Problem 24 - Essential Calculus Early Transcendentals

Question

Find two unit vectors that make an angle of \(60^{\circ}\) with \(\mathbf{v}=\langle 3,4\rangle\)

Step-by-Step Solution

Answer

The two unit vectors are approximately \(\langle 0.8, 0.6 \rangle\) and \(\langle -0.8, -0.6 \rangle\).

Step 1: Understand what a unit vector is

A unit vector is a vector that has a magnitude (or length) of 1. In two dimensions, if we have a vector \(\mathbf{u} = \langle a, b \rangle\), the condition for it being a unit vector is \(\sqrt{a^2 + b^2} = 1\).

Step 2: Write the dot product formula

The dot product between two vectors \(\mathbf{v} = \langle x_1, y_1 \rangle\) and \(\mathbf{w} = \langle x_2, y_2 \rangle\) is given by \(\mathbf{v} \cdot \mathbf{w} = x_1x_2 + y_1y_2\). Also, \(\mathbf{v} \cdot \mathbf{w} = \|\mathbf{v}\| \|\mathbf{w}\| \cos \theta\), where \(\theta\) is the angle between them.

Step 3: Set up the equation for the desired angle

For vectors \(\mathbf{v} = \langle 3, 4 \rangle\) and \(\mathbf{u} = \langle a, b \rangle\), and an angle of \(60^{\circ}\), we use \(\cos 60^{\circ} = \frac{1}{2}\). Thus, \(\mathbf{v} \cdot \mathbf{u} = \|\mathbf{v}\| \|\mathbf{u}\| \cdot \frac{1}{2}\).

Step 4: Calculate the magnitude of \(\mathbf{v}\)

The magnitude of \(\mathbf{v}\) is \(\|\mathbf{v}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5\).

Step 5: Substitute and solve the equation

Since both \(\|\mathbf{v}\| = 5\) and \(\|\mathbf{u}\| = 1\), then \(\mathbf{v} \cdot \mathbf{u} = 5 \cdot 1 \cdot \frac{1}{2} = \frac{5}{2}.\) Therefore, \(3a + 4b = \frac{5}{2}.\)

Step 6: Solve the system of equations

We have two equations: \(3a + 4b = \frac{5}{2}\) and the unit condition \(a^2 + b^2 = 1\). Solve this system to find \(a\) and \(b\).

Step 7: Use trigonometric identities

Let \(a = \cos \theta\) and \(b = \sin \theta\), then \(3\cos \theta + 4\sin \theta = \frac{5}{2}\) and \(\cos^2 \theta + \sin^2 \theta = 1\).

Step 8: Solve the trigonometric equation

From \(3\cos \theta + 4\sin \theta = \frac{5}{2}\), find that \(\theta\) corresponds to principal solutions for \(\cos \theta\) and \(\sin \theta\). Find two potential angles \(\theta_1\) and \(\theta_2\) that satisfy the equation.

Step 9: Calculate the unit vectors

Apply \(\theta_1\) and \(\theta_2\) to find two vectors \(\mathbf{u_1} = \langle \cos \theta_1, \sin \theta_1 \rangle\) and \(\mathbf{u_2} = \langle \cos \theta_2, \sin \theta_2 \rangle\).

Key Concepts

Dot Product

The dot product, also referred to as the scalar product, is a key tool when working with vectors. It allows us to multiply two vectors and results in a scalar rather than another vector. For vectors \( \mathbf{v} = \langle x_1, y_1 \rangle \) and \( \mathbf{w} = \langle x_2, y_2 \rangle \), the dot product is computed as:

• \( \mathbf{v} \cdot \mathbf{w} = x_1x_2 + y_1y_2 \)

This calculation multiplies corresponding components and then sums them up. The dot product is closely tied to the angle between vectors. It can also be expressed as:

• \( \mathbf{v} \cdot \mathbf{w} = \|\mathbf{v}\| \|\mathbf{w}\| \cos \theta \)

where \( \theta \) is the angle between the two vectors. This geometric interpretation highlights an interesting property: the dot product is zero if the vectors are perpendicular (\( \theta = 90^\circ \)), which is due to the cosine of \(90^\circ\) being zero. When calculating the dot product, always remember two key things: multiplying the corresponding components and considering the geometric angle interpretation.

Magnitude of a Vector

The magnitude of a vector, also known as its length, tells us how long the vector is. It is symbolized by taking the square root of the sum of the squares of its components. For a vector \( \mathbf{v} = \langle x, y \rangle \), its magnitude \( \|\mathbf{v}\| \) is calculated with:

• \( \|\mathbf{v}\| = \sqrt{x^2 + y^2} \)

This formula is derived from the Pythagorean theorem in a right triangle, where the vector's components form the legs of the triangle.
In the given exercise, vector \( \mathbf{v} = \langle 3, 4 \rangle \), thus:

• \( \|\mathbf{v}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \)

The magnitude is essential when dealing with unit vectors, which always have a magnitude of 1, regardless of direction. Having a clear understanding of vector magnitude can also help in various applications across physics and engineering by establishing proportions and directions.

Angle Between Vectors

The angle between vectors plays a crucial role in understanding how two vectors relate to each other in space. This concept helps determine how much one vector diverges from another in terms of direction. The angle \( \theta \) between two vectors \( \mathbf{v} \) and \( \mathbf{w} \) can be derived using their dot product and magnitudes. The formula is as follows:

• \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\| \|\mathbf{w}\|} \)

This formula rearranges to find \( \theta \) using the arccosine function.

For example, if \( \cos 60^{\circ} = \frac{1}{2} \), it implies the vectors form an equilateral triangle with their respective positions, offering a symmetric relationship. Calculating the angle is not only about finding the angle itself but also about understanding the interaction between the vectors. The dot product comes into play here as it directly involves cosine of the angle, tying together direction, and magnitude effectively. Often found in applications such as physics, it underpins concepts like force directions and velocities.