Problem 1 - Precalculus with Calculus Previews

Question

Sketch the given vector. Find the magnitude and the smallest positive direction angle of each vector. $$ \langle\sqrt{3},-1\rangle $$

Step-by-Step Solution

Answer

Magnitude = 2; Smallest positive direction angle = 330°.

Step 1: Understanding the Vector

The vector is given as \( \langle \sqrt{3}, -1 \rangle \), where \( \sqrt{3} \) is the horizontal component (x-component) and -1 is the vertical component (y-component).

Step 2: Sketching the Vector

To sketch the vector, start at the origin (0,0). Move \( \sqrt{3} \) units to the right (positive x-direction) and 1 unit downward (negative y-direction) to reach the point \( (\sqrt{3}, -1) \). Draw an arrow from the origin to this point.

Step 3: Finding the Magnitude of the Vector

The magnitude \( |\mathbf{v}| \) of the vector \( \langle \sqrt{3}, -1 \rangle \) can be found using the formula: \( |\mathbf{v}| = \sqrt{(\sqrt{3})^2 + (-1)^2} \). Calculate this value: \[ |\mathbf{v}| = \sqrt{3 + 1} = \sqrt{4} = 2. \]

Step 4: Calculating the Direction Angle

The direction angle \( \theta \) can be found using the formula: \( \theta = \tan^{-1} \left( \frac{-1}{\sqrt{3}} \right) \). This calculation gives us:\[ \theta = \tan^{-1} \left( \frac{-1}{\sqrt{3}} \right) = -30^\circ. \]

Step 5: Adjusting the Angle to a Positive Value

Since we want the smallest positive direction angle, we convert \(-30^\circ\) by adding 360° (since angles in standard position move counter-clockwise): \[ \theta = 360^\circ - 30^\circ = 330^\circ. \] This is the smallest positive direction angle.

Key Concepts

Magnitude of a Vector

The magnitude of a vector is a crucial concept in vector mathematics that measures the length or size of the vector. To find this, we use the Pythagorean Theorem, which relates the components of the vector to its length. For a vector given by \( \langle a, b \rangle \), the magnitude \( |\mathbf{v}| \) is calculated using the formula:

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

This formula is analogous to finding the hypotenuse of a right triangle, where \( a \) and \( b \) are the lengths of the other two sides. In our example, the vector \( \langle \sqrt{3}, -1 \rangle \) has a magnitude of \( 2 \). It’s essential because it tells us how far the vector extends from the origin in its specified direction. Finding the magnitude is also the first step in many vector applications, such as normalizing vectors or determining physical quantities like force or velocity.

Direction Angle

The direction angle of a vector indicates the angle it makes with the positive x-axis in a standard coordinate system. To find this angle \( \theta \), we use the inverse tangent function (also known as arctan or \( \tan^{-1} \)). This function helps in determining the angle based on the vector's components. The formula to find the direction angle is:

• \( \theta = \tan^{-1} \left( \frac{b}{a} \right) \)

For our vector \( \langle \sqrt{3}, -1 \rangle \), the calculation gives \( -30^\circ \), a negative angle because it points downward from the x-axis. However, angles are often expressed positively, measured counter-clockwise from the positive x-axis, so we adjust by adding 360 degrees, resulting in a smallest positive direction angle of \( 330^\circ \). This gives a complete understanding of the vector's orientation in the plane.

Coordinate System

Understanding the coordinate system is fundamental when working with vectors. It’s a framework used to locate points or objects within a given dimension. In two-dimensional space, we typically use the Cartesian coordinate system, with horizontal and vertical axes usually labeled as x and y, respectively.
A point in this system is described by an ordered pair \( (x, y) \), specifying horizontal and vertical positions. Vectors, which are directed line segments, start at one point and end at another, typically plotted starting from the origin (0, 0).

• The x-component defines horizontal movement.
• The y-component dictates vertical movement.

Our given vector \( \langle \sqrt{3}, -1 \rangle \) exists within this system, representing a shift of \( \sqrt{3} \) units along the x-axis and \(-1\) unit along the y-axis. Understanding the coordinate axes' directions and their relation to vector components is crucial for visualizing vector movements and calculating their relevant properties, such as magnitude and direction.

Trigonometry

Trigonometry plays a pivotal role in vector mathematics. It helps translate between angles and side lengths in triangles, a key part of understanding vectors in a plane. When dealing with vectors, trigonometric functions like sine, cosine, and tangent help find relationships between a vector's direction and its components.
In the context of vectors:

• \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \) is useful for finding the direction angle based on x and y components.
• Inverse trigonometric functions, such as \( \tan^{-1} \), assist in calculating angles when given side ratios.

For our vector \( \langle \sqrt{3}, -1 \rangle \), the tangent function was used to determine its direction angle. Understanding how trigonometric principles apply to vectors enables one to solve complex problems involving direction and magnitude. Thus, a solid grasp of trigonometry is indispensable when working with vectors in any field.