.

Tangent Function

The tangent function is a trigonometric function that relates the angles and sides of right triangles. It's crucial for understanding vector direction angles because it connects a vector's components to the angle it makes with a horizontal line.
In a right triangle:

• \( \tan \theta = \frac{\text{opposite side}}{\text{adjacent side}} \)

For vectors, this translates to:

• \( \tan \theta = \frac{b}{a} \)

where \(b\) is the vertical component and \(a\) is the horizontal component.
For the vector \(6 \mathbf{i} - 6 \mathbf{j}\), the tangent of the direction angle \(\theta\) is:

• \( \tan \theta = \frac{-6}{6} = -1 \)

This result means the angle \(\theta\) at which the vector resides is one where the components are equal in magnitude but opposite in direction.

Arctangent Computation

Arctangent is the inverse function of tangent, often used to find an angle when the tangent of the angle is known. It's symbolized as \( \arctan \) or sometimes \( \tan^{-1} \). This computation is key in determining the angle a vector forms relative to the axial lines.
Given that \( \tan \theta = -1 \), the arctangent calculation is:

• \( \theta = \arctan(-1) \)

In mathematics, \( \arctan(-1) \) typically returns an angle of \(-45^\circ\), reflecting the angle in standard position where the tangent is \(-1\).
The arctangent function alone may not directly indicate in which quadrant the actual vector lies, necessitating adjustment based on the signs of the vector's components.

Coordinate Quadrants

Understanding coordinate quadrants helps in correctly interpreting the calculated angle from trigonometric functions, especially when the vector lies in a specific quadrant due to its components.
In the Cartesian coordinate system, the plane is divided into four quadrants:

• Quadrant I: Both \(x\) and \(y\) are positive.
• Quadrant II: \(x\) is negative, \(y\) is positive.
• Quadrant III: Both \(x\) and \(y\) are negative.
• Quadrant IV: \(x\) is positive, \(y\) is negative.

For the vector \(6 \mathbf{i} - 6 \mathbf{j}\):

• \(x = 6\) (positive)
• \(y = -6\) (negative)

This places the vector in Quadrant IV.
While \(\theta = -45^\circ\) is the direct output of the arctangent function computation, we adjust by adding \(180^\circ\) to account for the correct quadrant direction, resulting in \(135^\circ\), ensuring the angle accurately reflects its directional position.