Problem 10 - Calculus: Early Transcendentals

Question

Find the cosine of the angle between \langle 2,0,0\rangle and \langle-1,1,-1\rangle ; use a calculator if necessary to find the angle.

Step-by-Step Solution

Answer

The cosine of the angle is \( \frac{-\sqrt{3}}{3} \) and the angle is approximately \( 125.26^{\circ} \).

Step 1: Understanding the Problem

We need to find the cosine of the angle between the vectors \( \langle 2,0,0 \rangle \) and \( \langle -1,1,-1 \rangle \). This requires using the dot product formula and the magnitudes of the vectors.

Step 2: Calculating the Dot Product

The dot product formula for vectors \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \) is \( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \). For the given vectors, \( \langle 2,0,0 \rangle \cdot \langle -1,1,-1 \rangle = 2(-1) + 0(1) + 0(-1) = -2 \).

Step 3: Finding the Magnitudes of the Vectors

Calculate the magnitude of each vector. The magnitude of a vector \( \vec{v} = \langle v_1, v_2, v_3 \rangle \) is \( ||\vec{v}|| = \sqrt{v_1^2 + v_2^2 + v_3^2} \). For \( \langle 2,0,0 \rangle \), the magnitude is \( ||\langle 2,0,0 \rangle|| = \sqrt{2^2 + 0^2 + 0^2} = 2 \). For \( \langle -1,1,-1 \rangle \), the magnitude is \( ||\langle -1,1,-1 \rangle|| = \sqrt{(-1)^2 + 1^2 + (-1)^2} = \sqrt{3} \).

Step 4: Applying the Cosine Formula

The cosine of the angle \( \theta \) between two vectors \( \vec{a} \) and \( \vec{b} \) is given by \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{||\vec{a}|| ||\vec{b}||} \). Substituting the values, we have \( \cos \theta = \frac{-2}{2 \cdot \sqrt{3}} = \frac{-1}{\sqrt{3}} \).

Step 5: Simplifying the Cosine Value

Rationalize the denominator to simplify. \( \cos \theta = \frac{-1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{-\sqrt{3}}{3} \).

Step 6: Interpreting the Result

Using a calculator, find the angle corresponding to \( \cos \theta = \frac{-\sqrt{3}}{3} \). The angle \( \theta \) is approximately \( 125.26^{\circ} \) or \( 2.186 \) radians.

Key Concepts

Dot Product

When learning about vector calculus, a fundamental operation to understand is the dot product. The dot product of two vectors measures their directional similarity and is computed as a scalar quantity. Given two 3-dimensional vectors \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \), the dot product is calculated as follows:
\[ \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \]
Essentially, this operation multiplies corresponding components of the vectors and adds them together.
One key use of the dot product is determining the angle between vectors. A positive dot product indicates that the angle is acute, a zero dot product signifies perpendicular vectors, and a negative dot product shows an obtuse angle.

Magnitude of a Vector

The magnitude of a vector represents its length or size and is crucial for normalizing vectors and computing angles between them. For a vector \( \vec{v} = \langle v_1, v_2, v_3 \rangle \), its magnitude, noted as \( ||\vec{v}|| \), is calculated using the formula:
\[ ||\vec{v}|| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]
This formula stems from the Pythagorean theorem, where each component squares and is summed, before taking the square root.
Knowing the magnitude of a vector is essential in various applications, such as finding unit vectors or computing the cosine of the angle between vectors. It transforms vector operations into more digestible scalar operations, making calculations straightforward.

Cosine of Angle Between Vectors

To find the cosine of the angle between two vectors, we leverage the dot product and their magnitudes. The formula used is:
\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{||\vec{a}|| \, ||\vec{b}||} \]
By substituting the values we get from previous calculations, this formula provides the cosine of the angle \( \theta \) between the vectors.
This expression simplifies the understanding by reducing it to basic arithmetic operations. Knowing the cosine of an angle is significant because it not only tells us about the alignment of these vectors but is also useful in projection formulas and many other vector applications.

Rationalization

Rationalization is a process used in mathematics to eliminate irrational numbers from a denominator. It typically involves multiplying both the numerator and the denominator by a conjugate or another form to make the denominator a rational number.
In the context of vector calculus, this is beneficial when finding angles between vectors, as it simplifies expressions such as \( \frac{-1}{\sqrt{3}} \) to a more manageable form, like \( \frac{-\sqrt{3}}{3} \).
The steps for rationalization involve:

• Identifying the irrational denominator.
• Multiplying both numerator and denominator by the appropriate term to rationalize it.
• Simplifying the resulting expression if possible.

Rationalizing expressions is helpful in problem-solving to ensure results are presented in an easily interpretable form, making analysis clearer.