Problem 33 - Precalculus Mathematics for Calculus

Question

Two vectors \(u\) and \(v\) are given. Find the angle (expressed in degrees) between \(\mathbf{u}\) and \(v\) $$\mathbf{u}=\langle 2,-2,-1\rangle, \quad \mathbf{v}=\langle 1,2,2\rangle$$

Step-by-Step Solution

Answer

The angle between the vectors is approximately 116.57 degrees.

Step 1: Calculate the Dot Product

The dot product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) can be calculated using the formula \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). For \( \mathbf{u} = \langle 2, -2, -1 \rangle \) and \( \mathbf{v} = \langle 1, 2, 2 \rangle \), the dot product is:\[ \mathbf{u} \cdot \mathbf{v} = (2)(1) + (-2)(2) + (-1)(2) = 2 - 4 - 2 = -4 \]

Step 2: Calculate the Magnitude of Vectors

The magnitude of a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) is given by \( ||\mathbf{a}|| = \sqrt{a_1^2 + a_2^2 + a_3^2} \).- For \( \mathbf{u} \): \[ ||\mathbf{u}|| = \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]- For \( \mathbf{v} \): \[ ||\mathbf{v}|| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

Step 3: Use Cosine Formula to Find the Angle

The angle \( \theta \) between two vectors can be found using the formula: \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \, ||\mathbf{v}||} \).Substitute the dot product and magnitudes from above:\[ \cos \theta = \frac{-4}{3 \times 3} = \frac{-4}{9} \]

Step 4: Calculate the Angle in Degrees

To find \( \theta \), take the inverse cosine (or arccos) of \( \frac{-4}{9} \) and then convert the angle from radians to degrees using the conversion factor \( \frac{180}{\pi} \).\[ \theta = \cos^{-1} \left( \frac{-4}{9} \right) \approx 116.57^{\circ} \]

Key Concepts

Dot Product

The dot product is a fundamental operation in vector algebra that helps in analyzing the relationship between two vectors. Imagine two vectors, \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \). To find their dot product, you use the formula:
\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]This expression essentially involves multiplying each corresponding component of the vectors and then summing the results.

• Example: For vectors \( \mathbf{u} = \langle 2, -2, -1 \rangle \) and \( \mathbf{v} = \langle 1, 2, 2 \rangle \), the dot product is calculated as \( (2)(1) + (-2)(2) + (-1)(2) = -4 \).
• The result tells you something about the orientation of the vectors; when the dot product is zero, the vectors are perpendicular.

Understanding the dot product helps in many fields such as physics and computer graphics where vector directions and magnitudes are key.

Magnitude of Vectors

The magnitude of a vector gives you an idea about its 'length' or 'size' in the space it occupies. For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), its magnitude is computed through:
\[||\mathbf{a}|| = \sqrt{a_1^2 + a_2^2 + a_3^2}\]This formula involves squaring each of the vector's components, summing these squares, and then taking the square root of the result.

• For vector \( \mathbf{u} = \langle 2, -2, -1 \rangle \), the magnitude is calculated as \( \sqrt{4 + 4 + 1} = 3 \).
• Similarly, for vector \( \mathbf{v} = \langle 1, 2, 2 \rangle \), its magnitude is also \( 3 \).

The magnitude is crucial when normalizing vectors or when calculating angles between vectors.

Angle Between Vectors

Finding the angle between two vectors is a powerful way to understand their spatial relationship. The angle \( \theta \) can be calculated by using the dot product and the magnitudes of the vectors. The formula used is:
\[\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \, ||\mathbf{v}||}\]This involves dividing the dot product of the two vectors by the product of their magnitudes.

• For vectors \( \mathbf{u} = \langle 2, -2, -1 \rangle \) and \( \mathbf{v} = \langle 1, 2, 2 \rangle \), using the previously calculated dot product \( -4 \) and magnitudes \( 3 \), you find \( \cos \theta = \frac{-4}{9} \).
• To find the actual angle \( \theta \), take the arccosine of \( \frac{-4}{9} \), which gives approximately \( 116.57^{\circ} \).

This calculation helps in applications such as determining alignment in physics or congruence in geometry.