Problem 1 - Algebra and Trigonometry

Question

1-8 Find \((a) \mathbf{u} \cdot \mathbf{v}\) and \((\mathbf{b})\) the angle between \(\mathbf{u}\) and \(\mathbf{v}\) to the nearest degree. $$\mathbf{u}=\langle 2,0\rangle, \quad \mathbf{v}=\langle 1,1\rangle$$

Step-by-Step Solution

Answer

\( \mathbf{u} \cdot \mathbf{v} = 2 \); the angle between is \( 45^{\circ} \).

Step 1: Calculate Dot Product

To find \( \mathbf{u} \cdot \mathbf{v} \), use the formula for the dot product of two vectors: \( \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 \). Here, \( \mathbf{u} = \langle 2, 0 \rangle \) and \( \mathbf{v} = \langle 1, 1 \rangle \). Compute: \( 2 \times 1 + 0 \times 1 = 2 + 0 = 2 \). So, \( \mathbf{u} \cdot \mathbf{v} = 2 \).

Step 2: Calculate Magnitude of \(\mathbf{u}\) and \(\mathbf{v}\)

The magnitude of a vector \( \mathbf{u} = \langle a, b \rangle \) is given by \( ||\mathbf{u}|| = \sqrt{a^2 + b^2} \). For \( \mathbf{u} = \langle 2, 0 \rangle \), \( ||\mathbf{u}|| = \sqrt{2^2 + 0^2} = \sqrt{4} = 2 \). For \( \mathbf{v} = \langle 1, 1 \rangle \), \( ||\mathbf{v}|| = \sqrt{1^2 + 1^2} = \sqrt{2} \).

Step 3: Calculate Cosine of the Angle

Use the formula \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||} \) to find the cosine of the angle \( \theta \) between the vectors. Here, \( \cos \theta = \frac{2}{2 \times \sqrt{2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \). Simplifying, \( \cos \theta = \frac{\sqrt{2}}{2} \).

Step 4: Calculate the Angle

Find the angle \( \theta \) by taking the inverse cosine of the cosine we found: \( \theta = \cos^{-1}\left( \frac{\sqrt{2}}{2} \right) \). This gives \( \theta = 45^{\circ} \). Thus, the angle between the vectors is approximately 45 degrees.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors resulting in a scalar (a single number). Unlike the regular multiplication of numbers, which gives another number of similar type, the dot product combines vectors to give a single value. The formula for the dot product of two vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \) is:\\[ \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 \].

For the given vectors \( \mathbf{u} = \langle 2, 0 \rangle \) and \( \mathbf{v} = \langle 1, 1 \rangle \), we calculate:

• \( 2 \times 1 = 2 \)
• \( 0 \times 1 = 0 \)

Adding these results gives the dot product \( 2\).

This scalar product informs us about the directional multiplicative interaction between the vectors. If the result is zero, the vectors are perpendicular, signifying that they are at right angles to each other.

Magnitude of a Vector

The magnitude of a vector is essentially its length, and it provides an understanding of the vector's size regardless of its direction. It's calculated using the Pythagorean theorem for the vector's components.

Every vector \( \mathbf{u} = \langle a, b \rangle \) has a magnitude \( ||\mathbf{u}|| \) given by:\\[ ||\mathbf{u}|| = \sqrt{a^2 + b^2} \].

For \( \mathbf{u} = \langle 2, 0 \rangle \):

• The magnitude is \( \sqrt{2^2 + 0^2} = \sqrt{4} = 2 \).

For \( \mathbf{v} = \langle 1, 1 \rangle \):

• The magnitude is \( \sqrt{1^2 + 1^2} = \sqrt{2} \).

The magnitude helps us understand the vectors' strengths and is vital when calculating distances or performing normalizations in various applications.

Angle Between Vectors

The angle between two vectors gives insight into how aligned or opposed they are to each other, which can be very useful, for example, in physics to understand forces or in graphics for determining light angles. This angle is determined using the concept of the dot product and vector magnitudes with the formula:

\[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||} \].

Substituting the known values:

• \( \cos \theta = \frac{2}{2 \times \sqrt{2}} = \frac{1}{\sqrt{2}} \)

Simplify this to \( \cos \theta = \frac{\sqrt{2}}{2} \).

To find the angle \( \theta \), apply the inverse cosine function:

• \( \theta = \cos^{-1}\left( \frac{\sqrt{2}}{2} \right) = 45^{\circ} \)

Therefore, the angle between the vectors is approximately 45 degrees. Knowing the angle helps us in understanding the orientation between the vectors, and a smaller angle indicates more similarity in direction.