Problem 14 - Precalculus Mathematics for Calculus

Question

Find \((a) u \cdot v\) and \((b)\) the angle between \(u\) and \(v\) to the nearest degree. $$\mathbf{u}=3 \mathbf{i}+4 \mathbf{j}, \quad \mathbf{v}=-2 \mathbf{i}-\mathbf{j}$$

Step-by-Step Solution

Answer

(a) \( u \cdot v = -10 \); (b) Angle = 152°.

Step 1: Calculate the Dot Product of Two Vectors

The dot product of two vectors \( \mathbf{u} = a\mathbf{i} + b\mathbf{j} \) and \( \mathbf{v} = c\mathbf{i} + d\mathbf{j} \) is given by the formula \( u \cdot v = ac + bd \). For the vectors \( \mathbf{u} = 3\mathbf{i} + 4\mathbf{j} \) and \( \mathbf{v} = -2\mathbf{i} - \mathbf{j} \), we calculate: \[ u \cdot v = (3)(-2) + (4)(-1) = -6 - 4 = -10. \]

Step 2: Compute Magnitudes of Vectors

The magnitude of a vector \( \mathbf{u} = a\mathbf{i} + b\mathbf{j} \) is calculated using the formula \( \|\mathbf{u}\| = \sqrt{a^2 + b^2} \). For \( \mathbf{u} = 3\mathbf{i} + 4\mathbf{j} \): \[ \|\mathbf{u}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5. \] For \( \mathbf{v} = -2\mathbf{i} - \mathbf{j} \): \[ \|\mathbf{v}\| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}. \]

Step 3: Calculate the Cosine of the Angle Between Vectors

To find the angle \( \theta \) between two vectors, use the cosine formula: \( \cos \theta = \frac{u \cdot v}{\|\mathbf{u}\| \|\mathbf{v}\|} \). Substitute known values: \[ \cos \theta = \frac{-10}{5 \times \sqrt{5}} = \frac{-10}{5\sqrt{5}} = \frac{-2}{\sqrt{5}} = -\frac{2\sqrt{5}}{5}. \]

Step 4: Determine the Angle Between the Vectors

Convert \( \cos \theta = -\frac{2\sqrt{5}}{5} \) to an angle. Use a calculator to find \( \theta = \cos^{-1}(-\frac{2\sqrt{5}}{5}) \). Computing this gives \( \theta \approx 152 \) degrees, rounded to the nearest degree.

Key Concepts

Vectors

Vectors are mathematical objects that have both a magnitude and a direction. They are often represented in a coordinate system using unit vectors like \( \mathbf{i} \) and \( \mathbf{j} \), which indicate directions along the x-axis and y-axis, respectively. For example, the vector \( \mathbf{u} = 3\mathbf{i} + 4\mathbf{j} \) has a component of 3 in the x-direction and 4 in the y-direction. This means the vector points towards the northeast, roughly speaking.

• The magnitude of a vector is its length. For a vector \( \mathbf{u} \) with components \( a \) and \( b \), the magnitude \( \| \mathbf{u} \| \) is found using the formula \( \| \mathbf{u} \| = \sqrt{a^2 + b^2} \).
• The direction of a vector is given by the angle it makes with a reference direction, typically the positive x-axis.

Understanding vectors is essential for solving problems involving motion and force, as they describe both where something is moving and how fast.

Angle Between Vectors

The angle between two vectors provides insight into how aligned the vectors are with each other. If two vectors point in the same direction, the angle between them is 0 degrees. If they point in opposite directions, the angle is 180 degrees. Calculating this angle involves understanding the dot product and magnitudes of the vectors.
Consider our vectors, \( \mathbf{u} = 3\mathbf{i} + 4\mathbf{j} \) and \( \mathbf{v} = -2\mathbf{i} - \mathbf{j} \). We've found:

• \( \mathbf{u} \cdot \mathbf{v} = -10 \), the dot product.
• The magnitudes \( \| \mathbf{u} \| = 5 \) and \( \| \mathbf{v} \| = \sqrt{5} \).

Using these, the cosine of the angle \( \theta \) is given by:
\[ \cos \theta = \frac{ \mathbf{u} \cdot \mathbf{v} }{ \| \mathbf{u} \| \| \mathbf{v} \| } \]
Substituting the values:\
\( \cos \theta = -\frac{2\sqrt{5}}{5} \). This value tells us that the vectors form an angle of approximately 152 degrees. A larger angle indicates that the vectors are mostly pointing in opposite directions.

Cosine Formula

The cosine formula is used to calculate the cosine of the angle between two vectors. This formula is especially useful in physics and engineering, where understanding the orientation of two vectors is critical. The formula is:
\[ \cos \theta = \frac{ \mathbf{u} \cdot \mathbf{v} }{ \| \mathbf{u} \| \| \mathbf{v} \| } \]

• The numerator, \( \mathbf{u} \cdot \mathbf{v} \), represents the dot product of the vectors which quantify how much of one vector projected along another.
• The denominator, \( \| \mathbf{u} \| \| \mathbf{v} \| \), is the product of the magnitudes of both vectors. It serves to normalize the dot product, ensuring that \( \cos \theta \) is between -1 and 1.

Through this relationship, we obtain the cosine value for the angle \( \theta \) between the vectors, allowing us to calculate the actual angle using the inverse cosine function \( \cos^{-1} \). This step leads us to the final calculation of the angle to the nearest degree, providing practical insights into the vectors' alignment.