Problem 26 - Precalculus : Building Concepts and Connections

Question

Find the magnitude and direction of each of the given vectors. Express the direction as an angle \(\theta\) in standard position, where \(0^{\circ} \leq \theta<360^{\circ},\) to tuo decimal places. $$\mathbf{w}=\langle 3,5\rangle$$

Step-by-Step Solution

Answer

The magnitude of the vector \(\mathbf{w}\) is \(\sqrt{34}\) and its direction is \(59.04^\circ\) in standard position.

Step 1: Compute the Magnitude

Calculate the magnitude (or length) of the vector \(\mathbf{w} = \langle 3,5 \rangle\). The formula for magnitude is \(\lVert\mathbf{w}\rVert = \sqrt{x^2 + y^2}\), where \(x\) and \(y\) are the coordinates of the vector. Thus, the magnitude becomes \(\lVert\mathbf{w}\rVert = \sqrt{3^2 + 5^2} = \sqrt{34}\).

Step 2: Compute the Direction

Next, compute the direction of the vector using the formula \(\theta = \tan^{-1}(y/x)\). Therefore, the direction becomes \(\theta = \tan^{-1}(5/3) = 59.04^\circ\). Standard position angles are usually between \(0^\circ - 360^\circ), so as our angle is in first quadrant, it does not require adjustment.

Key Concepts

Magnitude Calculation

Understanding the magnitude of a vector is crucial in fields such as physics and engineering where vectors are frequently used to represent quantities like force and velocity. The magnitude, often regarded as the 'length' of the vector, quantifies how much of a quantity there is.

For a two-dimensional vector given by its coordinates, \(\mathbf{v} = \langle x, y \rangle\), the magnitude can be computed using the Euclidean norm formula which is derived from the Pythagorean theorem. The formula is \( \lVert\mathbf{v}\rVert = \sqrt{x^2 + y^2} \). It involves squaring each component of the vector, adding these squares together, and then taking the square root of the sum.

Let's take the vector from our exercise \(\mathbf{w} = \langle 3,5 \rangle\) as an example. By applying the formula, we get the magnitude as \(\lVert\mathbf{w}\rVert = \sqrt{3^2 + 5^2} = \sqrt{34}\). This numerical value represents the vector's magnitude in the same units as its components.

Direction Angle of a Vector

The direction of a vector is the angle it makes with the positive direction of the x-axis, which can be described using the counter-clockwise standard from the axis in a two-dimensional space. For the vector \(\mathbf{v} = \langle x, y \rangle\), its direction angle \(\theta\) in standard position can be calculated using the arctangent function, represented by the inverse tangent \(\tan^{-1}\).

The formula used is \(\theta = \tan^{-1}(y/x)\), which yields the angle in radians or degrees. It's important to note the quadrant in which the vector lies, as this affects the angle's sign and value. For instance, in our exercise, the vector \(\mathbf{w}\) lies in the first quadrant where both x and y are positive. So, the direction angle is \(\theta = \tan^{-1}(5/3) = 59.04^\circ\).

Note on Quadrants:

Vectors in the second and third quadrants will yield negative x-coordinates, and those in the third and fourth quadrants will have negative y-coordinates. This must be taken into consideration when finding the true direction angle.

Standard Position Angle

The standard position angle of a vector is another term for the direction angle when measured counter-clockwise from the positive x-axis. It's a way to express the orientation of the vector in a consistent manner. The angle is restricted to a complete rotation, measured within the range of \(0^\circ \leq \theta < 360^\circ\).

In cases where the computed angle is negative because the vector points to the left of the y-axis, or beyond \(360^\circ\), an adjustment is needed to ensure it falls within the correct range. For the given vector \(\mathbf{w} = \langle 3,5 \rangle\), the angle \(59.04^\circ\) is already in standard position since it falls neatly within the defined range, and the vector lies in the first quadrant.

However, if a vector lies in another quadrant, the angle may require addition or subtraction of \(180^\circ\) or \(360^\circ\) to fit into the standard position range. In this way, we ensure consistency and clarity when communicating vector orientations in two dimensions.

Trigonometric Functions for Vectors

Vectors in two-dimensional space can be associated with trigonometric functions to describe their direction and relationship with the coordinate axes. The primary function used for this purpose is the tangent function, which relates the ratio of the vector's y-coordinate to its x-coordinate.

In context of vector \(\mathbf{w} = \langle 3,5 \rangle\) from the exercise, the use of the tangent function in the form of arctangent or \(\tan^{-1}\) helps us find the direction angle. If we were to explore further, sine and cosine functions could be used to find the components of a vector given its magnitude and direction.

For example:

• To find the x-component: \( x = \lVert\mathbf{w}\rVert \cdot \cos(\theta) \)
• To find the y-component: \( y = \lVert\mathbf{w}\rVert \cdot \sin(\theta) \)

It’s important to handle these functions with care as they can have different values depending on the quadrant in which the angle lies. In practical scenarios, understanding these trigonometric relationships is essential for breaking down vectors into their components or reconstructing them from magnitude and direction.