Problem 63 - Precalculus with Limits

Question

In Exercises 63-66, find the magnitude and direction angle of the vector \(\mathbf{v}\). \(\small{\mathbf{v} = 6\mathbf{i} - 6\mathbf{j}}\)

Step-by-Step Solution

Answer

The magnitude of the vector is \( 6\sqrt{2} \) and the direction angle is \( 315^\circ \)

Step 1: Calculate the Magnitude of the Vector

The magnitude of the vector \( \mathbf{v} = 6\mathbf{i} - 6\mathbf{j} \) is calculated using the formula \( \|\mathbf{v}\| = \sqrt{a^2 + b^2} \). Substituting the values \( a = 6 \) and \( b = -6 \) into the formula gives \( \|\mathbf{v}\| = \sqrt{6^2 + (-6)^2} = \sqrt{72} = 6\sqrt{2} \)

Step 2: Calculate the Direction Angle of the Vector

The direction angle \( \theta \) of the vector is found from the equation \( \tan(\theta) = \frac{b}{a} \). Substituting the values \( a = 6 \) and \( b = -6 \) gives \( \tan(\theta) = \frac{-6}{6} = -1 \). Use the arctangent function to find the angle, \( \theta = \arctan(-1) = -45^{\circ} \). Note that the direction angle is negative because the vector points downwards

Step 3: Adjusting for the correct quadrant

From the previous step, we have \( \theta = -45^\circ \). As the vector \(\mathbf{v} = 6\mathbf{i} - 6\mathbf{j}\) is in the fourth quadrant, we add 360 degrees to our negative result to get the proper equivalent positive angle which falls between 0 to 360 degrees. Hence, \( \theta = -45^\circ + 360^\circ = 315^\circ \) is the direction angle of vector \(\mathbf{v}\).

Key Concepts

Magnitude of a Vector

When we talk about the magnitude of a vector, we're referring to its length or size, irrespective of its direction. It gives us information about how far a point has moved from its original position if we think of the vector as a point in motion. To find a vector's magnitude, we utilize the Pythagorean theorem, akin to finding the hypotenuse of a right-angled triangle.

The general formula to calculate the magnitude of a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \) is \( \|\mathbf{v}\| = \sqrt{a^2 + b^2} \). In our exercise, the vector \( \mathbf{v} = 6\mathbf{i} - 6\mathbf{j} \) has a magnitude of \( \sqrt{6^2 + (-6)^2} \) which simplifies to \( 6\sqrt{2} \), indicating the vector’s length is \( 6\sqrt{2} \) units.

Direction Angle of a Vector

Moving beyond magnitude, the direction angle of a vector gives us its orientation in a plane. This angle is typically measured from the positive x-axis to the vector, going counter-clockwise. To find this angle, we use the tangent function, which compares the vertical and horizontal components of our vector, thought of as the opposite and adjacent sides of a right triangle respectively.

Specifically, for a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \), the direction angle \( \theta \) is obtained from the equation \( \tan(\theta) = \frac{b}{a} \). In our exercise, with \( a = 6 \) and \( b = -6 \) we find that \( \tan(\theta) = \frac{-6}{6} = -1 \), which leads us to \( \theta \) having a value of \( \arctan(-1) \) or \( -45^\circ \) initially.

Calculating Vector Components

Vectors can be dissected into their horizontal and vertical components, which are essential for understanding their overall behavior. The horizontal component corresponds to the x-axis, while the vertical one matches the y-axis. In our textbook example, the vector \( \mathbf{v} = 6\mathbf{i} - 6\mathbf{j} \) has a horizontal component of 6 and a vertical component of -6. These components are vital in not only establishing magnitude but also in analyzing vector addition, subtraction, and even multiplication by scalars.

Breaking down vector \( \mathbf{v} \) into \( a\mathbf{i} \) and \( b\mathbf{j} \) allows us to work with each dimension separately in calculations. This capability is exceptionally useful when solving for forces in physics or establishing velocity vectors in engineering.

Arctangent Function in Vectors

The arctangent function, denoted as \( \arctan \), is the inverse of the tangent function and is employed to deduce the angle when the tangent value is known. It’s an essential tool in vector mathematics for finding direction angles. The arctan function provides us with an angle whose tangent is the given value.

However, \( \arctan \) alone can sometimes lead to an angle that doesn't align with the original vector's quadrant. This happens because the tangent function has the same values for angles that are 180 degrees apart. So, for accuracy, we must adjust the angle based on the vector's components. For example, in the exercise \( \arctan(-1) \) indicates \( -45^\circ \) but considering that \( \mathbf{v} \) is in the fourth quadrant, we add 360 degrees to get \( 315^\circ \) as the correct direction angle.