Problem 10 - Algebra and Trigonometry

Question

Find (a) \(\mathbf{u} \cdot \mathbf{v}\) and (b) the angle between \(\mathbf{u}\) and \(\mathbf{v}\) to the nearest degree. $$ \mathbf{u}=2 \mathbf{i}+\mathbf{j}, \quad \mathbf{v}=3 \mathbf{i}-2 \mathbf{j} $$

Step-by-Step Solution

Answer

(a) The dot product is 4. (b) The angle is approximately 74°.

Step 1: Calculate the dot product

The dot product of two vectors \( \mathbf{u} = a_1 \mathbf{i} + b_1 \mathbf{j} \) and \( \mathbf{v} = a_2 \mathbf{i} + b_2 \mathbf{j} \) is calculated as \( \mathbf{u} \cdot \mathbf{v} = a_1a_2 + b_1b_2 \). For the vectors given, \( \mathbf{u} = 2\mathbf{i} + \mathbf{j} \) and \( \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} \), the dot product is \( 2 \times 3 + 1 \times (-2) = 6 - 2 = 4 \).

Step 2: Calculate magnitudes of the vectors

To find the angle between two vectors, we need their magnitudes. The magnitude of a vector \( \mathbf{u} = a \mathbf{i} + b \mathbf{j} \) is given by \( \| \mathbf{u} \| = \sqrt{a^2 + b^2} \). Thus, for \( \mathbf{u} \), \( \| \mathbf{u} \| = \sqrt{2^2 + 1^2} = \sqrt{5} \), and for \( \mathbf{v} \), \( \| \mathbf{v} \| = \sqrt{3^2 + (-2)^2} = \sqrt{13} \).

Step 3: Calculate the cosine of the angle between the vectors

The cosine of the angle \( \theta \) between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \| \| \mathbf{v} \|} \). Substituting the known values, \( \cos \theta = \frac{4}{\sqrt{5} \cdot \sqrt{13}} = \frac{4}{\sqrt{65}} \).

Step 4: Calculate the angle in degrees

To find the angle in degrees, we take the inverse cosine (arccos) and convert from radians to degrees. Use a calculator for \( \theta = \cos^{-1}(\frac{4}{\sqrt{65}}) \). This results in approximately \( 74^{\circ} \).

Key Concepts

Dot Product

The dot product of two vectors is a scalar value that gives us insight into the relationship between the two vectors. If you take two vectors, say \( \mathbf{u} = a_1 \mathbf{i} + b_1 \mathbf{j} \) and \( \mathbf{v} = a_2 \mathbf{i} + b_2 \mathbf{j} \), the dot product \( \mathbf{u} \cdot \mathbf{v} \) is calculated by:

• Multiplying corresponding components of the vectors.
• Adding these products together.

For the vectors \( \mathbf{u} = 2\mathbf{i} + \mathbf{j} \) and \( \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} \), the dot product becomes:

• \( 2 \times 3 = 6 \)
• \( 1 \times -2 = -2 \)
• So, \( \mathbf{u} \cdot \mathbf{v} = 6 - 2 = 4 \)

The dot product tells us that the vectors partially point in the same direction. If the dot product is positive, the vectors point somewhat in the same direction; if negative, somewhat in opposite directions.

Vector Magnitudes

To fully understand the relationship between two vectors, we need to know their magnitudes, which are also known as their lengths. You can think of the magnitude as the "size" or "length" of the vector in space.

• For any vector \( \mathbf{u} = a \mathbf{i} + b \mathbf{j} \), the magnitude is calculated using the formula \( \| \mathbf{u} \| = \sqrt{a^2 + b^2} \).

With \( \mathbf{u} = 2\mathbf{i} + \mathbf{j} \), the magnitude \( \| \mathbf{u} \| \) is:

• \( \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} \)

And for \( \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} \), the magnitude \( \| \mathbf{v} \| \) is:

• \( \sqrt{(3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \)

These magnitudes are essential when finding the angle between vectors, as they are used in the cosine formula.

Angle Between Vectors

The angle between two vectors tells us how divergent or convergent they are. To find this angle, we use the cosine of the angle formula:

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

Substituting in our values from the previous sections:

• Dot product \( = 4 \)
• Magnitude of \( \mathbf{u} = \sqrt{5} \)
• Magnitude of \( \mathbf{v} = \sqrt{13} \)
• \( \cos \theta = \frac{4}{\sqrt{5} \cdot \sqrt{13}} = \frac{4}{\sqrt{65}} \)

To find the angle \( \theta \), we use the arccosine function:

• \( \theta = \cos^{-1}\left( \frac{4}{\sqrt{65}} \right) \)

Converting the result from radians to degrees, the angle is approximately \( 74^{\circ} \). This angle shows how closely aligned or misaligned two vectors are, providing insights into their directional relationship.