Problem 36 - Precalculus Mathematics for Calculus

Question

Two vectors \(u\) and \(v\) are given. Find the angle (expressed in degrees) between \(\mathbf{u}\) and \(v\) $$\mathbf{u}=\mathbf{i}+2 \mathbf{j}-2 \mathbf{k}, \quad \mathbf{v}=4 \mathbf{i}-3 \mathbf{k}$$

Step-by-Step Solution

Answer

The angle between \( \mathbf{u} \) and \( \mathbf{v} \) is approximately \( 48.19^\circ \).

Step 1: Formula for the Angle between Vectors

The angle \( \theta \) between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) can be found using the dot product formula:\[\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}\]We'll use this formula to find the cosine of the angle.

Step 2: Calculate the Dot Product

The dot product \( \mathbf{u} \cdot \mathbf{v} \) is calculated as:\[(1 \cdot 4) + (2 \cdot 0) + (-2 \cdot -3) = 4 + 0 + 6 = 10\]

Step 3: Find the Magnitudes of the Vectors

The magnitude of \( \mathbf{u} \), \( \|\mathbf{u}\| \), is:\[\sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\]The magnitude of \( \mathbf{v} \), \( \|\mathbf{v}\| \), is:\[\sqrt{4^2 + 0^2 + (-3)^2} = \sqrt{16 + 0 + 9} = \sqrt{25} = 5\]

Step 4: Substitute into the Cosine Formula

Substitute the dot product and magnitudes into the cosine formula:\[\cos\theta = \frac{10}{3 \times 5} = \frac{10}{15} = \frac{2}{3}\]

Step 5: Calculate the Angle in Degrees

To find \( \theta \) in degrees, take the inverse cosine of \( \frac{2}{3} \):\[\theta = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.19^\circ\]

Step 6: Final Answer Verification

The calculated angle is approximately \( 48.19^\circ \). Double-checking the calculations confirms this is correct.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector algebra. It involves two vectors and returns a scalar. Here's the formula for calculating the dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\), represented in component form:

• \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \)
• \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \)

The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as: \[ a_1 b_1 + a_2 b_2 + a_3 b_3 \] The dot product helps determine the angle between two vectors and identifies if they are perpendicular (when the dot product is zero). For example, in the original exercise, you calculated \( \mathbf{u} \cdot \mathbf{v} = 10 \). This value is used in finding the angle between the vectors.

Magnitude of a Vector

The magnitude of a vector measures its length. For a vector \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \), the magnitude is calculated using the formula: \[ \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2 + a_3^2} \] Magnitude is always a non-negative value because it represents distance. In the exercise, the magnitudes of the vectors \( \mathbf{u} \) and \( \mathbf{v} \) were calculated as 3 and 5, respectively. These values indicate the "length" of the vectors in space, and they play an important role in determining the angle between them when using the cosine formula.

Cosine Formula

The cosine formula is pivotal in connecting vector operations to trigonometry. It helps us find the cosine of the angle between two vectors. Given vectors \( \mathbf{u} \) and \( \mathbf{v} \), the formula is: \[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \] By calculating the dot product and the magnitudes of the vectors, you can substitute these values into the formula to find \( \cos \theta \). In the exercise, you found: \[ \cos \theta = \frac{10}{15} = \frac{2}{3} \] This result provides the cosine of the angle, which is needed to determine the angle itself.

Inverse Cosine

The inverse cosine, or \( \cos^{-1} \), is used to calculate the angle from its cosine. This function is the reverse of the cosine function and is essential for solving vector angle problems. In this context, once we have \( \cos \theta \), we need to apply the inverse cosine to find \( \theta \): \[ \theta = \cos^{-1} \left( \frac{2}{3} \right) \] Most calculators have a special button for inverse cosine, typically labeled \( \cos^{-1} \) or "acos". It outputs the angle in radians or degrees, depending on the settings. For the original problem, using this function gave the angle \( \theta \approx 48.19^\circ \). This is the final step in determining the direction between the two vectors.